Show that the polynomial $x^2+y^2-1$ is irreducible over $\mathbb{Q}[x,y]$

Question

Show that the polynomial $x^2+y^2-1$ is irreducible over $\mathbb{Q}[x,y]$.

Theorem : Let $A$ an integral domain and $I$ a proper ideal of $A$. If $f(x) \not \equiv a(x)b(x) \pmod I$ for any polynomials $a(x)$, $b(x)$ $\in A[x]$ of degree $\in [1, \deg(f))$, then $f(x)$ is irreducible in $A[x]$

I think I have to use this theorem, maybe in using the ideal $(y^2)$

The fact that $\mathbb{Q}[x,y]=\mathbb{Q}[x][y]$ and $\mathbb{\mathbb{Q}[y]}$ is a unique factorization domain could it help me here? Is anyone could help me at this point?

Out of respect for the users of the site, you could try to think these problems for a longer while... — YoTengoUnLCD, Mar 10 '16 at 01:39
I have only 13 years old; sometime I need help to solve some problems. It is not always evident at this level. — Taj Mohamed Bandalandabad, Mar 10 '16 at 01:41
@TajMohamedBandalandabad wow a 13 year old doing this stuff!! almost unbelievabl :-). — Anurag A, Mar 10 '16 at 01:45

score 1 · Answer 1 · answered Mar 10 '16 at 01:37

Hint: you're on the right track in considering the isomorphism $\mathbb{Q}[x, y] \cong (\mathbb{Q}[y])[x]$. Viewing $x^{2}+y^{2}-1$ as a polynomial in $x$ with coefficients in $\mathbb{Q}[y]$, observe that $y^{2}-1$ belongs to the ideal $\langle y-1\rangle \in \mathbb{Q}[y]$. Is this ideal prime? Does $y^{2}-1$ belong to $\langle (y-1)^{2} \rangle$?

score 1 · Answer 2 · answered Mar 10 '16 at 01:38

1

As you observed, $\mathbf{Q}[x,y] = \mathbf{Q}[x][y]$ so viewing $x^2 + y^2 - 1$ as a polynomial with coefficients in $\mathbf{Q}[x]$, we can apply Eisenstein with, for example, $(x+1)$, which is a prime ideal in $\mathbf{Q}[x]$.

answered Mar 10 '16 at 01:38

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Show that the polynomial $x^2+y^2-1$ is irreducible over $\mathbb{Q}[x,y]$

2 Answers2