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

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.


1 Answer 1


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 $A/\mathfrak{m}$.

  • $\begingroup$ Thank you for this elegant proof. $\endgroup$
    – Didier
    Feb 22, 2022 at 19:42
  • 1
    $\begingroup$ Just a comment: there is a typo in the definition of $x_p^s$: the first condition should be "if $x_p \in (\Bbb F_p)^2$. $\endgroup$
    – Didier
    Feb 22, 2022 at 20:35

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