# If $x$ and $y$ are integers, then at least one element in $\{x,y,xy\}$ is a square in $(\prod_{p \text{ prime}} \Bbb F_p)/\mathfrak{m}$.

$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\P}{\cal P} \newcommand{\F}{\Bbb F} \newcommand{\m}{\mathfrak{m}}$$

I am currently teaching exercises sessions for undergraduate students (third year) in algebra, and I am testing my limits on the subject in order to know what I can confidently give them in the exercises sheets. In a former exam of this course, the very last question was this exercise:

Let $$A = \prod_{p\in \P} \F_p$$ endowed with the component-wise ring structure ($$\P$$ is the set of primes), $$I$$ the ideal of eventually vanishing elements in $$A$$, and $$\m$$ a maximal ideal containing $$I$$. Let $$K = A/\m$$, which is a field of characteristic $$0$$. Show that for $$x$$ and $$y$$ integers, at least one element of $$\{x,y,xy\}$$ is a square in $$K$$.

I know that in a finite field $$\F_q$$, the product of two non-square elements is a square since either $$q$$ is even and any element is a square (the Frobenius is an isomorphism), either $$q$$ is odd and the squares of $$\F_q^{\times}$$ form a subgroup of index 2.

For this exercise, it would be sufficient to show that if $$x$$ and $$y$$ are both non-squares in $$K$$, then the set of primes $$p \in \P$$ such that $$xy$$ is not a square in $$\F_p$$ is finite. However, all my attempts failed to show this last claim and I can't figure out another way. I would really appreciate any hint.

We only need that $$\mathfrak{m}$$ is a prime ideal.
Let $$x^s \in A, \qquad x^s_p=\begin{cases}x_p \text{ if } x_p\in (\Bbb{F}_p)^2\\ 0 \text{ otherwise }\end{cases}$$ Construct $$y^s$$ similarly.
• If $$x-x^s\in \mathfrak{m}$$ then $$x=x^s$$ is a square in $$A/\mathfrak{m}$$
• If $$y-y^s\in \mathfrak{m}$$ then $$y=y^s$$ is a square in $$A/\mathfrak{m}$$
• Otherwise $$x^s(x-x^s)=0,y^s(y-y^s)=0$$ give that $$x^s,y^s\in \mathfrak{m}$$ so $$(x-x^s)(y-y^s)=xy\bmod \mathfrak{m}$$ is either $$0$$ or a product of two non-squares at every $$p$$, it is a square in $$A/\mathfrak{m}$$.
• Just a comment: there is a typo in the definition of $x_p^s$: the first condition should be "if $x_p \in (\Bbb F_p)^2$. Feb 22, 2022 at 20:35