Question: Let $(R,\mathfrak m,k)$ be an Artinian Gorenstein local ring such that $$\mu(\mathfrak m)=2, \quad\mathfrak m^2\ne 0,\quad\text{and}\quad \mathfrak m^3=0.$$
Then, is it true that every non-maximal ideal of $R$ is principal?

*Thoughts*: We have $\mathfrak m^2 \subseteq (0:\_R \mathfrak m)$. Since $R$ is Artinian Gorenstein, so $\dim\_k (0:\_R \mathfrak m)=1$, hence $\dim\_k \mathfrak m^2=1$. Thus, $\mathfrak m^2, (0:\_R \mathfrak m)$ are principal ideals. For an arbitrary ideal $I\subsetneq \mathfrak m$, we have
$$\begin{array}{r@{}l}
\mu(I)
&=\dim\_k(I/\mathfrak mI)\\
&\le \dim\_k(\mathfrak m/\mathfrak m I)-1\\
&=\dim\_k(\mathfrak m/\mathfrak m^2)+\dim\_k(\mathfrak m^2/\mathfrak m I)-1\\
&\le \mu(\mathfrak m)+1-1\\
&=2,
\end{array}
$$
which is just one off from being principal.

[Also note that since $\mu(\mathfrak m) -\dim R=2$ and $R$ is Gorenstein, so a result of Serre implies $R$ is complete intersection]

Answer: 
There is a [standard basis](https://en.wikipedia.org/wiki/Angular_momentum_operator#Orbital_angular_momentum_in_spherical_coordinates) for finite-dimensional $SO(3)$ (and hence $\mathfrak{su}(2)$) representations, denoted by $\left| j, m\right\rangle$; the total spin $j\ge 0$ labels the representation, the magnetic quantum number $m=-j,\ldots,j$ labels the basis. If you look at the first few equations in the Wikipedia article on [$3j$-symbols](https://en.wikipedia.org/wiki/3-j_symbol), one can immediately realize that they give the following formula for the (unique up to a scalar multiple) intertwiners in this basis:
$$ I\_{j\_1,j\_2}^{j\_3} \colon \left| j\_3, m\_3\right\rangle \mapsto \sum\_{m\_1=-j\_1}^{j\_1} \sum\_{m\_2=-j\_2}^{j\_2} (-1)^{-j\_1+j\_2+m\_3} \sqrt{2j\_3+1} \begin{pmatrix} j\_1 & j\_2 & j\_3 \\ m\_1 & m\_2 & -m\_3 \end{pmatrix} \left|j\_1, m\_1\right\rangle \otimes \left|j\_2, m\_2\right\rangle.$$
You can find a quick review of the relevant definitions of $3j$, $6j$, $9j$, symbold in Section 7 of my article with Chipalkatti ["The higher transvectants are redundant"](https://aif.centre-mersenne.org/item/AIF_2009__59_5_1671_0/), Annales de l'Institut Fourier, Tome 59 (2009) no. 5, pp. 1671-1713.

I think the best best way to see the vector space corresponding to the spin $j$ representation (instead of spherical harmonics etc.) is the space of homogeneous polynomials with complex coefficients in two variables and of degree $2j$. The group $SU(2)$ obviously acts on $\mathbb{C}^2$ and thus has a natural action on functions on $\mathbb{C}^2$, in particular polynomial functions, and even more in particular, homogeneous polynomial functions. These are called binary forms.

Finally, please correct the use of "Clebsch-Gordon" to "Clebsch-Gordan" with an "a". Half a century before physicists got interested in these coefficients etc. [Paul Gordan](https://mathshistory.st-andrews.ac.uk/Biographies/Gordan/) and [Alfred Clebsch](https://mathshistory.st-andrews.ac.uk/Biographies/Clebsch/) figured out the decomposition into irreducibles of a tensor product of irreducibles for $SU(2)$, with explicit intertwiners. The analogous result in higher dimension is pretty much an open problem. For $SU(3)$, this was done recently by Böhning and Graf von Bothmer in [this article](https://www.sciencedirect.com/science/article/pii/S0001870809003648).
For higher dimension, and for tensor products of symmetric powers only, there are such explicit results, most notably by Jerzy Weyman and Peter Olver. See [this article](https://arxiv.org/abs/1911.11045) and references therein.