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I want to show that $p(x,y)=x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$. There is already a similar problem posted:

$\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle$ is an integral domain, and its field of fractions is isomorphic to $\mathbb Q(t)$

but this question uses the irreducibility of $p(x)$ to show that it is prime and therefore an integral domain. The part of the problem I am stuck on is showing that $x^2+y^2-1$ is irreducible over $\mathbb{Q}$ in the first place.

I have considered taking the approach used in the above link except for backwards, but I am unsure of whether or not the converses of the theorems used still apply.

Edit: I see that this question has been marked as a duplicate, but I think I have explained how I am asking a different question than the one that was posted previously.

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  • $\begingroup$ Try a degree argument. It's simple. $\endgroup$
    – Pedro
    Apr 10, 2017 at 2:18
  • $\begingroup$ I understand that concept, but I have never worked with a multi-variable polynomial before, and am unsure of how to approach it $\endgroup$ Apr 10, 2017 at 2:19
  • $\begingroup$ In the same thread there are two answers (but who has time to read them?). In this one it's proved that your polynomial is irreducible. $\endgroup$
    – user26857
    Apr 10, 2017 at 7:02
  • $\begingroup$ @user26857 I have already read these answers and they are not what I was looking for- I am also unable to make comments on answers so I cannot ask for clarification on questions asked by others. $\endgroup$ Apr 10, 2017 at 15:55
  • $\begingroup$ @MathStudent1324 You definitely didn't read the non-accepted answer and keep claiming that your question is different when in fact you asked for a proof that $X^2+Y^2-1$ is irreducible over $\mathbb Q$ and this is proved there e.g. by using the Eisenstein's criterion. $\endgroup$
    – user26857
    Apr 10, 2017 at 16:10

1 Answer 1

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Suppose that polynomial is a product $f(x,y)g(x,y)$. Then $f,g$ both must be linear because $p = x^2+y^2-1$ is of largest degree $2$. Now write down what $p = fg $ means and see what $f,g$ can actually be.

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    $\begingroup$ So does it suffice to let $p(x,y)=(ax+by+c)(mx+ny+r)$, $p(x)=amx^2+bny^2+(an+bm)xy+(ar+mc)x+(br+nc)y+cr$, set up a system of equations (e.g. $am=1$, $bn=1$, $ar+mc=0$, $br+nc=0$, etc.), and find a contradiction? $\endgroup$ Apr 10, 2017 at 4:19
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    $\begingroup$ Exactly, @mathstudent1324 $\endgroup$
    – Pedro
    Apr 10, 2017 at 4:37

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