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This question comes from Vakil's notes on algebraic geometry:

Describe the maximal ideals of $\mathbb{Q}[x,y]$ corresponding to a) $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$ b) $(\sqrt{2},-\sqrt{2})$ and $(-\sqrt{2},\sqrt{2})$ and their residue fields.

I figured that the ideals are $(x-y, y^2-2)$ and $(x+y, y^2-2)$ with residue fields $\mathbb{Q}(\sqrt{2})$. I was initially considering $(x^2-2,y^2-2)$ but that doesn't work since we don't get the relationship between $x$ and $y$ that we need in both cases, but Ravi mentions in his notes (top of p. 108) that the first sets of points are glued together and so are the second sets since they are Galois conjugates. But I'm confused about two things:

  1. Is this ideal maximal? I believe it is, since $\mathbb{Q}[x,y]/(x^2-2,y^2-2) \cong \mathbb{Q}(\sqrt{2})$ - i.e. a field. Though I might be mistaken.

  2. What does 'Galois conjugate' mean here? Because all four points seem to correspond to the ideal I wrote above, so shouldn't they all be glued together? (I understand what Galois conjugates mean over 1 variable, but I'm a bit confused with this 2 variable example)

KReiser
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User20354
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    1 is mistaken: $\mathbb{Q}[x,y]/(x^2-2,y^2-2) = \mathbb{Q}[x]/(x^2-2) \otimes \mathbb{Q}[y]/(y^2-2)=\mathbb{Q}(\sqrt{2}) \otimes \mathbb{Q}(\sqrt{2})$ which isn't a field. If you want to see this directly, note that $(x-y)(x+y)$ is zero in $\mathbb{Q}[x,y]/(x^2-2,y^2-2)$ but that neither $x-y$ nor $x+y$ is zero. – Aphelli Jul 27 '21 at 22:13
  • I see. Aside from just constructing the isomorphism, is there a more principled way of seeing the quotient is isomorphic to the tensor product? I don't see how one would notice that. I see what you mean about that product being zero but not the individual polynomials, but I don't see how you extrapolated that the quotient is the tensor product from there. Is there some universal property idea I'm missing? – User20354 Jul 27 '21 at 22:23
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    I guess the simplest way is to see this in terms of zero varieties. If $X,Y$ are affine complex varieties (in the “simplest meaning”) with coordinate rings $A$ and $B$, then the coordinate ring of $X \times Y$ is $A \otimes B$. Now, $K[x,y]/(x^2-2,y^2-2)$ is the coordinate ring of ${(x,y)|x=\pm \sqrt{2},y=\pm \sqrt{2}}$={\pm \sqrt{2}} \times {\pm \sqrt{2}}$ – it’s a product, and thus the coordinate ring is a tensor product. – Aphelli Jul 27 '21 at 22:41
  • That's a good way to view it. Thanks! – User20354 Jul 27 '21 at 22:42

1 Answers1

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  1. We may write $\Bbb Q[x,y]/(x^2-2,y^2-2)\cong (\Bbb Q[x]/(x^2-2))\otimes_{\Bbb Q} (\Bbb Q[y]/(y^2-2))$ which is isomorphic to $\Bbb Q(\sqrt{2})\otimes_{\Bbb Q} \Bbb Q[y]/(y^2-2)$, or $\Bbb Q(\sqrt{2})[y]/(y^2-2)\cong \Bbb Q(\sqrt{2})\times\Bbb Q(\sqrt{2})$. Another way to check that your guess is wrong is to compute $(x+y)(x-y)$ and note that neither $x+y$ or $x-y$ are zero.

  2. The Galois action on the coordinate algebra $\Bbb Q(\sqrt{2})[x,y]$ by is by acting on the coefficients of the polynomials. This gives that the Galois action on coordinates is the diagonal action (and that there's an inverse hiding somewhere - but since this group is of order two, we don't really have to think about that). To be specific, the two ideals $(x-\sqrt{2},y-\sqrt{2})$ and $(x+\sqrt{2},y+\sqrt{2})$ are Galois conjugate because the Galois action swaps $\sqrt{2}$ and $-\sqrt{2}$, so the points $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$ are conjugate. The other pair of points works the same way.

KReiser
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