Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Problem: let $m$ be a positive integer. Find a necessary and sufficient condition on $m$ so that $I=(m, x^2+y^2)$ is a prime ideal in $R=\mathbb{Z}[x,y]$.

An easy necessary condition is: $m$ is a prime number, and $m\neq 2$. (in fact, if $m=2$, then $x^2+2xy+y^2\in I\Rightarrow (x+y)^2\in I$ but $x+y\notin I$). I am stuck proving sufficiency (I don't know if this is really a sufficient condition).

share|cite|improve this question
up vote 4 down vote accepted

The conditions $m\neq2$ and $m$ prime are indeed necessary but not sufficient.
So, conversely, given an odd prime $m=p$ what is the condition for $I$ to be prime or equivalently for the ring $(\mathbb Z/p\mathbb Z)[X,Y]/(X^2+Y^2)= \mathbb F_p[X,Y]/(X^2+Y^2)$ to be a domain ?
Since $\mathbb F_p[X,Y]$ is a unique factorization domain, the condition is exactly that the polynomial $X^2+Y^2$ be irreducible in $\mathbb F_p[X,Y]$.
A little calculation (that I'll leave to you) shows that this is the case exactly if $-1$ is not a square in $\mathbb F_p$. And finally, deciding whether $-1$ is a square modulo $p$ is a very classical question that you can look up in a textbook or solve for yourself, using the result that the multiplicative group $\mathbb F_p^*$ is cyclic (the answer involves the residue modulo $4$ of $p$) .

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.