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I stuck in the following question.

Prove that $ \mathbb{C}[x,y]/\langle x^2+y^2+1 \rangle $ is an integral domain, using the following:

Let $\mathbb{F}$ be a field, $c \in \mathbb{F} $. Then $ \mathbb{F}[s,t]/\langle st + c \rangle $ is an integral domain if and only if $c \neq 0$.

I have proven that $ \mathbb{C}[x,y]/\langle x^2+y^2+1 \rangle $ is isomorphic to $ \mathbb{C}\left[x,\sqrt{-1-x^2} \right]$ using the homomorphism:

$$ \alpha \in \mathbb{C} \ \ \mapsto \alpha \in \mathbb{C} $$

$$ x \ \ \mapsto x$$

$$ y \ \ \mapsto -1-x^2.$$

Now what I think is I should proof that $ \mathbb{C}[x,\sqrt{-1-x^2} ]$ is isomorphic to some $ \mathbb{C}[s,t]/\langle st + c \rangle $, defining $s$ by $x$ and $t$ by $\sqrt{-1-x^2} $ but I can't find any way to do that.

Am I in the right way or my way of proof is wrong?

Thanks in advance.

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2 Answers

up vote 7 down vote accepted

Hint: Try $s=x+iy, t=x-iy$. ${}$

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Okay, but the next question is the same but instead of $\mathbb{C}$ with $\mathbb{R}$.what now? –  Shai Alkoby Dec 17 '13 at 19:17
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I would try to talk myself into believing that $\Bbb{R}[x,y]/(x^2+y^2+1)$ is a subring of $\Bbb{C}[x,y]/(x^2+y^2+1)$. –  Jyrki Lahtonen Dec 17 '13 at 19:21
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If it was not mandatory to follow the hint, you could also say that $x^2+y^2+1$ is irreducible in $\mathbb{C}(x)[y]$, and it is because $x^2+1$ is not a square. Hence you have that $\mathbb{C}(x)[y]/(x^2+y^2+1)$ is a field. Now, you know that $\mathbb{C}[x][y]/(x^2+y^2+1)$ is a free $\mathbb{C}[x]$-module with basis $\{1,y\}$, hence in particular it's flat. Now consider the injection $\mathbb{C}[x]\subset\mathbb{C}(x)$ and tensor it over $\mathbb{C}[x]$ with $\mathbb{C}[x][y]/(x^2+y^2+1)$. Hence you still get the injection $\mathbb{C}[x,y]/(x^2+y^2+1)\subset\mathbb{C}(x)[y]/(x^2+y^2+1)$, but the right side is a field, hence the left side has to be a domain. I think that this proof, even if more conceptual, is more algebraic, meaning as an example, that you can follow the same lines to show the result for a generic field $\mathbb{K}$ instead of $\mathbb{C}$, at least for characteristic different from 2.

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We didn't learn about irreducibles yet. Thanks! –  Shai Alkoby Dec 19 '13 at 9:59
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