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Let $k$ be an algebrically closed field, then $\operatorname{Spec}(k[x,y]/(xy))$ can be viewed as the union of two affine lines, but I cannot link this idea with the algebrical formalism. In particular I would like to find the points (closed, dense, etc...) and their residue fields.

Does exploiting the immrsion map $\operatorname{Spec}(k[x,y]/(xy))\to\operatorname{Spec}(k[x,y])$ induced by the projection make sense? Another thing which suprise me is that apparently there is no dense point...

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  • $\begingroup$ Your inclusion map should go the other way, probably. $\endgroup$
    – xyzzyz
    Apr 3, 2017 at 18:56
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    $\begingroup$ Are you looking for an explicit description of all prime and maximal ideals in $k[x,y]/(xy)$? $\endgroup$
    – Andrew
    Apr 3, 2017 at 20:29
  • $\begingroup$ @xyzzyz I edited the typo thx ;) $\endgroup$
    – Dinisaur
    Apr 4, 2017 at 19:30
  • $\begingroup$ @Andrew Yes that would be a good step. There are also other things that leave me a bit confused, for example the "origin" should be either (x) or (y) but these look to me like different points. In particular I would be able to compute the dimension of the fiber in the origin (x) of the projection map to the affine line Spec(K[x]), but I think that if this post doesn't give me enought insight that would fit well another question. $\endgroup$
    – Dinisaur
    Apr 4, 2017 at 19:37
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    $\begingroup$ If $f \in k[x]$ is irreducible then $k[x,y]/(y,xy,f(x))$ is a field, the same for $k[x,y]/(x,xy,f(y))$, so you have twice more maximal ideals than in $k[x]$ $\endgroup$
    – reuns
    Apr 4, 2017 at 19:43

1 Answer 1

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Let's start by understanding points in $\operatorname{spec}(k[x,y]/(xy))$. First examine closed points. A closed point is a maximal ideal in $k[x,y]/(xy)$. But by the correspondence theorem for rings, these are in bijection with maximal ideals $\mathfrak{m}\subset k[x,y]$ that contain the ideal $(xy)$. We know maximal ideals in $k[x,y]$ are just $\mathfrak{m}=(x-a,y-b)$, and for $(xy)$ to be contained in one of these we need either $a=0$ or $b=0$. In other words, $\mathfrak{m}$ must correspond to a point on either the $x$ or $y$-axis.

As for non-closed points, you just want prime ideals that aren't maximal in $k[x,y]/(xy)$. Again by correspondence, these are prime ideals $\mathfrak{p}$ in $k[x,y]$ containing $(xy)$. These are fully described in Miles Reid's book "Undergraduate Commutative Algebra," page 22. When $k$ is algebraically closed, prime ideals in $k[x,y]$ are just $0$, $(f)$ for irreducible $f(x,y)\in k[x,y]$, and the maximal ideals. So, prime ideals in $k[x,y]/(xy)$ are of the above form and must contain $(xy)$. This only leaves the maximal ones and the ideals $(f)$ where $(xy)\subset (f)$. That is, $f=x$ or $f=y$ (since $f$ is irreducible). Thus, there are only two prime non-maximal ideals: $(x)$ and $(y)$ in $k[x,y]/(xy)$, each corresponding to the generic points of the irreducible components, namely the $y$ and $x$-axes respectively.

As for residue fields: since $k$ is algebraically closed, the residue fields at any closed point will just be $k$. At the two prime ideals, you'll pick up either $k(y)$ or $k(x)$. Here's an example computation. Let $L$ be the reside field at the prime ideal $(x)$. We get $$ \begin{align*} L = [k[x,y]/(xy)]_{(x)} &\cong \frac{k[x,y]_{(x)}}{(xy)_{(x)}}\\ &\cong \frac{k[x,y]_{(x)}}{(x)_{(x)}}\\ &\cong \left(\frac{k[x,y]}{(x)}\right)_{(0)}\\ &\cong k[y]_{(0)} = k(y). \end{align*} $$


Explanation of the isomorphism $\frac{k[x,y]_(x)}{(x)_{(x)}} \cong (k[x,y]/(x))_{(0)}$. In general, if $I\subset R$ with $I$ an ideal and $R$ a ring, and $S\subset R$ a multiplicative set, there is an isomorphism $$ S^{-1}R/S^{-1}I \cong \bar{S}^{-1}(R/I), $$ where $\bar{S}$ is the image of $S$ in $R/I$. To see this, start with the exact sequence $0\to I\to R \to R/I\to 0$. Then since localization is exact, we have the sequence $$ 0 \to S^{-1}I \to S^{-1}R \to S^{-1} (R/I)\to 0. $$ When we form the localization $S^{-1}(R/I)$ we are really forming $\bar{S}^{-1}(R/I)$, since we only care about how the elements of $S$ act modulo $I$. (There may be a more formal argument for this, but this is at least how I think about it. This statement is also exercise 4 in Section 3 of Atiyah & Macdonald.)

In our example, $I=(x)$ and $S=k[x,y]-(x)$, and so $\bar{S} = (k[x,y]-(x))/(x)$. But if we take a polynomial $f(x,y)$ not divisible by $x$, then the map $k[x,y]/(x) \to k[y]$ sending $f\mapsto f(0,y)$, which won't be identically the zero polynomial. That is, we can identify $\bar{S}$ with all nonzero elements of $k[y]$, and whence $\bar{S}=k[y]-(0)$. This explains the subscript $(0)$.

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  • $\begingroup$ Hi Andrew, could you help me see the third isomorphism in the last display? (when you change from the ideal $(x)$ to $(0)$) I can show the isomorphism with $k(y)$ by hand, but there seems to be something clever happening here. $\endgroup$
    – user347489
    Apr 20, 2017 at 19:27
  • $\begingroup$ Yes, I'll edit my answer with clarification. $\endgroup$
    – Andrew
    Apr 20, 2017 at 19:27
  • $\begingroup$ It's clear now, thank you! $\endgroup$
    – user347489
    Apr 20, 2017 at 19:58

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