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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?

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marked as duplicate by user26857 abstract-algebra Mar 10 '16 at 11:09

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    $\begingroup$ Out of respect for the users of the site, you could try to think these problems for a longer while... $\endgroup$ – YoTengoUnLCD Mar 10 '16 at 1:39
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    $\begingroup$ I have only 13 years old; sometime I need help to solve some problems. It is not always evident at this level. $\endgroup$ – Taj Mohamed Bandalandabad Mar 10 '16 at 1:41
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    $\begingroup$ @TajMohamedBandalandabad wow a 13 year old doing this stuff!! almost unbelievabl :-). $\endgroup$ – Anurag A Mar 10 '16 at 1:45
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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 $\langle (y-1)^{2} \rangle$?

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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 $\mathbf{Q}[x]$.

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