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I am studying a first course in commutative algebra and I'm currently working through some exercises on calculating $Spec(R_P)$, where $R_P = R[(R\backslash P)^{-1}]$ is the localization of $R$ at a prime ideal $P$. Unfortunately, I'm not sure if I'm making much progress.

Here's one example:

$R = K[x,y]/(xy)$ ($K$ a field), and $P = (x-1, y)$.

My thinking is to creating a homomorphism $\phi : R_P \rightarrow R$ (which I think is injective), from which, I can induce a homomorphism of sets $\phi^* : Spec(R) \rightarrow Spec(R_P)$, which I also believe is injective, since $\phi$ is injective.

I tried calculating $Spec(R)$ and I think it is $\{ (x), (y) \}$. This would suggest to me that $Spec(R_P)$ is formed of two ideals, and I think it would be $\{ (0), (x-1) \}$.

As you can probably gather, I'm very unsure on all of this, but it's the best attempt I've got so far. Any advice would be greatly appreciated; thanks!

Mystery_Jay
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2 Answers2

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In general there's not a homomorphism $R_P \to R$. There is a homomorphism $R \to R_P$ and this does induce a map $\mathrm{Spec} \ R_p \to \mathrm{Spec} \ R$.

For computing $\mathrm{Spec} \ R_P$ the following is an important theorem about localizations which you should try to prove yourself:

Theorem. If $S \subseteq R$ is any multiplicative set then the prime ideals of $S^{-1}R$ are in one-to-one correspondence with the prime ideals of $R$ that don't meet $S$.

(The maps for the correspondence are given by extending and contracting ideals through the natural map $R \to S^{-1}R$ given by $a \mapsto \frac{a}{1}$.)

Applying this to your situation it says that the prime ideals of $R_P$ correspond to the prime ideals of $R$ that are contained in $P$. In your example you computed $\mathrm{Spec} \ k[x, y]/xy = \{(x), (y)\}$. In fact it's $\mathrm{Spec} \ k[x, y]/xy = \{(x, f(y)), (y, g(x))\}$ where the $f$ and $g$ range over all irreducible polynomials as well as $0$.

To see this note that $xy = 0$ is contained in every prime ideal so every prime ideal contains either $x$ or $y$. If it contains $x$ then it corresponds to a prime ideal of $$(k[x, y]/xy)/x \simeq k[x, y]/(x, xy) = k[x, y]/(x) \simeq k[y]$$ and the prime ideals of $k[y]$ are all of the form $(f)$ where $f$ is irreducible. The argument for when $P$ contains $y$ is symmetric.

Now, $x$ is not contained in $P$ and $y$ is. Note that $g(x)$ is contained in $P$ if and only if $x - 1$ divides $g(x)$. As $g(x)$ is either $0$ or irreducible this means $g(x)$ is either $0$ or $x - 1$.

Thus $\mathrm{Spec} \ (k[x, y]/xy)_P = \{(y), (y, x - 1)\}$.

Jim
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  • @user143137: Fyi, my original answer was wrong, I've edited it to be correct. – Jim Feb 18 '15 at 22:08
  • Dear Jim: you obviously understand what the spectrum is but your description is not correctly written from the point of view of set-theory. I suggest you write the spectrum as the union of two infinite sets (parametrized by irreducible polynomials) and a two-point set. (By the way, I have deleted my previous comment) – Georges Elencwajg Feb 18 '15 at 22:11
  • @Jim Thanks for editing! I noticed the comment about the mistake too and was trying to calculate $Spec(R)$ whilst you were editing. – Mystery_Jay Feb 18 '15 at 22:12
  • @GeorgesElencwajg: I changed one "exactly" to a "corresponds to". Other than that I think there are no set theoretic problems, save possibly the objection that I use $x$ to mean both an element of $k[x, y]$ and of $k[x, y]/I$ for some ideal $I$. I think that's pretty standard though, and am unwilling to bear the ugliness of what working around it looks like. – Jim Feb 18 '15 at 22:15
  • Dear Jim, the spectrum as you wrote it would be a set with two elements: adding a description "where the $f$ and $g$ can be either $0$ or any irreducible polynomial" outside the braces is not allowed. You must write $\operatorname {Spec} (k[x,y]/(xy))={(x),(y)}\cup{(x,f(y))|f(y) \operatorname {an irreducible polynomial}}\cup{(y,g(x))|g(x) \operatorname {an irreducible polynomial}} $ – Georges Elencwajg Feb 18 '15 at 22:27
  • @GeorgesElencwajg: I changed the language to make it clear there is an implicit quantifier. I disagree that sets must be written out in such a painstaking way. Explaining quantifiers outside the notation is common, it's much like writing $\sum_n$ and then explaining the range of $n$ elsewhere. – Jim Feb 18 '15 at 22:29
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$R_P=k[x,y]_P/(xy)=k[x,y]_P/(y)\simeq k[x]_{(x-1)}$, and the last ring has clearly only two prime ideals: $(0)$ and $(x-1)k[x]_{(x-1)}$.

user26857
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