Subject (1). Let $f: \mathbb{Q}^{*} \rightarrow \mathbb{Q}^{*}$ be an endomorphism of the group $\left(\mathbb{Q}^{*}, \cdot\right)$.

a) Determine $f(-1)$.

b) Determine all endomorphisms $f$ with the property that $f(p)=p$, for any $p>0$, $p$ a prime number.

MARCEL ŢENA
Solution.

a) We have $f(-1) f(-1)=f((-1)(-1))=f(1)=1$ and thus $f(-1)= \pm 1$. Since both $f(x)=x$ and $f(x)=|x|$ are endomorphisms, both values can be taken. I thought that "to determine" had a stricter connotation... it might have been more appropriate to ask for "finding" all possible values of $f(-1)$.

b) Since any positive rational number can be written as a product of distinct prime numbers raised to integer exponents, and any endomorphism commutes with products, it follows that $f(r)=r$ for any $r \in \mathbb{Q}_{+}^{*}$. Since then we also have $f(-r)=f(-1) f(r)=f(-1) r$, in accordance with part a) we can only have $f(x)=x$ or $f(x)=|x|$ (exactly the two examples given above).

The official solution is poorly written. It suffers from the use of $\mathbb{Q}^{*}$ when it was clearly intended to use $\mathbb{Q}_{+}^{*}$; it then directly admits the values $f(-1)= \pm 1$ from algebraic operations, without verifying that both values are allowed (this is partially "fixed" at the end, when the two possible endomorphisms are constructed); finally, it juggles undeclared symbols, such as $q$ and $r$. The problem is remarkably trivial, the endomorphisms $f$ being "fixed" over $\mathbb{Q}_{+}^{*}$, with the only possible variation given by the value at the point -1.