text:
\section{Introduction}
Let $\mathcal X$ be a genus $g\geq 2$ smooth, irreducible projective curve, defined over an algebraically closed  field $k$ and $\pi_i : \mathcal X \to \mathcal X_i$, $1 \leq i \leq s$ be coverings to genus $g_i$ curves $\mathcal X_i$, respectively.  Denote by $\Jac  (\mathcal X)$ and $\Jac  (\mathcal X_i)$ the Jacobians of these curves respectively.  A classical question is to determine when the Jacobian $\Jac  (\mathcal X)$ is isogenous to a product of $\Jac  (\mathcal X_i)$.
The problem has been studied in the XIX century by Legendre, Jacobi, Hermite, Klein, Kovalevskaya, Hecke, et al.  In this context, these curves were of interest due to the fact that split Jacobians give relations between abelian and elliptic integrals. Most of these efforts were focused on the case when the coverings $\pi_i: \mathcal X \to \mathcal X_i$ are Galois coverings.

The case when genus $g(\mathcal X_i)=1$ for some $i$ and $\pi_i : \mathcal X \to \mathcal X_i$ not necessarily Galois  is especially interesting due to its implications in number theory and relations to  Lang's conjecture; see \cite{faltings-1, faltings-2} and \cite{martens}. When $g=2$ such coverings  have different ramification structures when $\deg \pi_i$ is odd or even; see \cite{b-sh} for a summary of results in genus two.   In both cases the moduli space of such covers with  fixed ramification structure can be embedded as a subvariety of ${\mathcal M}_2$.

The Jacobian $\Jac  (\mathcal X)$  is said to (completely) split if it is isogenous to the product of elliptic curves.  Ekedahl and Serre  \cite{Se} posed the following questions:
i) Is it true that for every integer $g > 1$ there exists a curve of genus $g$ with split Jacobian?
ii) Is there a bound on the genus of a curve with split Jacobian?
Towards answering these questions they used modular curves, and coverings of curves of genus 2 and 3.  The maximal genus of their examples was 1297. Many other authors have considered these questions as well, and in recent work, Shaska \cite{arxiv-1, arxiv-2} and  Yamauchi \cite{Ya}, have produced somewhat unusual families of curves with decomposable Jacobians.

In this paper we will focus on the cases when the coverings $\pi_i : \mathcal X \to \mathcal X_i$ are cyclic Galois coverings. Hence, these coverings induce automorphisms on $\mathcal X$.  By focusing on such curves  we have a more organized way to study the decompositions of such Jacobians starting from the automorphism group $\Aut (\mathcal X)$ of $\mathcal X$.  While we can determine the full list of groups that occur as full automorphism groups of genus $g \geq 2$, the corresponding equations of curves for each group are not known; see \cite{dec_jac}.   There is, however, a nice family of curves when such decomposition can be fully explored and the factors of such Jacobians can be fully determined. This is the class of superelliptic curves, namely the curves which can be written as affine varieties with equation $y^n = f(x)$ for some $n\geq 2$ and discriminant $\Delta (f, x) \neq 0$.  For such curves we know precisely the structure of the automorphism group $G$, the signature, the equations of the curves, and their invariants; see \cite{beshaj-2}.
Hence, it is possible  to determine explicitly the components of such Jacobians and in some cases the moduli space of such curves in the moduli space ${\mathcal M}_g$.
A complete description of curves for which such  decompositions is based on their automorphisms is intended in \cite{dec_jac}.

The second goal of the paper is to extend the family of curves introduced by Yamauchi in \cite{Ya}.  The family of curves introduced in \cite{Ya} are non-hyperelliptic curves with arbitrary large genus such that the Jacobian has hyperelliptic components.   We attempt to extend this family to curves whose Jacobians have superelliptic components. Our proof   is based on the automorphisms of curves.  We show that such decomposition based on the automorphisms of the curve induces some arithmetic condition among the orders of such automorphisms.  This condition implies that the family $F_{m, n}$ constructed in \cite{Ya} agrees with our family of curves only for the values $F_{m, 1}$ and $F_{m, 2}$.

This paper is organized as follows. In section 2 we briefly define the Jacobian of a curve and give two classical results of Accola; see \cite{Ac1, Ac2}.  In these results it is described how one can start from a partition of a group of automorphisms of the curve $\mathcal X$  and get a decomposition of the  $\Jac  (\mathcal X)$.  Further in this section we describe results of Kani and Rosen were the previous results were generalized; see \cite{KR}.  We use such results   in our decomposition of Jacobians of the superelliptic curves  for the family of curves described in section 4.

In section 3 we study the decomposition of Jacobians of superelliptic curves.  A superelliptic curve $\mathcal X$ is a curve with equation $y^n = f(x)$ defined over a field $k$, where $\left( \mbox{char } k, n \right)=1$ and $k$ is algebraically closed.   Such curves have the \emph{superelliptic} automorphism $\tau : (x, y) \to (x, \zeta_n y)$, where $\zeta_n$ is an $n$-th primitive root of unity and $\tau$ is central in $G:=\Aut (\mathcal X)$.
Let $\sigma$ be another automorphism of $\mathcal X$ such that its projection $\bar \sigma \in \bG:=G/\< \tau \>$ has order $m$.   Then $\sigma$ and $\sigma\tau$ fix two subfields of the function field $k (\mathcal X)$ and therefore there are two quotient curves $\mathcal X_1:= \mathcal X /\< \sigma\>$ and $\mathcal X_2:= \mathcal X/ \< \sigma\tau\>$.  We determine equations of $\mathcal X_1$ and $\mathcal X_2$ and a necessary and sufficient condition in terms of $m$ and $n$ such that $\Jac  (\mathcal X)$ is isogenous to $\Jac  (\mathcal X_1) \times \Jac  (\mathcal X_2)$.

In section 4 we generalize a construction of Yamauchi of a family $F_{m, n}$ of non-hyperelliptic curves with decomposable Jacobians, where all components are hyperelliptic Jacobians.  Instead we construct a family $\mathcal X_{r, s}$ of curves whose Jacobians are superelliptic Jacobians.  We prove that the automorphism group of these \emph{component curves} are  cyclic or dihedral groups. Moreover, we find a necessary and sufficient condition in terms of $r$ and $s$ for this decomposition to occur. Our decomposition is based solely on the decomposition induced by the automorphisms of the curves.

\medskip

\noindent \textbf{Notation:} Throughout this paper by $g$ we denote an integer $\geq 2$ and  $k$ denotes an algebraically closed field.  Unless otherwise noted,  by a \textit{curve} we always mean the isomorphism class of an algebraic curve defined over $k$.   The automorphism group of a curve always means the full automorphism group of the curve over $k$.  We denote the cyclic group of order $n$ by $C_n$ and the dihedral group of order $2n$ by $D_{2n}$.  $V_4$ denotes the Klein 4-group and $\zeta_n$ denotes an $n$-th primitive root of unity.

\section{Preliminaries}

Let $\mathcal X$ be a genus $g \geq 2$ algebraic curve defined over ${\mathbb C}$. We choose a symplectic homology basis for $\mathcal X$, say $ \{ A_1, \dots, A_g, B_1, \dots , B_g\},$ such that the intersection products $A_i \cdot A_j = B_i \cdot B_j =0$ and
$A_i \cdot B_j= {\delta }_{i j},$ where ${\delta }_{i j}$ is the Kronecker delta. We choose a basis $\{ w_i\}$ for the space of holomorphic 1-forms such that $\int_{A_i} w_j = {\delta }_{i j}$. The matrix $\Omega= \left[ \int_{B_i} w_j
\right] $ is  the \emph{Riemann matrix} of $\mathcal X$ and the matrix $\left[ I \ | \Omega \right]$ is called the \textit{period matrix}.  The columns of the matrix $\left[ I \ | \Omega \right]$ form a lattice $L$ in  ${\mathbb C}^g$.  The complex torus  ${\mathbb C}^g/ L$ is called the Jacobian  of $\mathcal X$ is denoted by $\Jac  (\mathcal X)$.

Let $\mathcal H_g$ be the \emph{Siegel upper-half space}. Then $\Omega \in \mathcal H_g$ and there is an injection
\[ {\mathcal M}_g \hookrightarrow \mathcal H_g/ Sp_{2g}({\mathbb Z}) =: \mathcal A_g \] where $Sp_{2g}({\mathbb Z})$ is the \emph{symplectic group}.
A non-constant morphism  $f : A \to B$ between two Abelian varieties which is surjective and of finite kernel is called an \textbf{isogeny}. An Abelian variety is called \textit{decomposable} if it is isogenous to a product of Abelian varieties, it is \textit{simple} if it has no non-trivial Abelian subvarieties.
An Abelian variety is called \textbf{completely decomposable} or \textbf{completely split}   if it is isogenous to a product of elliptic curves.

A map of algebraic curves $f: \mathcal X \to \mathcal Y$ is called a \textbf{maximal covering} if it does not factor over a nontrivial isogeny.  A map $f: \mathcal X \to \mathcal Y$ induces maps between their Jacobians $f^*: \Jac  (\mathcal Y) \to \Jac  (\mathcal X)$ and $f_*: \Jac  (\mathcal X) \to \Jac  (\mathcal Y)$. When $f$ is maximal then $f^*$ is injective and $\ker (f_*)$ is connected, see \cite[p. 158]{Se-book} for details.  Hence, $\Jac  (\mathcal X) \cong \Jac  (\mathcal Y) \times A$, where $A$ is some Abelian variety.  Thus, coverings $f: \mathcal X \to \mathcal Y$ give factors of the Jacobian $\Jac  (\mathcal X)$.   Such methods have been explored for genus 2 curves by Shaska et al. in \cite{sh_01, deg3, deg5}.  They are the only examples that we know when explicit computations have been performed and the corresponding locus has been computed for non-Galois coverings.

If the covering $f: \mathcal X \to \mathcal Y$ is a Galois covering then its monodromy group is isomorphic to   a subgroup $H$ of the automorphism group $G=\Aut (\mathcal X)$.  Hence a common procedure to produce decompositions of Jacobians is to explore the automorphism group of the curve.

\def\textbf{C}{\textbf{C}}

Fix an integer $g\ge2$ and a finite group $G$. Let $C_1, \dots ,C_r$ be conjugacy classes $\ne\{1\}$ of $G$. Let $\textbf{C}=(C_1, \dots ,C_r)$, be an unordered tuple, repetitions are allowed. We allow $r$ to be zero, in which case $\textbf{C}$ is empty.

Consider pairs $(\mathcal X,\mu)$, where $\mathcal X$ is a curve and $\mu: G \to \Aut(\mathcal X)$ is an injective homomorphism. Mostly we will suppress $\mu$ and just say $\mathcal X$ is
a curve with $G$-action, or a $G$-curve, for short. Two $G$-curves $\mathcal X$ and $\mathcal X'$ are called equivalent if
there is a $G$-equivariant isomorphism $\mathcal X\to \mathcal X'$.

We say a $G$-curve $\mathcal X$ is \textbf{of ramification type} $(g, G,\textbf{C})$ if  $g$ is the
genus of $\mathcal X$ and  the points of the quotient $\mathcal X/G$ that are ramified in the cover $\mathcal X\to \mathcal X/G$ can be
labelled as $p_1, \dots , p_r$ such that $C_i$ is the conjugacy class in $G$ of distinguished inertia group
generators over $p_i$ (for $i=1, \dots ,r$).

Let $\mathcal X$ be a $G$-curve and $H < G$. Then there is a covering $\mathcal X \to \mathcal X/H$. Let the genus of $\mathcal X/H$ be denoted by $g_H$.  How is $g_H$ determined in terms of $g$ and $G$?
Consider the following problem: let $H_1, \dots , H_r$ be subgroups of $G$ and $g_1, \dots , g_r$ the genera of the $\mathcal X/H_1 , \dots , \mathcal X/H_r$ respectively.  Is there any arithmetic relation between $g, g_1, \dots , g_r$?

Accola proved the following results which provide a method for decomposing Jacobians, see \cite{Ac1}.
\begin{thm}[\cite{Ac1}]
Let $\mathcal X_g$ be a $G$-curve, $H_i < G$, $g_i$ the genus of $\mathcal X_g/H_i$ for $i=1, \dots s$, and $g_0 = g(\mathcal X_g/G)$. Assume that $H_i \cap H_j = \{ e \}$ for all $i\neq j$.  Then,
\[  g_0 \, |G| = g - sg + \sum_{i=1}^s |H_i | \, g_i.   \]
\end{thm}
\noindent Moreover, we have the following relation among the genera; see \cite{Ac2}.
\begin{thm}[\cite{Ac2}]
Let $\mathcal X_g$ be a $G$-curve, and $H_i$, for $i=1, \dots s$, subgroups of  $G$ such that $G= \cup_i^s H_i$.  Denote by  $g_i$ the genus of $\mathcal X_g/H_i$ and by $H_{ij...k}=H_i \cap H_j \cap \dots \cap H_k$.

 Then,
\[ g_0 \, |G| = \sum_{i=1}^s |H_i | \cdot g_i - \sum \left|H_{ij} \right| \cdot g_{ij} + \sum \left|H_{ijk}\right| \cdot g_{ijk} - \cdots - (-1)^s \sum \left|H_{12...s} \right| \cdot  g_{12..s}. \]

\end{thm}

See \cite{Ac2} for details.

\subsection{Decomposing the Jacobian by group partitions}

Let $\mathcal X$ be a genus $g$  algebraic curve with automorphism group $G:=\Aut (\mathcal X)$. Let $H \leq G$ such that $H = H_1 \cup \dots \cup H_t$ where the subgroups $H_i \leq H$ satisfy $H_i \cap H_j = \{ 1\}$ for all $i\neq
j$.  Then,
\[ \Jac^{t-1} (\mathcal X ) \times \Jac^{|H|} (\mathcal X / H)\,  \cong \, \Jac^{| H_1 |} (\mathcal X / H_1) \times \cdots \Jac^{| H_t | } (\mathcal X / H_t).\]
The group $H$ satisfying these conditions is called a group with partition. Elementary Abelian $p$-groups, the projective linear groups $PSL_2 (q)$, Frobenius groups, dihedral groups are all groups with partition.

Let $H_1, \dots , H_t \leq G$ be subgroups with $H_i \cdot H_j = H_j \cdot H_i$ for all $i, j \leq t$, and let $g_{ij}$ denote the genus of the quotient curve $\mathcal X/(H_i\cdot H_j)$. Then, for $n_1, \dots , n_t \in {\mathbb Z}$
the conditions    $ \sum n_i n_j g_{ij} =0$, $\sum_{j=1}^{t} n_j g_{ij}=0$,  imply the isogeny relation
\begin{equation} \label{dec_1}
\prod_{n_i > 0} \Jac^{n_i} (\mathcal X / H_i) \cong \prod_{n_j < 0} \Jac^{n_j} (\mathcal X / H_j).
\end{equation}
In particular, if $g_{ij}=0$ for $2\leq i < j \leq t$ and if   $ g = g_{\mathcal X/ H_2} + \dots + g_{\mathcal X / H_t}$,  then
\begin{equation} \label{dec_2}
\Jac  (\mathcal X) \cong \Jac  (\mathcal X /H_2) \times \cdots \times \Jac  (\mathcal X / H_t).
\end{equation}
The proof of the above statements can be found in \cite{KR}.

\section{Jacobians of superelliptic curves}

A curve $\mathcal X$ is called superelliptic if there exist an element $\tau \in G:=\Aut( \mathcal X)$ such that $\tau$ is central and $g \left(\mathcal X / \< \tau \> \right) =0$. Let $\mathcal X_g$ have affine equation  given by some polynomial in terms of $x$ and $y$ and denote by $K=k(x, y)$ the function field of $\mathcal X_g$.  Let $H=\< \tau \>$ be a cyclic subgroup of $G$ such that $| H | = n$ and $H \triangleleft G$, where $n \geq 2$. Moreover, we assume that the quotient curve $\mathcal X_g / H$ has genus zero.   The \textbf{reduced automorphism group of $\mathcal X_g$ with respect to $H$} is called the group  $\G \, := \, G/H = \bAut (\mathcal X)$.

Assume  $k(x)$ is the  genus zero subfield of $K$ fixed by $H$.   Hence, $[ K : k(x)]=n$. Then, the group  $\G$ is a subgroup of the group of automorphisms of a genus zero field.   Hence, $\G <  PGL_2(k)$ and $\G$ is finite. It is a classical result that every finite subgroup of $PGL_2 (k)$  is  isomorphic to one of the following: $C_m $, $ D_{2m}$, $A_4$, $S_4$, $A_5$.

The group $\G$ acts on $k(x)$ via the natural way. The fixed field of this action is a genus 0 field, say $k(z)$. Thus, $z$
is a degree $|\G|$ rational function in $x$, say $z=\phi(x)$.

It is obvious that $G$ is a degree $n$ central extension of $\G$ and $\G$ is a finite subgroup of $PGL_2(k)$.  Hence, if we know all the possible groups that occur as $\G$ then we can determine  $G$ and the equation for $K$; see \cite{Sa2}.
In the next Lemma we establish some basic properties of superelliptic curves, the proof of the part i)  can be found in \cite{tw1} and part ii)  in \cite{beshaj-2}.

\begin{lemma} \label{lem_1}
Let $\mathcal X_g$ be a superelliptic curve  with affine equation $y^n= f(x)$ where $\Delta (f,x) \neq 0$ and $\deg f =d >n$.  Then the following hold:

i) $\mathcal X_g$ has genus \[ g = 1 + \frac 1 2 \, \left( nd -n -d - \gcd (d, n) \right). \]
If $d$ and $n$ are relatively prime then $g = \frac {(n-1)(d-1)} 2$.

ii) Let  $\sigma \in \Aut (\mathcal X_g)$ such that its projection $\bar \sigma \in \bAut (\mathcal X_g)$ has order  $m\geq 2$.  Then its equation is given as $y^n= g(x^m)$ or $y^n=x g(x^m)$ for some $g \in k[x]$.
\end{lemma}

Let $\mathcal X_g$ be a superelliptic curve and $\sigma \in \Aut (\mathcal X_g)$ such that its projection $\bar \sigma \in \bAut (\mathcal X_g)$ has order  $m\geq 2$.  We can choose a coordinate in $\mathbb P^1$ such that $\bar \sigma (X) = X^m$. Since $\sigma$ permutes the Weierstrass points of $\mathcal X$ and it has two fixed points then the equation of the curve will be $y^n= f(x^m)$ or $y^n=x f(x^m)$  as claimed in ii) above.

Assume that $\mathcal X$ has equation
\begin{equation}\label{first_eq}
 Y^n = f(X^m) := X^{{\delta } m } + a_1 X^{({\delta }-1) m}  \dots + a_{{\delta }-1} X^m + 1 .
\end{equation}
We assume that $\bar \sigma$ lifts to $G$ to an element of order $m$.   Then, $\sigma( X, Y) \to (\zeta_m X, Y)$.  Denote by $\tau: (X, Y) \to (X, \zeta_nY)$ its superelliptic automorphism. Since $\tau$ is central in $G$ then $\tau\sigma=\sigma\tau$.
We will denote by $\mathcal X_1$ and $\mathcal X_2$ the quotient curves $\mathcal X/\<\sigma\>$ and $\mathcal X/\< \tau \sigma\>$ respectively.  The next theorem determines the equations of $\mathcal X_1$ and $\mathcal X_2$.  We denote by $K$ the function field of $\mathcal X_g$ and by  $F$ and $L$ the function fields of $\mathcal X_1$ and $\mathcal X_2$ respectively.

\begin{thm}\label{thm_1}
Let $K$ be a genus $g\geq 2$  level $n$ superelliptic field and $F$ a  degree $m$ subfield fixed by $\sigma: (X, Y) \to (\zeta_m X, Y)$.

i) Then,  $K=k(X, Y)$ such that
\begin{equation}\label{first_eq}
 Y^n = f(X^m) := X^{{\delta } m } + a_1 X^{({\delta }-1) m}  \dots + a_{{\delta }-1} X^m + 1 .
\end{equation}
for    $\Delta (f, x)  \neq 0$.

ii) $F = k(U, V) $ where $U=X^m$,    and $V=Y$  and
\begin{equation}\label{ell_curve}
 V^n= f(U).
 \end{equation}

iii) There is another subfield $L= k(u, v) $ where  \[ u=X^m,  \quad v=X^i Y, \] and
\begin{equation}\label{gen_2}
 v^n = u \cdot f(u),
 \end{equation}
for $m=\lambda n$ and $i=\lambda (n-1)$.
\end{thm}

\proof
 The proof of i) follows from the above remarks.  To show that the   subfield  $F$ is generated by $X^m$ and $ Y$ it is enough to show that it is  fixed by $\sigma$.  This is obvious.

In  iii) we have to show that $u=X^m$ and $ v=X^i Y$ are fixed by $\sigma\tau : (X, Y) \to (\zeta_m X, \zeta_n Y)$.  We have that
$ \sigma \tau (v) =  \sigma \tau (X^i Y) = \zeta_m^{\lambda (n-1)} \zeta_n \cdot X^i Y = X^i Y$.
It is easily checked that $\zeta_m^{\lambda (n-1)} \zeta_n=1$.
 \qed

\begin{center}
\begin{figure}[htpb]
\[
\xymatrix{
                     & K=k(X, Y) \ar@{->}[d]^{\, \, \,n}  \ar@{->}[ld]^{\, \, \,m}  \ar@{->}[rd]^{\, \, \, \, \lcm(m, n)}           &                  \\
 %
F=k(X^m, Y)   \ar@{->}[d]^{\, \, \,n}   & k(X)  \ar@{->}[ld]^{\, \, \,m}  \ar@{->}[rd]^{\, \, \, \lcm(m, n)}  & L=k(X^m, X^iY)   \ar@{->}[d]^{\, \, \,n}   \\
k(X^m, Y^n)       &     & k( X^m, (X^i Y)^n)       \\
}
\]
\caption{The lattice of a genus $g$ superelliptic field with an extra automorphism}
\end{figure}
\end{center}

In the rest of this section we want to find necessary and sufficient conditions on $n$ and $m$ such that the Jacobian $\Jac  (\mathcal X_g)$ is isogenous to the product $\Jac  (\mathcal X_1) \times \Jac  (\mathcal X_2)$. First we focus on  hyperelliptic curves.

\subsection{Hyperelliptic curves}

\begin{thm} \label{lem_2}
Let $\mathcal X_g$ be a hyperelliptic curve and its reduced automorphism group $ \bAut (\mathcal X_g) \cong  C_m = \<\sigma \>$. Then $\mathcal X_g$ is isomorphic to a curve with equation
\[\mathcal X_g: Y^2=x^{{\delta }  m} + a_1 x^{({\delta }-1)  m} + \dots + a_{{\delta }-1} x^m + 1.\]
There exists subcovers $\pi_i : \mathcal X_g \to \mathcal X_i$, for $i=1, 2$ such that
 \[
 \begin{split}
&  \mathcal X_1 : \quad Y^2= X^{{\delta }} + a_1 X^{{\delta }-1} + \dots + a_{{\delta }-1} X + 1, \\
& \mathcal X_2 : \quad Y^2=X(  X^{{\delta }} + a_1 X^{{\delta }-1} + \dots + a_{{\delta }-1} X + 1  ). \\
\end{split}
\]
The  Jacobian of $\mathcal X$ is isogenous to the product
\[ \Jac  (\mathcal X) \cong  \Jac  (\mathcal X_1)   \times   \Jac  (\mathcal X_2) \]
if and only if the full automorphism group $\Aut (\mathcal X)$ is isomorphic to the Klein 4-group $V_4$.
\end{thm}

\proof Assume $\mathcal X_g$ is a hyperelliptic curve and $\sigma \in \bar G \cong C_m$.  The equation of the hyperelliptic curve $\mathcal X_g$ is
\[\mathcal X_g: Y^2=x^{{\delta }  m} + a_1 x^{({\delta }-1)  m} + \dots + a_{{\delta }-1} x^m + 1.\]
Denote with  $\tau$ the hyperelliptic involution. We have an extra involution $\sigma \in \bar G \cong C_m$ and since extra involutions come in pairs we have $\sigma \tau \in \bar G$. The two fixed spaces of $\sigma$ and $\sigma \tau$ have equations respectively
 \[ \mathcal X_1 : \quad Y^2= X^{{\delta }} + a_1 X^{{\delta }-1} + \dots + a_{{\delta }-1} X + 1\]
and
\[ \mathcal X_2 : \quad Y^2=X(  X^{{\delta }} + a_1 X^{{\delta }-1} + \dots + a_{{\delta }-1} X + 1  ),\]
where $X=x^m$.  The genera $g_1, g_2$ are  respectively $[ \frac{{\delta }-1}2]$ and $[ \frac{{\delta }}2].$ Assume ${\delta }$ is even, say ${\delta }=2k$, then we have $g= g_1+g_2=  {\delta }-1$. When ${\delta }$ is odd, say ${\delta }=2k+1$ we have  $g= g_1+g_2 ={\delta }-1$ and from \cite[Table 3]{beshaj-2}, ${\delta }$ is as follows
$ \frac{2(g+1)}{m}$,  $\frac{2g+1}{m}$, $ \frac{2g}{m}-1$
and $m$ respectively   $\frac{2(g+1)}{{\delta }}$, $ \frac{2g+1}{{\delta }+1}$, $ \frac{2g}{{\delta }+1}$.

Considering each case we get $m=2$ in the first case and $m$ not an integer in the other two.  Therefore, $G \cong V_4$.

Conversely, assume $G \cong V_4$. The equation of the hyperelliptic curve is
\[\mathcal X_g: Y^2=x^{2g+2} + a_1 x^{2g} + \dots + a_g x^2 + 1. \]
There are two extra involutions $\sigma$ and $\sigma \tau$ in $G$ such that $|\sigma|= |\sigma \tau| = 2$. They fix the following curves:
 \[ \mathcal X_1 : \quad Y^2= X^{g+1} + a_1 X^{g} + \dots + a_{g} X + 1\]
and \[ \mathcal X_2 : \quad Y^2=X(  X^{g+1} + a_1 X^{g} + \dots + a_{g} X + 1  ),\]
where $X=x^2$. If we evaluate the genera we have respectively    $g_1=\left[\frac{(g+1)-1}2 \right] = \left[\frac{g}2 \right]$ and  $g_2=\left[\frac{(g+2)-1}2 \right] = \left[\frac{g+1}2 \right]$.  Thus $g=g_1 +g_2$.   Therefore,   Jacobian is isogenous to
\[ \Jac  (\mathcal X) \cong  \Jac  (\mathcal X_1)   \times   \Jac  (\mathcal X_2). \]
This completes the proof.

\qed

\subsection{Non-hyperelliptic curves}

\noindent Next, we generalize the previous theorem.

\begin{thm}\label{thm:superelliptic}
 Let $\mathcal X_g$ be a level $n$ superelliptic curve and $C_m = \< \bar \sigma \> \hookrightarrow \bAut (\mathcal X_g)$, where  $m \geq 2$ and  the equation of $\mathcal X_g$ is  $ y^n = f(x^m)$,
with $\deg (f) = d = {\delta }\, m$,  $d > n$.  Then there exist degree $m$ coverings $\pi : \mathcal X_g \to \mathcal X_i$, $i=1, 2$  where
\[  \mathcal X_1 : \quad Y^n=f(X) \quad and  \quad \mathcal X_2 : \quad  Y^n= X f(X).  \]
Then,  \[ \Jac  (\mathcal X) \cong \Jac  (\mathcal X_1 )  \times \Jac  ( \mathcal X_2 ) \]
if and only if
\begin{equation}\label{eq_m}
  {\delta } (n-1) (m-2) = 1 -   \left(\gcd({\delta }+1, n) + \gcd({\delta }, n) - \gcd({\delta } m, n)    \right).
\end{equation}
\end{thm}

\proof
 Let $\mathcal X_g$ be a superelliptic curve with and extra automorphism of order $m \geq 2$ and equation $y^n = f(x^m)$.  There is the superelliptic automorphism
\[ \tau :   \, \,   (x, y) \to (x, \zeta_n y), \quad and \quad \bar \sigma :   \, \,  (x, y) \to (\zeta_m x, y) .  \]
   We denote by $\sigma$ the lifting of $\bar \sigma$ in $\Aut (\mathcal X)$.   Then, $\sigma\tau=\tau\sigma$.

Let  $H_1 : =\<  \sigma \> $ and  $H_2 : = \< \sigma \tau \>$ be subgroups in $G$.  Then, $| H_1 | = n$ and $ | H_2 | = \lcm (n, m)$.    Thus, we have $H:=H_1 \times H_2 \hookrightarrow G$. It is easy to check that $g \left(\mathcal X_g / (H_1 H_2) \right)=0$.

Moreover,    $\sigma $ and $\sigma \tau$ fix the curves
 \[ \mathcal X_1 : \quad Y^n= X^{{\delta }} + a_1 X^{{\delta }-1} + \dots + a_{{\delta }-1} X + 1\]
and \[ \mathcal X_2 : \quad Y^n=X(  X^{{\delta }} + a_1 X^{{\delta }-1} + \dots + a_{{\delta }-1} X + 1  ).\]
Let $g_1$ and $g_2$ denoted their genera respectively.  From Lemma 3.1 we have  that
\[ g_1 = 1 + \frac 1 2 \left( n {\delta } - n - {\delta } - \gcd({\delta }, n)   \right)  \]
and
\[ g_2 = 1 + \frac 1 2 \left( n ({\delta }+1) - n - ({\delta } + 1)  - \gcd({\delta } + 1, n)   \right).  \]
Then we have
\[ g_1 + g_2 =  \frac 3 2 + n{\delta } - \frac n 2 - {\delta } - \frac 1 2 \left( \gcd( {\delta }, n) + \gcd({\delta }+1, n) \right). \]
The genus of $\mathcal X$ is
 %
 \[ g = 1 + \frac 1 2 \left(  n {\delta } m - n - {\delta } m - \gcd(m {\delta }, n)  \right). \]
Then, $g=g_1+g_2$ implies that
\[  {\delta }  (n-1) (m-2) = 1 -   \left(\gcd({\delta }+1, n) + \gcd({\delta }, n) - \gcd({\delta } m, n)    \right). \]
Thus,  from Eq.~\eqref{dec_2} we have that
\[ \Jac  (X_g) \cong \Jac  (\mathcal X / H_1) \times \Jac  (\mathcal X / H_2) \]
which completes the proof.

\endproof

\begin{cor}
 Let $\mathcal X_g$ be a level $n$ superelliptic curve as in Theorem~\ref{thm:superelliptic}.  Furthermore, suppose $n$ is prime.  Then $m=2$ or $m=3$.

In particular, one of the following situations is true:
\begin{itemize}
\item $n$ is any prime, $m=2$, and $\delta \equiv 0 \text{ (mod $n$)}$;
\item $n=2$, $m=2$, and $\delta$ is odd;
\item $n=3$, $m=3$, $\delta=1$;
\item $n$ is any odd prime, $m=2$,  $\delta \not\equiv 0,-1 \text{ (mod $n$)}$.
\end{itemize}
\end{cor}

\begin{proof}
If $n$ is prime, then the gcd's on the right hand side of the Eq.~\eqref{eq_m} are each either $1$ or $n$.  We consider cases below.  Note that at least one of $\gcd(n,\delta)$ and $\gcd(n,\delta+1)$ equals 1 because $n$ is prime.

If $\gcd(n,\delta)=n$, then $\gcd(n,m\delta)=n$, so \[1 + \gcd(n,m\delta) - \gcd(n,\delta) - \gcd(n,\delta+1) = 1+n-n-1=0.\]  Thus, $\delta(m-2)=0$.  But $\delta\neq0$, so $m=2.$  And since $\gcd(n,\delta)=n$, then $\delta\equiv0\text{ (mod $n$)}$.

If $\gcd(n,\delta+1)=n$, then $\gcd(n,\delta)=1$, so $\gcd(n,m\delta)=\gcd(n,m)$.  Thus, \[1 + \gcd(n,m\delta) - \gcd(n,\delta) - \gcd(n,\delta+1) = \gcd(n,m)-n.\]  Since we have $m\geq2$, we know $\delta(n-1)(m-2)\geq0$, so $\gcd(n,m)-n\geq0$, so $\gcd(n,m)\geq n$, so $\gcd(n,m)=n$.  Thus, \[1 + \gcd(n,m\delta) - \gcd(n,\delta) - \gcd(n,\delta+1) = 0,\] so $\delta(n-1)(m-2)=0$, so we again conclude $m=2$.  Since $\gcd(n,m)=n$ with $n$ prime and $m=2$, we get $n=2$.  And since $\gcd(n,\delta+1)=n$, $\delta$ is odd.

If $\gcd(n,\delta)=\gcd(n,\delta+1)=1$, then we consider $\gcd(n,m)$. If $\gcd(n,m)=1$, then again \[1 + \gcd(n,m\delta) - \gcd(n,\delta) - \gcd(n,\delta+1) = 0,\] so $m=2$ and $\delta\not\equiv 0,-1\text{ (mod $n$)}$.
If $\gcd(n,m)=n$, then \[1 + \gcd(n,m\delta) - \gcd(n,\delta) - \gcd(n,\delta+1) = n-1,\] so $\delta(m-2)=1$, so $\delta=1$ and $m=3$.  Since $\delta=1$ and $\gcd(n,\delta+1)=1\neq2$, this implies that $2\nmid n$, so $n$ is an odd prime.
\end{proof}

\begin{exa}
For triagonal curves, $n=3$.  Then either i) $m=3$ and $\delta=1$; or ii) $m=2$ and $\delta\equiv0,1\text{ (mod 3)}$.
\end{exa}

\section{Jacobians with superelliptic components}
In this section we study a family of non-hyperelliptic curves introduced in \cite{Ya}, whose    Jacobians decompose into factors which are  hyperelliptic Jacobians.  We will extend this family of curves and investigate if we can obtain in this way curves of arbitrary large genus having decomposable Jacobians.

In \cite{Ya}  were introduced a family of curves  in $\mathbb P^{s+2}$ given by the equations
\begin{equation}\label{curve}
\left\{
\begin{split}
zw & = c_0x^2+c_1xw+c_2w^2 \\
y_1^r & = h_1 (z, w) := z^r+ c_{1, 1} z^{r-1} w + \cdots + c_{r-1, 1} zw^{r-1} + w^r,\\
 & \dots  \\
y_s^r & = h_s (z, w) :=  z^r+ c_{1, s} z^{r-1} w + \cdots + c_{r-1, s} zw^{r-1} + w^r,  \\
\end{split}
\right.
\end{equation}
where  $c_i \in k$, $i=0, 1, 2,$ and   $c_{i,j} \in k$ for $i=1, \cdots, r$, $ j=1, \cdots , s$.
The variety $\mathcal X_{r, s}$ is an algebraic curve since the function field of $\mathcal X_{r, s}$ is a finite extension of $k(z)$. $\mathcal X_{r, s}$ is a complete intersection.
For a proof of the following facts see  \cite{Ya}.

\begin{rem}Let $\mathcal X_{r, s}$ be as above. Assume that $\mathcal X_{r, s}$ is smooth and $c_0\neq 0$. Then  the following hold:

i) The genus of  $\mathcal X_{r, s}$ is
\[ g(\mathcal X_{r, s}) = (r-1)(rs\cdot2^{s-1}-2^s+1).\]

ii) If $r\geq3$ and $s\geq1$, then $\mathcal X_{r, s}$ is non-hyperelliptic.
\end{rem}

Fix $r\geq 2$.  Let $\lambda$ be an integer such that $1  \leq \lambda\leq s$.    Define the superelliptic curve $C_{r, \lambda, m}$ as follows
\[ C_{r, \lambda, m}:  \qquad Y^r=\prod_{i=1}^{\lambda} h_{i}(X^m, 1),\]
for some $m\geq 2$.  The right side of the above equation has degree $d = rm \lambda$. Using  Lemma~\ref{lem_1} we have that
\[ g( C_{r, \lambda, m} ) = 1 + \frac 1 2 \left(  r^2 m\lambda - r- m\lambda r - \gcd(\lambda r m, r)\right). \]
Hence,
\begin{equation} \label{eq_g}
g( C_{r, \lambda, m} ) = 1 + \frac r 2 \left((r-1)\lambda m -2  \right).
\end{equation}

\begin{thm}
Let $C_{r, \lambda, m}$ be a generic curve as above. Then the following hold

i) $\bAut \left( C_{r, \lambda, m} \right) \cong C_m$.

ii) $\bAut \left( C_{2, \lambda, m} \right) \cong D_{2m}$.

\end{thm}

\proof  i) The equation of $C_{r, \lambda, m}$ is $y^r= g(x^m)$. Then the curve has the following two automorphisms
\[ \tau (X, Y) \to (X, \zeta_r Y), \quad \sigma (X, Y) = (\zeta_m X, Y).\]
Since $\tau$ commutes with all automorphisms then $\< \tau \> \triangleleft \Aut (C_{r, \lambda, m} )$ and $C_{r, \lambda, m} / \< \tau \>$ is a genus 0 curve.  Then $\< \sigma \>  \hookrightarrow \bAut (C_{r, \lambda, m}) $. If $C_{r, \lambda, m}$ is a generic curve then
$\< \sigma \>  \cong \bAut (C_{r, \lambda, m}) $.

ii)  When $r=2$ then the equation of $C_{2, \lambda, m}$ is given by
\[ y^r = \prod_{i=1}^\lambda (X^{2m} + a_i X^m + 1). \]
From \cite{beshaj-2} such curves have reduced automorphism group isomorphic to $D_{2m}$. This completes the proof.
\qed

The following theorem determines the full automorphism group of such curves.

\begin{thm
Let $C_{r, \lambda, m}$ be a generic curve as above and G:=$\Aut \left( C_{r, \lambda, m} \right)$. Then the following hold \\

\noindent a) If    $\bAut \left( C_{r, \lambda, m} \right) \cong C_m $, then $G \cong C_{mn}$ or   $G$ is isomorphic to
\begin{center}
$\left\langle \gamma, \sigma \right|\gamma^n=1,\sigma^m=1,\sigma\gamma\sigma^{-1}=\gamma^l \rangle$
\end{center}
where (l,n)=1 and $l^m\equiv 1$ (mod n). But if $(m,n)=1$, then $l=n-1$. \\

\noindent  b) If    $\bAut \left( C_{r, \lambda, m} \right) \cong D_{2m} $, then  \\

(1) If n is odd then $G \cong D_{2m} \times C_n$.

(2) If n is even and m is odd then $G \cong D_{2m} \times C_n$ or $G$ is isomorphic to the group with presentation
\[ \left\langle \gamma, \sigma, \tau \right|\gamma^n=1,\sigma^2=\gamma,\tau^2=\gamma^{n-1},(\sigma\tau)^m=\gamma^{\frac{n}{2}},\sigma\gamma\sigma^{-1}=\gamma,\tau\gamma\tau^{-1}=\gamma \rangle. \]

(3) If n is even and m is even then $G $ is isomorphic to one of the following groups $D_{2m} \times C_n$, $D_{2mn}$, or one of the following
\begin{align*}
\begin{split}
G_1=& \left\langle \gamma, \sigma, \tau \right|\gamma^n=1,\sigma^2=\gamma,\tau^2=1,(\sigma\tau)^m=1,\sigma\gamma\sigma^{-1}=\gamma,\tau\gamma\tau^{-1}=\gamma^{n-1} \rangle,\\
G_2=& \left\langle \gamma, \sigma, \tau \right|\gamma^n=1,\sigma^2=\gamma,\tau^2=\gamma^{n-1},(\sigma\tau)^m=1,\sigma\gamma\sigma^{-1}=\gamma,\tau\gamma\tau^{-1}=\gamma \rangle,\\
G_3=& \left\langle \gamma, \sigma, \tau \right|\gamma^n=1,\sigma^2=\gamma,\tau^2=1,(\sigma\tau)^m=\gamma^{\frac{n}{2}},\sigma\gamma\sigma^{-1}=\gamma,\tau\gamma\tau^{-1}=\gamma^{n-1} \rangle, \\
G_4=& \left\langle \gamma, \sigma, \tau \right|\gamma^n=1,\sigma^2=\gamma,\tau^2=\gamma^{n-1},(\sigma\tau)^m=\gamma^{\frac{n}{2}},\sigma\gamma\sigma^{-1}=\gamma,\tau\gamma\tau^{-1}=\gamma \rangle.
\end{split}
\end{align*}
\end{thm}

\proof The full automorphism group $G$ of such curves is a degree $n$ central extension of $C_m$ or $D_{2m}$.  Such extensions were determined
in \cite[Thm.~3.2]{Sa1} and \cite[Thm.~3.3]{Sa1}.

\qed

\subsection{Decomposition of Jacobians}

In \cite{Ya} it is proved that the Jacobians of $C_{2, \lambda, 2}$, defined over any algebraic number field $k$,  curves are isogenous to a product of superelliptic Jacobians. A similar theorem was suggested for curves $C_{r, \lambda, 2}$ and was remarked that the proof was similar to the case $C_{2, \lambda, 2}$ curves. The proof is of arithmetic in nature and is based on K\"uneth's formula, the Frobenius map on $Gal (\bar k / k)$, Chebotarev's theorem, and Faltings theorem.

We will generalize such theorems for curves $C_{r, \lambda, m}$ over an algebraically closed field $k$ of characteristic relatively prime to $r$. Our proof is based solely on automorphisms of curves and it is much simpler than in \cite{Ya}.

Let  $\mathcal X_{r, s}$ be a generic algebraic  curve  defined over an algebraically closed field  $k$   and $C_{r, \lambda, m}$ as above. Then we have the following.

\begin{thm}
The Jacobian $\Jac  (\mathcal X_{r, s})$ is isogenous to the product of the $C_{r, \lambda, m}$, for $1 \leq \lambda \leq s$, namely
\[  \Jac (\mathcal X_{r, s}) \cong  \prod_{1 = \lambda}^s   \Jac (C_{r, \lambda, m}),   \]
if and only if
\begin{equation} r=4\cdot\frac{1+s-2^s}{ms(s+1)-s\cdot2^{s+1}}\label{num-thy-eqn}.
\end{equation}
\end{thm}

\proof
We denote by $\sigma_i (x, y_i, z) \to (x, \zeta_r y_i, z)$, for $i=1, \dots , s$.  Then the quotient spaces $\mathcal X_{r, s} / \< \sigma_i \>$ are  the curves $C_{r, i, s}$, for  $i=1, \dots , s$.  Since $\sigma_i$ is a central element in $G = \Aut (\mathcal X_{r, s} )$ then $H_i := \< \sigma_i \> \triangleleft G$, for all $i=1, \dots , s$.  Obviously, for all $i \neq j$ we have $H_i \cap H_j = \{ e \}$.  Hence, $H_1, \dots , H_s$ forms a partition for $G$.

The genus for every $C_{r, i, s}$, by Lemma~\ref{lem_1} is given by Eq.~\eqref{eq_g}.   Then we have
\[
\begin{split}
\sum_{\lambda=1}^s \, g \left( C_{r, \lambda, m} \right)  & = \sum_{\lambda=1}^s \left( 1 + \frac r 2 \left((r-1)\lambda m -2  \right)  \right)  \\
& = s(r-1) \left( \frac r 4 m (s+1) -1    \right).   \\
\end{split}
 \]
From the results of Eq.~\eqref{dec_2}  we have that
\[  \frac r 4  m s (s+1) - s   =  rs \cdot 2^{s-1} - 2^s +1.  \]
Hence,
\[ r = 4 \cdot \frac {1+s-2^s} {ms(s+1)- s \cdot 2^{s+1}}.\]
This completes the proof.
\qed

\begin{rem}
For $m=2$ this result is the case of Theorem~4.2 in \cite{Ya}. We get  $r= \frac  2 s$. Hence, $s=1$ or $s=2$.
Therefore,    Theorem~4.2 in \cite{Ya} is true only for curves $F_{m, 1}$ or $F_{m, 2}$.
\end{rem}

Next we determine integer combinations of $r,m$, and $s$ that satisfy   Eq.~\eqref{num-thy-eqn}.  First we need the following lemma.

\begin{lemma}
For integers $a, n$ with $n>1$, suppose $a^n\equiv1\text{ (mod }n)$.  Let $p$ be the smallest prime divisor of $n$.  Then $a\equiv1\text{ (mod }p)$.
\end{lemma}

\begin{proof}
Suppose $a^n\equiv1\text{ (mod }n)$ for integers $a,n$ with $n>1$.  Let $p$ be the smallest prime divisor of $n$.  Since $a^n\equiv1\text{ (mod }n)$, then $a^n\equiv1\text{ (mod }p)$.  Let $d$ be the order of $a$ modulo $p$.  Since $a^{p-1}\equiv1\text{ (mod }p)$, $d$ divides $p-1$, so $d<p$.  And since $a^{n}\equiv1\text{ (mod }p)$, $d$ divides $n$.  However, since $p$ is the smallest prime divisor of $n$, the only divisor of $n$ which is less than $p$ is 1.  Hence, $d=1$, so the order of $a$ modulo $p$ is 1, so $a\equiv1\text{ (mod }p)$, as desired.

\end{proof}

\begin{prop}Suppose $r,m,s\in\mathbb{N}$ satisfy Eq.~\ref{num-thy-eqn}.  Then $mrs=4k$ for some odd integer $k$. Moreover,

i) If $s\equiv1\text{ (mod 2)}$, then $s=1$.

ii) If $s\equiv2\text{ (mod 4)}$, then $s=2t$ for some odd integer $t$ which satisfies $4^t\equiv1\text{ (mod }t)$.  Furthermore, $t$ is a multiple of 3.

iii) If $s\equiv0\text{ (mod 4)}$, then $s=4u$ for some odd integer $u$ which satisfies $16^u\equiv1\text{ (mod }u)$.  Furthermore, $u$ is a multiple of 3 or 5.
\end{prop}

\begin{proof}
We first determine possible values of $s$ for which $r$, which is given by the equation \[ r = 4 \cdot \frac {1+s-2^s} {ms(s+1)- s \cdot 2^{s+1}},\] can be an integer.  We proceed in cases, considering the highest power of 2 that divides $s$.  In particular, the powers that we consider are 0, 1, 2, and then at least 3.

First, suppose $s$ is an odd integer.  Since $s$ divides the denominator, it follows that $s$ divides $4(1+s-2^s)$.  Since $\gcd(s,4)=1$, we conclude that $2^s\equiv 1\text{ (mod $s$)}$.  If $s=1$, then this congruence is satisfied.  Now, suppose $s>1$ and let $p$ be the smallest prime divisor of $s$.  By the Lemma above, one has $2\equiv 1\text{ (mod $p$)}$, so $p$ divides 1, which is impossible.  Thus, if $s$ is odd, then $s=1$.

Also, note that in this case if $s=1$, then one finds that $m=2$ and $r=2$.

Next, suppose $s=2t$ for some odd integer $t$.  Substituting in, we get \[ r = 4 \cdot \frac {1+2t-2^{2t}} {m(2t)(2t+1)- 2t \cdot 2^{2t+1}}=2\cdot\frac{1+2t-4^t}{mt(2t+1)-2t\cdot4^{t}}.\]  As above, since $t$ divides the denominator, $t$ divides $2(1+2t-4^t)$.  Since $\gcd(t,2)=1$, we conclude that $4^t\equiv 1\text{ (mod $t$)}$.  If $t=1$, then this congruence is satisfied.  Now, suppose $t>1$ and let $p$ be the smallest prime divisor of $t$.  By the Lemma above, one has $4\equiv1\text{ (mod $p$)}$, which implies $p$ divides 3, so $p=3$.  Thus, $s=2t$ for some integer $t$ which is either $1$ or is a multiple of 3 and which satisfies $4^t\equiv1\text{ (mod $t$)}$.

Next, suppose $s=4u$ for some odd integer $u$.  Substituting in, we get \[ r = 4 \cdot \frac {1+4u-2^{4u}} {m(4u)(4u+1)- 4u \cdot 2^{4u+1}}=\frac{1+4u-16^u}{mu(4u+1)-2u\cdot16^{u}}.\]  As above, since $u$ divides the denominator, $u$ divides $(1+4u-16^u)$.  Thus, $16^u\equiv 1\text{ (mod $u$)}$.  If $u=1$, then this congruence is satisfied.  Now, suppose $u>1$ and let $p$ be the smallest prime divisor of $u$.  By the Lemma above, one has $16\equiv1\text{ (mod $p$)}$, which implies $p$ divides 15, so $p=3$ or $p=5$.  Thus, $s=4u$ for some integer $u$ which is either $1$ or is a multiple of either 3 or 5 and which satisfies $16^u\equiv1\text{ (mod $u$)}$.

Finally suppose $s=8v$ for some (even or odd) integer $v$.  Substituting in, we get \[ r = 4 \cdot \frac {1+8v-2^{8v}} {m(8v)(8v+1)- 8v \cdot 2^{8v+1}}=\frac{1+8v-2^{8v}}{2mv(8v+1)-2v\cdot2^{8v+1}}.\]  Thus, since $2$ divides the denominator, $2$ divides $(1+8v-2^{8v})$, so $2^{8v}\equiv 1\text{ (mod 2)}$, which occurs only if $v=0$.  Thus, $s=0$.

To show $mrs=4k$ for some odd integer $k$, we have two cases to consider.  If $s$ is odd, then $s=1$, so $m=r=2$ and thus $mrs=4$.  If $s$ is even, then we clear denominators of Eq.~\ref{num-thy-eqn} and consider the equation modulo 8 to get
\[rms(s+1)-rs2^{s+1}\equiv 4\cdot(1+s-2^s)\text{ (mod $8$)}.\]
Since $s$ is even and thus at least $2$, this simplifies to $rms\equiv4\text{ (mod $8$)}$, so $mrs=4k$ for some odd integer $k$.
\end{proof}

\begin{rem}
i) When $s=2t$ as in the proof of the theorem, then the first few integer values of $t$ are $1, 3, 9, 21, 27, 63, 81, 147, 171, 189, 243$.  This is sequence A014945 in the Online Encyclopedia of Integer Sequences.

ii) When $s=4u$ as in the proof of the theorem, then  the first few integer values of $u$ are $1, 3, 5, 9, 15, 21, 25, 27, 39, 45, 55, 63, 75, 81, 105, 117$.  This is sequence A014957 in the Online Encyclopedia of Integer Sequences.
\end{rem}

We then search for values of $s$ that satisfy the above proposition.  Searching among $1\leq s < 500$, we find the following possible values for $s$.  \begin{align*}\{1, 2, 4, 6, 12, 18, 20, 36, 42, 54, 60, 84, 100, 108, 126, 156, 162, \\ 180, 220, 252, 294, 300, 324, 342, 378, 420, 468, 486, \dots \}.\end{align*}  We now check each of these to find corresponding integer values of $m$ and $r$.

\begin{cor}
Suppose $r,m,s\in\mathbb{N}$ satisfy Eq.~\ref{num-thy-eqn}, and suppose $1\leq s<500$ with $s\neq 300, 420, 468$.  Then $s\in\{1,2,6,18,42,126,162,294,378,486\}.$  We display the combinations of integers $s,m$, and $r$ in the table below.  Note that $m$ and $r$ grow quickly relative to $s$, hence the scientific notation for the cases where $s>100$.

\begin{center}
\begin{tabular}{|c|c|c|}\hline
$s$ & $m$ & $r$ \\ \hline \hline
$1$ & $2$ & $2$ \\ \hline
$2$ & $2$ & $1$ \\ \hline
$6$ & $18$ & $19$ \\ \hline
$18$ & $27594$ & $29125$ \\ \hline
$42$ & $204560302842$ & $209430786241$ \\ \hline
$126$ & $1.3397\times10^{36}$ & $1.3503\times10^{36}$ \\ \hline
$162$ & $7.1730\times10^{46}$ & $7.2173\times10^{46}$ \\ \hline
$294$ & $2.1579\times10^{86}$ & $2.8391\times10^{62}$ \\ \hline
$294$ & $2.1579\times10^{86}$ & $3.3025\times10^{31}$ \\ \hline
$294$ & $2.1579\times10^{86}$ & $2.2665\times10^{27}$ \\ \hline
$378$ & $3.2488\times10^{111}$ & $3.2574\times10^{111}$ \\ \hline
$486$ & $8.2050\times10^{143}$ & $8.2219\times10^{143}$ \\ \hline
$486$ & $8.2050\times10^{143}$ & $1.4596\times10^{20}$ \\ \hline
\end{tabular}
\end{center}
\end{cor}

Note that there are multiple combinations of integers $r$ and $m$ when $s=294$ or $486$.  Also note that we were unable to get results for $s=300, 420$, or $468$.  This is because of the time required to factor $(1+s-2^s)$ in those cases.

Interestingly, we do not have any cases where $s=4u$ for some odd integer $u$.

\medskip

\noindent \textbf{Acknowledgments:} The authors want to thank the anonymous referee for helpful comments and remarks.

\begin{bibdiv}
\begin{biblist}

\bibselect{bibl-1}

\end{biblist}
\end{bibdiv}

\end{document}

redpajama_set_name: RedPajamaArXiv


I was able to purchase upgrades to my inventory from Hestu from the road on the way to Kakariko village. He said he is leaving to go to the woods, but I can't find him there. Where has Hestu gone?
After Hestu leaves the road near Kakariko village, he can appear at any of the stables in the game and shortly after that he will show up at the town in the Korok forest where the Deku tree lives.


text:
Archive for the 'Painting' Category
ART AUCTION RESULTS SOTHEBY'S HONG KONG PRESS RELEASE
We present you with the latest press release from Sotheby's Hong Kong on their autumn sale of modern and contemporary Southeast Asian paintings:
SOTHEBY'S HONG KONG
MODERN AND CONTEMPORARY SOUTHEAST ASIAN PAINTINGS 2010 AUTUMN SALE
TOTALS HK$78 MILLION / US$10 MILLION
(high estimate: HK$45 million / US$5.7 million*)
THE HIGHEST TOTAL FOR A VARIOUS-OWNER SALE
IN THIS CATEGORY AT SOTHEBY'S HONG KONG
ACHIEVING NUMEROUS ARTIST RECORDS AT AUCTION
"FATHER OF INDONESIAN MODERNISM" –
S.SUDJOJONO'S A NEW DAWN SOLD FOR AN IMPRESSIVE
HK$10.7 MILLION / US$1.4 MILLION
OVER 4 TIMES THE HIGH ESTIMATE
FILIPINO ARTIST RONALD VENTURA 'S NATURAL-LIES FETCHED
HK$2.5 MILLION / US$326,000
9 TIMES THE HIGH ESTIMATE
Other artist records were set for works by Indonesian artists including
Gede Mahendra Yasa, Ay Tjoe Christine, Samsul Arifin, Hendra Gunawan and Filipino artist Andres Barrioquinto, among others
Following the tremendous success of the Spring sale, Sotheby's Autumn sale of Modern and Contemporary Southeast Asian Paintings today commanded a stunning total of HK$78 million / US$10 million (high estimate: HK$45 million / US$5.7 million*), the highest sale total for a various owners sale in this category at Sotheby's Hong Kong. Today's sale provoked active participation in the room and over the phone. There was particularly strong interest in top-end Southeast Asian contemporary paintings, which led to two auction records set for two artists – S. Sudjojono , Father of Indonesian Modernism, and Filipino artist Ronald Ventura.
MOK Kim Chuan, Sotheby's Head of Southeast Asian Paintings, commented: "Top end Contemporary works fetched strong prices today with many pieces bringing multiples of their pre-sale high estimates. Among Modern works, the supreme highlight was the S. Sudjojono, a museum-quality example of the artist's work which spurred a fierce bidding battle among nine bidders before selling for HK$10.7 million, a price which was four times the top estimate and set a record for the artist at auction. These results confirm the strategy of using conservative estimates to attract competition and let the market set the price level."
The sale of 20th Century Chinese Art and Contemporary Asian Art continue in the evening.
Attached please find the relevant press releases, top-ten list as well as an image of the saleroom for your use. Should you need further information, please do not hesitate to contact Sotheby's Hong Kong Press Office on +852 2868 6755 /Winnie.tang@sothebys.com.
* Estimates do not include buyer's premium
Sotheby's Hong Kong Press
Related Topics: Southeast Asian art, market watch – auctions, business of art, collectors
Australian modern and contemporary arts gain momentum: top five auctioned works listed – September 2010 – the latest auction results reveal a strong preference for paintings
Sotheby's to hold first even international auction house sale of calligraphy in Doha – July 2010 – an unprecedented sale of ancient and new calligraphy works in the Middle East
Saffronart once again pioneers new technology within art industry – July 2010 – details this auction houses' successes with mobile phone bidding
Korean art hit and miss at Seoul Auction Hong Kong: New York Times – July 2010 – explains how Asian collectors favour local artists
Surprising sell-through rates for Chinese art recorded in London's February auctions – March 2010 – details the rise and rise of the Chinese art market
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Posted in Artist Nationality, Auctions, Business of art, China, Collectors, Hong Kong, Market watch, Medium, Painting, Southeast Asian, Venues | Tagged: Andres Barrioquinto, art auction results, auction, auctions, Ay Tjoe Christine, Gede Mahendra Yasa, Hendra Gunawan, MOK Kim Chuan, Ronald Ventura, S. Sudjojono, Samsul Arifin, Sothebys, Sothebys Hong Kong, Southeast Asian art, Southeast Asian paintings | 1 Comment »
How is Chinese ink painting explored in contemporary art? RedBox Review in discussion with Liang Quan
CHINESE INK CHINESE ARTISTS ARTIST INTERVIEWS
In a conversation with Chinese-based art blog RedBox Review the artist Liang Quan (b. 1948), living and working in Shenzhen, China, explains how ink painting is used in contemporary art and how this exploration continues to follow the philosophy of traditional Chinese painting.
Liang Quan is considered as one of the pioneers of contemporary ink painting.
"Ink painting", also known as "wash painting", was developed in China during the Tang Dynasty. Ink painting or shui-mo hua in Chinese (水墨畫) is composed of water, shui and Chinese ink, mo. In Western art, using similar techniques, it is known as drawings.
Liang Quan, 'Tea Stain No. 3', 2008, ink and paper, 63.8x48 cm.
In this conversation Liang Quan highlights to RedBox the difference between ink painting and ink art:
The exploration of using ink and referring to the tradition of Chinese painting is part of a greater narrative to define a cultural identity.
American contemporary artists like Brice Marden and Cy Twombly inspired Liang Quan while he was living and working abroad. On top of using ink painting and water, Liang incorporates paper into his works.
Liang's ink painting seems abstract but in reality he follows the philosophy of this art. He aims to capture the soul of the subject rather that trying to reproduce the exact appearance of it. As he relates to RedBox,
My use of collage, combining strips of ink and/or tea stained paper, may seem abstract to the unknowing eye, and without direct correlation to a depiction of reality. But my works, collages, are actually diagrams of traditional Chinese landscape paintings and the Chinese still life painting genre of birds and flowers.
Interesting difference between ink painting in West and East: perspective
Having explored ink painting in Western art, Liang Quan observed a major difference between it and Chinese landscape painting: multiple points of perspective are used where Western painting uses only one or two. As he relates to RedBox,
To view a Chinese painting, one's eye usually follows the flow of water from the bottom of the mountains as it meanders farther into the hills and up the composition of the painting.
Following this philosophy and adding paper strips and color makes Liang's painting abstract.
After exploring the multiple points of perspective in Chinese landscape painting, Liang Quan combined this concept with the ideals of Nan Pai, also known as Southern School. As said in the RedBox article,
By addressing the theme of Chinese tradition, he is distinguished from his contemporaries choosing to use painting as a depiction of or social response to modern society.
SB/KN/KCE
Related Topics: Chinese artists, definitions, ink
Wilson Shieh revitalises ancient Chinese painting techniques – video – September 2010 – an overview of a ChooChoo TV interview with the artist
Gao Minglu's maximalist exhibition blurs boundaries between traditional and contemporary Chinese art – June 2005 – an excellent profiling of a little-known movement in Chinese art
Liu Kuo-sung London retrospective inspires potential British Museum collection – May 2010 – a brief post on the works that inspired this interest
Chinese ink artist Nobel Laureate Gao Xingjian shows 80 works in Spain – Guardian, Int Literary Quarterly – January 2009 – an exhibition alluding to a number of contemporary art world trends
New York gallery launches series 15 New Chinese Ink artist solo shows – Yishu – October 2008 – outlining a number of solo shows with details on a special show by Wei Ligang
Subscribe to Art Radar Asia for more on contemporary Chinese ink
Posted in Artist Nationality, Calligraphy, China, Chinese, Classic/Contemporary, Collage, Drawing, Ink, Interviews, Landscape, Painting, Shenzhen, Styles, Themes and subjects, Trends | Tagged: Brice Marden, Chinese art, Chinese art blogs, Chinese contemporary art, Chinese ink, Chinese tradition, collage, contemporary art, Cy Twombly, drawings, ink painting, Liang Quan, Nan Pai, perspective, Redbox Review, Saloua Bougajdi, Shenzhen, shui-mo hua, Southern School, Tang Dynasty, Tea Stain No. 3, traditional Chinese Landscape, Western art | 1 Comment »
Artpartment a Hong Kong space for experimental art – video
ARTIST-RUN SPACES VIDEOS PERFORMANCE ART VIDEO ART STOP MOTION
We bring you another summary of an [art]attack show by ChooChooTV, this one profiling C&G Artpartment, founded by Clara Cheung, who studied art in the United States for four years, and Cheng Yee Man (Gum), an HKAPA and RMIT graduate. Artpartment is a gallery and studio space in Hong Kong dedicated to the production and exhibition of experimental art.
Artists Clara Cheung and Cheng Yee Man (Gum) discuss their Hong Kong studio and gallery C&G Artpartment on ChooChooTV.
We set up Artpartment for two reasons. Firstly we wanted a place to exhibit artworks, like an art gallery or a space for experimental art, and secondly we wanted to create a studio to teach painting. Clara Cheung on [art]attack
The artists own collaboration lies in performance art pieces, mostly conducted on the streets of Hong Kong. Says Gum,
"I totally disagree that an exhibition doesn't require an audience;… for any exhibition, the more audience you have the better it is. We want to do things that attract people and performing art can provide that. You are forced to view it since we are on location in front of you."
The video focuses on art created by the pair for the stop motion art group exhibition, "No Money for Art vs. No Time for Art", held at Artpartment. They use video, drawing and painting to create videos expressing the social aspirations behind their work.
"We went to Poland in September for an art camp, it's similar to an artist residency programme, and there were a lot of artists from different countries. Our work that we are exhibiting was inspired during that programme." Clara Cheung on [art]attack
Both artists have strong views about the job of an artist and these are expressed in the video.
"The direction of our artwork is firstly, about our society and secondly, about the art society…. Art should create awareness, it should also be something we've not seen before, so the way we should approach art is to use it to reflect the society and political issues." Cheng Yee Man (Gum) on [art]attack
"Different art media should all be part of the art scene. We need to unite and strengthen the art scene." Clara Cheung on [art]attack
Watch the video here (length 6:39 minutes)
KN/HH
Related Topics: videos, video art, performance art, Hong Kong artists, artist-run spaces
Tsong Pu discusses six artworks: Part II – Installations and art funding – August 2010 – Pu talks about the financial issues facing one of Taipei's longest-running non-commerical spaces
Sa Sa Gallery and Art Project, new artist-run initiatives in Cambodia – July 2010 – an interview with Vuth Lyno, manager of Sa Sa Gallery
Nindityo Adipurnomo talks with Art Radar on "+Road" collaboration with Myanmar artists, "gambling spirit" of Indonesian collectors – July 2010 – Burmese artists generally inexperienced painters, strength lies in performance art
Interventions explore art industry relationships in curator Meenakshi Thirukode's Guild Art Gallery show: interview – June 2010 – Thirukode gives an in-depth explanation of the numerous performance pieces
Contemporary Indian art meets tradition at inaugural 'Sowing Seeds' artist workshop in Rajasthan, Indian – interview – March 2010 – an interview with founder Mr. Vagaram Choudhary
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Posted in Art spaces, Artist Nationality, Artist-run, China, Drawing, Emerging artists, Hong Kong, Hong Kong Artists, Medium, Painting, Performance, Social, Stop motion, Venues, Video, Videos | Tagged: art camps, art galleries, art society, art studios, artist residency, artist residency programmes, artist-run space, artist-run spaces, artist-run spaces China, Artpartment, C&G Artpartment, Cheng Yee Man (Gum), ChooChooTV, Clara Cheung, experimental art, HKAPA, hong kong, Hong Kong art galleries, Hong Kong art studios, Internet TV, No Money for Art vs. No Time for Art, performance art, performance artists, Poland, political art, RMIT, societal art, united states, Video, Video art, [art]attack | Leave a Comment »
Australian modern and contemporary arts gain momentum: top five auctioned works listed
Posted by artradar on September 6, 2010
AUSTRALIAN MODERN AND CONTEMPORARY ART PRICES AND TRENDS
As the 17th Sydney Biennale drew to a close, a recent article published on artprice.com reported on the improvement of the Australian modern and contemporary art market since 2007, despite its confinement to Sydney and Melbourne. There is a strong preference among Australian collectors for paintings, oil, acrylic and figurative work.
The article provides a list of the top ten Australian works which have been sold at the highest price between 2000 and 2010. Here is the list of the top five:
First-Class Marksman (1946) by Sidney Robert Nolan (1917-1992): sold at USD4,103,100 by Menzies Fine Art Auctioneers and Valuers in March 2010
'First-Class Marksman' depicts a square-helmeted Ned Kelly pointing his gun into the Australian bushes to protect himself from the police. Picture taken from deutschermenzies.com.au.
The Olgas for Ernest Giles (1985) by Brett Whiteley (1939-1992): sold at USD2,445,280 by Menzies Fine Art Auctioneers and Valuers in June 2007
"It's a highly charged, erotic painting and the landscape itself is depicted as having the qualities of flesh," said Adrian Newstead, managing director of Deutscher-Menzies, talking to the 'Sydney Morning Post' in 2007 about 'The Olgas for Ernest Giles'. Picture taken from deutschermenzies.com.au."
The Old Time (1969) by John Cecil Brack (1920-1999): sold at USD2,301,320 by Sotheby's in May 2007
'The Old Time' is a painting of a ballroom dancing couple. Picture taken from Art News Blog.
Opera House (1971-1982) by Brett Whiteley (1939-1992): sold at USD1,972,560 by Sotheby's in May 2007
This painting of the Sydney Opera House was owned by Qantas Airline. It hung in the club travellers lounge in Sidney. Picture taken from artquotes.net.
The Bar (1954) by John Cecil Brack (1920-1999): sold at USD1,893,060 by Sotheby's in April 2006
Modelled on Manet's A Bar at the Folies-Bergère, this painting mocks the Six-o'clock swill - the last minute rush to buy drinks in bars due to their early closing. Picture taken from Brookston Beer Bulletin.
Sidney Robert Nolan's First-Class Marksman, fetching over USD4,000,000 in 2010, tops the list. This is against the price trend of Nolan's works, which has been downward since 2007.
Brett Whiteley, named the "most sought Australian artist during the decade" by the article, produced The Olgas for Ernest Giles which has fetched over USD 2,400,000. It has been reported that "100 euros invested in one of his works in 1998 were worth an average of 555 euros by February 2010".
Among the best results of 2009 and 2010 are the sales of works by Norman Alfred Williams Linsay which went for between USD100,000 and 235,000.
In the affordable USD10,000-40,000 price range are the best works by Frederick Cress and large watercolors by John Henry Olsen and Frederick Ronald Williams. In the higher USD40,000-120,000 price range are the still-lifes by Grace Cossington SMITH and tranquil landscapes by Lloyd Frederic REES.
Representing the young generation of artists loyal to the Australian figurative tradition are Rick Amor, Lin Onus and Vincent Fantauzzo. Rick Amor broke the USD100,000 line with The Waiter which fetched USD100,300 at Menzies Fine Art Auctioneers and Valuers in May 2010. The value of Lin Onus' Reflections, Barmah Forest leapt from USD100,600 less than seven years ago to USD200,600 in March 2010 at Deutscher & Menzies. Some of her oils on cardboard from the 1970s can be picked up at less than USD10,000. Vincent Fantauzzo's portrait Brandon fetched USD 43,580 in June 2010 at Menzies Art Brands, Sydney.
While the purchase of contemporary art in Australia is picking up speed, the performance of Aboriginal art has been in serious decline since its peak in 2007. This may be because the buying spree of best works by Aboriginal art masters who have died in the last decade is gradually coming to an end.
Related Topics: Australian artists, lists, trends
Balgo Hills art: Indigenous Australian art by renowned masters in rare tour through Asia – March 2010 – works by a small community of indigenious Australians in a touring exhibition
The 17th Sydney Biennale – Art Radar rounds up highlights, disappointments and critiques – August 2010 – a round-up on this year's Sydney Biennale which showcased a lot of Australian works
The Problem of Asia: Para/Site art exhibition explores Asian identity in Sydney – May 2010 – a list of works in an exhibition in Sydney that tells about Asian identity
ART HK 10 reports strong sales figures, lists major artworks sold – June 2010 – a list of major works sold in the HK international artfair this year
Top 8 Art Radar search terms: contemporary art trends revealed – August 2010 – evaluate results with regard to contemporary art trends
Subscribe to Art Radar Asia for more contemporary art market trends articles
Posted in Artist Nationality, Auctions, Australian, Business of art, Collectors, Individual, Lists, Market watch, Medium, Painting | Tagged: art prices, auction sales, Australian art, Brett Whiteley, Carmen Bat Ka Man, contemporary art, Deutscher & Menzies, First-Class Marksman, Frederick Cress, Frederick Ronald Williams, Grace Cossington Smith, John Cecil Brack, John Henry Olsen, Lin Onus, Lloyd Frederick Rees, modern art, Opera House, Rick Amor, Sidney Robert Nolan, Sothebys, The Bar, The Old Time, The Olgas for Ernest Giles, Vincent Fantuazzo | 2 Comments »
Wilson Shieh revitalises ancient Chinese painting techniques – video
HONG KONG ARTISTS CHINESE INK PAINTING VIDEO
Art Radar Asia brings you another video (length 5:22 minutes) from Internet channel ChooChooTV's show [art]attack. This one features Wilson Shieh, one of the few full-time professional artists working in Hong Kong. Specialising in gong-bi, or fine-line, Shieh is admired for creatively merging traditional Chinese painting techniques with modern art elements. In the video on ChooChooTV, Shieh talks about how he develops his unique style of painting and creates intriguing works with it.
"Before I learned the fine-brush technique, I considered this style as just a kind of antique craftsmanship. But after all, as you can see, I have adopted the fine-brush manner in my work. The ancient sense of beauty looks fresh to contemporary eyes." Wilson Shieh, as quoted on Crown Point Press
Wilson Shieh at work.
While pursuing bachelor's and master's degrees in fine arts at the Chinese University of Hong Kong, Shieh became interested in gong-bi paintings, which inspired him to create an unusual style of work.
Experimenting with the combination of old and new
Since gaining his bachelor's degree in 1994, Shieh has been working on figure paintings. One of the traditional Chinese painting techniques which Shieh has been using in his works is lin-mo, which involves paper-scanning figures onto traditional gong-bi paintings.
To introduce modern elements in his works, Shieh replaces the ancient costumes of the scanned figures with modern clothing. He also experiments with nude bodies, taking the work "back to basics and nature" and removing the sense of time. An example of this is "Musical Family" (2008), a set of paintings in which nude bodies imitate instruments.
Click here to view more videos from ChooChooTV show [art]attack and read our summaries of them.
Related topics: Hong Kong artists, ink painting, videos
Anti "commercial" art, Luk Tsing Yuen comments on corporate greed: video – August 2010 – also from ChooChoo TV's [art]attack, watch the Hong Kong artist share his anti-commercial art
Your interpretation or mine? Hong Kong artist Chow Chun Fai reflects in video interview – July 2010 – also from ChooChoo TV's [art]attack, watch the Hong Kong artist share the paradox of meaning of art
Gao Minglu's maximalist exhibition blurs boundaries between traditional and contemporary Chinese art – June 2010 – discusses maximalist art and artists in the context of an exhibition
Liu Kuo-sung London retrospective inspires potential British Museum collection – May 2010 – reports British interest in the works of one of the founders of the New Ink Painting Movement
gallery EXIT a new Hong Kong gallery to watch – May 2010 – introduces a new high quality Hong Kong gallery
Subscribe to Art Radar Asia to for more videos on comtemporary Asian artists
Posted in Artist Nationality, Hong Kong Artists, Ink, Medium, Painting, Videos, Wilson Shieh, Z Artists | Tagged: Carmen Bat Ka Man, Chinese painting, Chinese University of Hong Kong, ChooChooTV, contemporary artist, Exercise, fine arts, fine-line, gong-bi, Hong Kong artist, ink painting, lin-mo, Musical Family, paper scanning, traditional Chinese art, Video, Video Interview, Wilson Shieh | Leave a Comment »
Comic art of Popok Tri Wahyudito portrays scenes of transport calamity
GALLERY SHOWS COMIC ART DRAWING INDONESIA
In July this year, Valentine Willie Fine Art (VWFA) partnered with Kuala Lumpur's The Annexe Gallery to bring "BERGERak" to Malaysia. In his first Malaysian solo, Indonesian artist Popok Tri Wahyudi, uses "Jogja comic style" to create paintings which narrate the experiences of "cattle-class" airline travellers and other mass transport users. His work is accessible to a wide audience because of its familiar subject matter and simple, colorful presentation.
'Please Let Me Go', 2010, acrylic on canvas, 40 x 188 cm. Image courtesy of VWFA.
"Popok Tri Wahyudhi's works in his first Malaysian solo exhibition are stories about commuting, travelling, human mobility and migration. Presented in a wide range of media, from paintings and drawings to woodblock prints, silkscreen on canvas and mini sculptures, these bittersweet and sometimes macabre narratives negate the glamorous images of the jet set…" Valentine Willie Fine Art
The artist is one of the founding members of Apotik Komik, an artist group formed in 1997 by thirteen students from Indonesian Institute of the Arts, Yogyakarta. The group first created mural work and then moved into printing comics, publications more visual and alternative than what was available in Indonesia at that time. Their style, influenced heavily by popular culture, is known as "playful".
'...oops!!!', 2010, woodcut on paper, 79.5 x 54.5 cm. Image courtesy of VWFA.
He is most well known for portraying Indonesian life and political situations in a sinister comic light. However he has worked with international subject matter, most notably during artist residencies at California's 18th Street Art Center in 2001 and the Akademie Schloss Solitude in Stuttgart in 2007. In addition to making paintings in his signature comic style, he has also worked on large scale wall art and created and exhibited three-dimensional pieces.
Popok Tri Wahyudi was born in Mojokerto, East Java, in April, 1973.
Related Topics: Indonesian artists, Southeast Asian artists, gallery shows, drawing
Pop culture references abound in Indonesian art: curator Eva McGovern discusses Indieguerillas' Happy Victims and the Southeast Asian art climate – June 2010 – McGovern explains the differences between the Indonesian, Malaysian and Singaporean art communities
Is Singapore threatening Hong Kong as next Asian art mecca? Wall Street Journal – November 2009 – we link to and summarise this article by Alexandra A. Seno, a Hong Kong-based writer
Inspiring art in important Indonesian art shows Spring 2009 – April 2009 – Adeline Ooi talks about her tour of some of the most important exhibitions of Indonesian art around Southeast Asia
Subscribe to Art Radar Asia for more on Southeast Asian gallery shows
Posted in Artist Nationality, Comic, Drawing, Events, Gallery shows, Indonesian, Malaysia, Painting, Political, Pop Art, Popok Tri Wahyudi, Sculpture, Southeast Asian, Urban, Vehicles, Venues | Tagged: ...oops!!!, 18th Street Art Center, Ad poster #3, airline travel, Akademie Schloss Solitude, Apotik Komik, art exhibitions, art gallery shows, BERGERak, California, cattle-class, comic art, comic style, comics, drawings, East Java, Indonesian art, Indonesian artists, Indonesian Institute of the Arts, Kate Nicholson, life art, Malaysia, Malaysian art galleries, mini sculpture, Mojokerto, mural, narration, paintings, playful art, Please Let Me Go, political art, political artist, pop culture, Popok Tri Wahyudi, popular culture, silkscreen on canvas, social art, Southeast Asian art, Southeast Asian art galleries, Southeast Asian artists, storytelling, Stuttgart, The Annexe Gallery, transport, valentine willie fine art, VWFA, wall art, woodblock prints, Yogyakarta | Leave a Comment »
Japanese artist Hiroshige Fukuhara reappears after 8 year absence – Art Radar interview
JAPANESE ARTIST INTERVIEW PAINTING DRAWING EMERGING ARTISTS
Eight years ago, Japanese artist Hiroshige Fukuhara was building up a successful career as a promising contemporary artist. He showed work at P.S.1 Contemporary Art Center in 2001 then disappeared from the contemporary art world. Then, in 2009, he reappeared at Tokyo's ULTRA002 art fair, and in March this year exhibited work at NYC PULSE. Last month, Art Radar Asia spoke with Fukuhara in a special interview in which he talked about artwork from his recent solo exhibition "Binary" and explained what he has been doing in the eight years that he withdrew from the art world.
Fukuhara is represented by Ai Kowada Gallery in Ebisu, Japan, where he had a solo exhibition, "Binary", earlier this year. His artwork from this exhibition features a series of drawings on which he sketches images of flora and fauna onto a black background with pencil, making the image difficult to see in certain lighting or at certain angles. We interviewed Fukuhara at this gallery, surrounded by his most recent work. Here he explained the reasons he chose this new medium and talked about his inspiration for the title of the exhibition. We discussed his background, what drives him and his art and the challenges that face young artists working today.
Artist Hiroshige Fukuhara next to his piece, 'The Night' at his latest solo exhibition at Ai Kowada Gallery in Japan. Image property of Art Radar Asia.
When did you first feel that you were an artist?
Something like, when I felt different from other people? (laughs)
You mean you wanted to do something different?
Yes, when I thought that, and also when there are judgment calls between something that's supposed to be "good", and "bad", and I felt that, even though I know what's socially right, morally right, I want to take those social and moral judgments separately. For example, with morals, morally, something could be bad, but it can still be good. I sometimes made that kind of judgment, but people around me, if it was morally bad they would always consider it bad. So there's a difference there.
After exhibiting at P.S.1 in 2001, you quit all activity as an artist to pursue other work until ULTRA002 in 2009. What influenced your decision to quit activity as an artist?
When I was in university, what I made was the same as now, fine art. Then, when I was there, Phillip Morris [International] did things like Art Hour. I was remaining as a candidate until the finals – well there were twenty, thirty of us – but I was one of the finalists. Then, that became a trigger that led to some small exhibitions.… That time it was already the final. There were people from other countries in the finalists, so there were a quite lot of people, but the Finalist Award pretty much triggered other things, several other things, but after that, I sort of got tired of it…. And then, what I mean by 'tired of it' is that I sort of grew tired of what I was making at the time as well, and then from there, I went more towards media art.
Things like graphic design?
No, more interactive than that. There's some programming, then projection, like that…. Then, when I was making interactive art work, places like Sendai, Mediatheque from Sendai, and New York, and there was talk of Kyuushu at one point, although that didn't end up happening, there was talk of going to these places … and when I participated in that, then I really ended up tired of what I was doing. What I mean is, I like media art, but I don't think I can do it.
So what did you do after you grew bored of new media art? Why hadn't you been creating art until recently?
In 2001, I did one exhibition, but then I started to question whether there was a point in doing art without the thought, without the creativity. And then I really began to think, was there a point in doing art? Is there a point in making, say, a sculpture? Who would it be for? And what manner of creating art would satisfy me? Keeping these things in mind I made some simple test pieces…. Samples. For example, making something without a shape. Not exactly design… just the idea. Just the philosophy behind it. And so the period of time that I spend just focusing on the philosophy part, the philosophy regarding art, the creative part is open. And so I subdivided my brain a little, separated creative as creative, and that part I used when I was doing design, which I don't consider fine art. In my head, therefore I had space to consider what I should do with the "art" side of things. I kept thinking. I mean really, I tell everybody this but, I spent at least six years thinking about this.
What have you learnt during your absence from the art world?
I realised there's no need to make things that are already visible. For example, let's say you go somewhere, travel somewhere, maybe. You see a very beautiful landscape. I think you can leave that for a photograph. So I decided not to recreate things that exist in the first place … I think that it's best to draw something that uses imagination and inspiration as a way to consolidate your own philosophy.
Tell me about how you came to participate in ULTRA002 (2009) and NYC PULSE (2010).
That's because I'm part of this gallery of artists. The artists associated with this gallery … can speak with the directors and discuss the possibility of entering the next ULTRA art fair, and it's not certain you'll pass, but you know, you apply for it.
Why did you want to become an artist?
I think that art is like a subject. It's academic … the basis of art is quite academic. But the viewer has freedom. That's why, when I make my work, it's more philosophical. I like to have philosophical ideas and make pieces…. The point is that the people who critique art often have very academic backgrounds, but I think even children and people who don't know anything about art should be able to see the art, and freely feel what it means to them. I feel that is the most pure, somehow. And so, for fine art, there aren't any restrictions. For example, the big difference between 'design' and 'art' is whether or not it's been requested. The thing with design is that, after all, it's somebody else's intention, or somebody's … desire…. There's a purpose, very clearly. And so, for fine art, the purpose is in the self, so it remains extremely pure…. For example, nobody is going to be sad as a result, or maybe they won't be happy either, and maybe they will be sad, but, even so, it might make them happy. Thus it's really quite a … place where one can face new challenges.
So would you consider yourself a fine artist? What do you consider your main line of work?
Myself? I would like to keep being an artist.
What major influences have you had in your life?
I suppose books…. I don't really read novels much. Other than novels, documentaries, philosophical books, chemistry books, things like that. Especially books that might change one's perspective, thoughts. Or else something that changes one's thoughts, one's mind. How should I explain this? To 'dephase'…. And so, I'm always trying to find opportunities for change, so yes, perspective. What kind of perspective to have each time.
What was it that changed you as an artist?
Maybe books. I suppose books. For example, even people you can never meet, people who you really respect, even if you've never met them, that person's words are written down. The words affect us, and make us consider things like, maybe there's no value in that, or that's not quite right. In the end it's yourself thinking, but the trigger for that, what gives it initiative, are the words of those people you respect.
What has challenged you as an artist? Why? What kinds of things have been challenges for you as an artist?
Everyday is a challenge (laughs). There's a kind of fulfillment when you finish a piece, but at the end, that's it; and so little by little, I try to find something I don't like about it. Even if I'm pleased with it, I look for something I find displeasing, and next time, try to make it better. Whether it's the technique, or the philosophy behind it, or the surface, that [makes it] good. And so I don't know what it is, but I try to improve it, even if it's just a little bit.
What do you like about art?
After all, we don't have to have art, but it's better to have it. We can have art, or not, but it's definitely better to have it; the strangeness in that! The fact that we don't know if there is or isn't value; it's unclear. I think it's obvious that it's better to have it, so that's what's fun.
What makes your work different from other artists' in your generation?
I think they're all very accomplished (laughs)…. The difference is that they are Fukuhara, or they aren't. What I do, only I can do; when I'm doing art, I think like that…. For example, I consider the boundary between something existing and not existing. I like that boundary … I pay a lot of close attention to that, so, for example, the medium can be pencil, or oil paints, or metal, it can be anything. And so, if something does exist, or it doesn't … I pay attention to that, I want to express that. And so I consider how to convey that, I look for that. And even if there's someone who's thinking the same thing, that person and I will probably come up with different ideas. And since we have different knowledge, that's only possible for me to do, there's only me.… My priority is not for the expression. I'm more inclined towards the philosophy involved.
What are your plans for the future? Do you have any future projects?
A big art fair. It hasn't been decided yet but either in Miami, PULSE in Miami, or next year in New York … or a show. There are also some shows that want to exhibit my work, they're pretty far ahead, but there are some exhibitions.
What challenges do you see for young artists working in contemporary art today?
In Japan? The circumstances are bad. Business is bad. Right now, it's so. And, yes, the Japanese economy is very … the Japanese arts, arts scene? The arts scene I suppose, or more like, the custom here, is very bad. It depends on the culture. In Europe, and probably at PULSE as well, probably everyone is quite understanding, so they say, 'Mr. Fukuhara, I can support you.' As in, before they ask 'How much?' or things like that, they tell me, 'I can support you.' I've never heard of it in Japan. There were some people like that up till now, and so, yes, there are some, but they're few. Overseas, in the USA, what I learned when I went to PULSE was that in fact, both very rich people, and people not quite so rich are willing to buy a piece of art if they like it. Because if they support a young artist, and since they like the piece itself, maybe they'll become really well-known later on. There's sort of a feeling like that. And also, they know that if they buy this piece, the artist can go on to make their next piece; they'll approach artists in a sensibility like this, even if they don't say it to this extent. Japan likes modern art. In Japan, there's a kind of feeling that there's a tendency towards it. I think that's because the value is already determined, like: 'This is good'; 'That there, that is worth about this much.' But, there isn't much of a sense of supporting young artists … Japan has lots of really amazing young artists, but I think it's very difficult in Japan.
Hiroshige Fukuhara, 'The Night With a Clouded-over Moon (Carp)", 2010, pencil on aluminum panel, 50 x 35 cm. Image courtesy of the artist.
Tell me about your current exhibition, "Binary".
The color is black, but, color depends on the light. The color is determined by the light. This here, what the color white means, is that it's reflecting white light back at us. And here the light is getting absorbed and so it's black in color. This looks silver, but all it is is that the carbon in the pencil lead is shiny. If you put black on black, well, you can't see it very well, but it shines, and sometimes you can see it well, sometimes you can't. And if you see it at night, it's almost completely dark. If you see it in a room at night the surface is completely black.
What was the philosophy behind "Binary"?
As I said earlier, I'm trying to reach the borderline between what exists and what doesn't. That's why, for example, from here it's impossible to see this painting because of the glare, yes? Because of the acrylic board, because it's darkened. But if you put your eye close to it, you can see that there's quite a bit drawn on it. 'Then don't draw on it!' someone might say. But I want to draw on it. …When you interpret a computer, the data formats are, for example… there's a thing called text, and text goes on forever. But with binary code, it tells you in the first row how many letters there are. And you can't have any more than that. In that lies a big difference between so-called binary code and text. With binary you know the end from the very start, and in text it goes on forever so the end is unknown. Text has no limit but binary defines the end in its first row of numbers.
How did you use this idea of "binary" in your artworks?
In how I incorporated a limit into my work process. To start with, drawing lines in a way that makes them invisible is in itself limiting. And also, binary is in two states, so it's '0' and '1'… there's no middle point, I didn't draw any middle tones. They're all drawn as solid lines, and it's not in grey scale, but it's black on white. What is it? Gradation? Gradation is hard to reproduce. If someone says, 'Here is some gradation, go copy that exactly onto here,' it's really difficult to do that…. If's it's only two colors, if there's a line in exactly the same place, it can be reproduced. …It's just the placement. And so, this is somehow maybe worthless in value in terms of creativity. The act of purposely making something that can be re-created easily, that's somehow important, the value. The easiness to re-create and the difficulty to re-create. Maybe it should be the priority to make things that are harder to re-create, but I deliberately want to express what's easy to re-create.
What do you like most about this exhibition, "Binary"? What do you like most about this series?
The fact that it's black (laughs)… It can be black or it can be white, but to have none… The good part is that it's clear if it is or isn't there … if you go in what you notice in the moment you enter is that there are black squares. And then in that, there's a, what do you call it, in minimalism they made black panels, or red panels, but I can't get that stoic, and I do want to express…. I want to express something animated, something pulsated, but part of me also doesn't want to express it…
And so you make it harder to see.
Yes that's it. And also in a picture, you try to fill it up; this goes here…. And so in order to not do a layout, I start drawing from an area.
Is that easier to do if it's black?
No, that doesn't affect it. In order to make the layout quieter… this isn't fixated. And so if you take the acrylic board off and touch it, it'll come off.
Why did you decide to use black gesso?
That's because black holds a lot of different meanings. For example, it's very still, it has a sense of immense quiet, and also a strong sense of night and also darkness… And it's possible to see a highly dense something in black. White things are the opposite and they're pure, there's cleanliness. Black for me is a mysterious color. In order to fully expose the good qualities of the color black, I wanted to make it black on black. In the end it looks more like black on silver than black on black, but the act was to put black on black.
Is there a reason you decided to use pencil?
I think it's the freshness?… For example that piece there, I've fixated. It's more like a CD. If you compare it to music, it's more like a CD. And this is more like a live show. It's possible to do black on black by using a brush, for example, to place a transparent medium on the black, that would make it black on black as well. But if you do that, I think that makes it more like laying it out. I think that once you start deciding the composition, the picture will get more like, well industrial arts, or arts and crafts.
And it will get harder to see.
Right. Also I don't intend to do arts and crafts, so, for lines that I can only draw in this instant, I want to draw them in that instant as much as possible, and with pencil it's fast.
Do you draw directly onto the gesso or the aluminium?
Yes. As a piece…. I'm repeating myself a bit, but the relationship between the pencil and the gesso is that, it's ultimately about being able to adjust the image, and I suppose how to deal with the lighting, because I'm not using colour. And so, it's all about how much you control the light, and so I don't question the medium. And this acrylic case protects it, but the piece is actually the whole thing, case and all, so it's okay. It's fine if the surroundings are reflected on the acrylic board. It's all included in it.
Gallery view of Hiroshige Fukuhara's latest solo exhibition at Ai Kowada Gallery in Japan. Image courtesy of the artist.
Hiroshige Fukuhara, 'The Night', 2010, pencil, black gesso on wooden panel, 900 x 630 mm. Image courtesy of the artist.
This artwork shaped like a horse is a little different? Could you tell me more about this one?
It is different. I'm starting to do these recently, but as I said earlier about layouts, pictures tend to mostly be rectangular shaped. I want to be able to connect the philosophy and the technique as directly as possible onto the square, the surface. However, somehow various… um, it has to go through, a certain way, and so it's inevitable that the expression becomes more …angled than what was being imagined. For example, just drawing instinctively… without making a draft and drawing in real-time, directly, means that the lines aren't pre-determined. After drawing a strawberry-like image… a flower appeared, and then below that are some moss-like things … and in each of those instants, there's something that's alive, and I try to draw them, picking up these random images from the library in my head and placing them onto the canvas. And so, when it's square, I can't help placing an object on it. For example, the butterfly, I put the butterfly down. …This shape here, because the shape is intentional, internally it is tremendously free. I'm thinking I'd rather continue to do this sort of thing. And then when you do that, the place where it's displayed? There might be more freedom in where you hang it and, if it's square, for instance, often, pictures are something I want to be displayed in houses,… or museums and such as well. And so, with things like that, you feel an urge to place it bang in the middle. For example… there's a horse drawn on that one. It's just that a horse is there, but I drew the horse by accident. But, instead, if the canvas is a horse, then isn't there no need to draw a horse? I can draw the pattern more freely from within, because if the tableau is square, I'm compelled to draw a horse. It's a way that I strategise, but if the canvas is already shaped like a horse, there's no more room to place one, and it makes it easier to make a direct connection between my head and my hand…. It's impossible to remember what I drew (laughs).
You mean, the order that they're hung is decided?
Yes, like maybe you want some more space between them. But if they're shaped like this, and for instance, if there's a small picture then maybe it might be good to put them in a spot like this. Yes, you might be freer to put them where you want, and maybe the meaning of the piece might change depending on where you place it. Also, outside? Having an association with a silhouette, also, makes the interior extremely … there's a feeling of my own sense of alive-ness, and so for me I'd really like to continue to do this sort of thing.
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