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I'm considering the ideal $I=\langle x,y \rangle \subseteq \mathbb{Q}[x, y]$. We have $\mathbb{Q}[x, y]/I \cong \mathbb{Q}$. The polynomial $x^{2}+y^{2}-1 \equiv -1 \pmod I$ is irreducible in $\mathbb{Q}$, so $x^{2}+y^{2}-1$ must be irreducible in $\mathbb{Q}[x, y]$?

Also, how can I prove that the polynomial is irreducible using Eisenstein's criteria?

I can see now that it's possible to Eisenstein's criteria when we look at $x^{2}+y^{2}-1 \in \mathbb{Q}[x][y]$.

If I go with the prime ideal then $\langle y-1 \rangle$, then certainly $y-1 \mid y^{2}-1$ and $(y-1)^{2} \nmid y^{2}-1$, hence the polynomial is irreducible.

John Smith
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    Does this answer your question? $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$. – morrowmh Dec 08 '22 at 21:53
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    Wouldn't your argument also apply to $(x+1)(y+1)$? – Daniel Schepler Dec 08 '22 at 21:53
  • @DanielSchepler: You're right, but then what did I do wrong? – John Smith Dec 08 '22 at 21:55
  • @morrowmh I saw that question but I didn't find the solution worked out using Eisenstein's criteria and I couldn't figure it out myself. – John Smith Dec 08 '22 at 21:57
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    The issue with your proposed method is that non-units can become units after taking a quotient by an ideal. – Karl Kroningfeld Dec 08 '22 at 22:04
  • 'What did I do wrong?' You seem to apply a general theorem of the form: if $R$ is a ring, $I$ an ideal and $a$ an element of $R$ such that $a mod I$ is irreducible in the ring $R/I$ then $a$ is irreducible in $R$ as well. Why do you think such a theorem applies? I can think of one answer, but it it makes perhaps more sense if you answer this question yourself. Anyway, my guess is that your intuition goes something like this: suppose that $a$ is not irreducible, so it can be written $bc$, then we can detect this decomposition already in $R/I$ as then $a \mod I = (b \mod I)(c \mod I)$. – Vincent Dec 08 '22 at 22:07
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    Your argument flounders because the image of $x^2+y^2-1$ is not "irreducible in $\mathbb{Q}$"; it is a unit. Units are not irreducible, by definition. – Arturo Magidin Dec 08 '22 at 22:10
  • (ctd from previous comment) This sounds plausible. What goes wrong is nicely visual in Daniel's example: 'downstairs' in $R/I$ the decomposition $a \mod I = (b \mod I)(c \mod I)$ might reduce to something innocous like $1 = 1 * 1$ which does not mean that the left hand side is reducible. So going down to $R/I$ fails to detect the reducibility of $a$ in that case, and as a result we cannot treat irreducibility in $R/I$ as evidence for irreducibility in $R$. This is also what Karl said, I see now – Vincent Dec 08 '22 at 22:12
  • @Vincent: Thank your for your explanation. I just missed seeing that $-1$ is a unit and not irreducible. Arturo Magidin used the right word to describe the situation. – John Smith Dec 08 '22 at 22:15
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be an open-ended proof checking machine. – Bill Dubuque Dec 09 '22 at 03:07
  • @JohnSmith, I agree that -1 is not irreducible because of being a unit, but I am not entirely convinced that that is at the heart of the problem. Look at this example. Suppose you were looking at $x^2z + y^2z + z$ in $\mathbb{Q} [x, y, z]$. Then modding out by $I = \langle x, y \rangle$ you get $z$ which is irreducible in the quotient $\mathbb{Q}[z]$, but that doesn't of course mean that $x^2z + y^2z + z$ was irreducible to begin with! – Vincent Dec 09 '22 at 09:12
  • Solution: $p(x, y)=y^2-(1-x^2) = y^2 - (1-x)(1+x) \in \mathbb{Q}[x][y].$ Then $1-x$ is prime in $\mathbb{Q}[x]$ and irreducibility follows by Eisenstein. – John Smith Aug 14 '23 at 16:19

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