Maximal Ideals/Domination/Local Rings

**topspin1617** · June 22nd 2011, 01:25 PM

If anyone is interested in the context of this question, I am reading through the proof of Lemma 6.5 in Hartshorne's Algebraic Geometry.

We have fields $k\subseteq K$ such that $\mathrm{tr.deg}_k(K)=1$ . Also, assume $k$ is algebraically closed.
Let $y\in K\setminus k$ , and let $B$ be the integral closure of the ring $k[y]$ in $K$ . It can be shown that $B$ is a Dedekind domain which is finitely generated as a $k$ -algebra (and I'm fine with that).

Now, suppose $R$ is a discrete valuation ring between $k$ and $K$ such that $y\in R$ . Since $R$ is integrally closed (being a DVR), we must have $B\subseteq R$ .

Here's where I get lost: they say that if $\mathfrak{m}_R$ is the (unique) maximal ideal of $R$ , then $\mathfrak{n}=B\cap \mathfrak{m}_R$ is a maximal ideal of $B$ , and that $B$ is dominated by $R$ . What I don't see is (1) how we know that $\mathfrak{n}$ is a maximal ideal of $B$ (instead of just prime), and (2) why they are saying that $B$ is local. They seem to be claiming this, as the relation of domination is defined only for two local rings.

**topspin1617** · June 22nd 2011, 06:04 PM

Anyone? Sorry if it seems like there are too many details; I just didn't know how many details would be necessary to solve this. Hartshorne just skips right over this as if it's blatantly obvious (who knows, maybe it is... but I don't see it).

**tonio** · June 22nd 2011, 06:58 PM

Originally Posted by topspin1617

Anyone? Sorry if it seems like there are too many details; I just didn't know how many details would be necessary to solve this. Hartshorne just skips right over this as if it's blatantly obvious (who knows, maybe it is... but I don't see it).

In my humble opinion, Hartshorne's book is a terrible, awful, anguishing, sickening one for learning algebraic geometry. It though can be a very good one for advanced students in the subject and/or as a reference book.
Somewhere in the net (google it) solutions to some of this book's ridiculously hard (sometimes) exercises. give it a shot.

Tonio

**NonCommAlg** · June 22nd 2011, 07:33 PM

Originally Posted by topspin1617

If anyone is interested in the context of this question, I am reading through the proof of Lemma 6.5 in Hartshorne's Algebraic Geometry.

We have fields $k\subseteq K$ such that $\mathrm{tr.deg}_k(K)=1$ . Also, assume $k$ is algebraically closed.
Let $y\in K\setminus k$ , and let $B$ be the integral closure of the ring $k[y]$ in $K$ . It can be shown that $B$ is a Dedekind domain which is finitely generated as a $k$ -algebra (and I'm fine with that).

Now, suppose $R$ is a discrete valuation ring between $k$ and $K$ such that $y\in R$ . Since $R$ is integrally closed (being a DVR), we must have $B\subseteq R$ .

Here's where I get lost: they say that if $\mathfrak{m}_R$ is the (unique) maximal ideal of $R$ , then $\mathfrak{n}=B\cap \mathfrak{m}_R$ is a maximal ideal of $B$ , and that $B$ is dominated by $R$ . What I don't see is (1) how we know that $\mathfrak{n}$ is a maximal ideal of $B$ (instead of just prime), and (2) why they are saying that $B$ is local. They seem to be claiming this, as the relation of domination is defined only for two local rings.

well, suppose that $\mathfrak{n}$ is not a maximal ideal of $B$ and let $\mathfrak{n}'$ be a maximal ideal of $B$ which contains $\mathfrak{n}$ . now, we get a chain of prime ideals $(0) \subset \mathfrak{n} \subset \mathfrak{n}'$ of $B$ , contradicting this fact that the Krull dimension of a Dedekind domain is at most $1$ .

by the way, understanding Hartshorne without having a good commutative algebra background is impossible. the author assumes that background.

**topspin1617** · June 22nd 2011, 07:49 PM

Originally Posted by NonCommAlg

well, suppose that $\mathfrak{n}$ is not a maximal ideal of $B$ and let $\mathfrak{n}'$ be a maximal ideal of $B$ which contains $\mathfrak{n}$ . then we will have a chain of prime ideals $(0) \subset \mathfrak{n} \subset \mathfrak{n}'$ of $B$ , contradicting this fact that the Krull dimension of a Dedekind domain is at most $1$ .

by the way, understanding Hartshorne without having a good commutative algebra background is impossible. the author assumes that background.

Oh wow..... thank you. Now I feel like a complete moron lol, this is so obvious.

I do have a very good commutative algebra background. I just wasn't even thinking about Krull dimension. I was thinking more in the geometric direction, trying to figure out if there was something about the geometry here that forced that ideal to be maximal.

That still leaves my other question though... do we know that $B$ must be local? Again, it's just the way that the author mentions " $R$ dominates $B$ " which is only defined for local rings. Though I SUPPOSE that could be a typo/oversight/mistake.

**NonCommAlg** · June 22nd 2011, 08:05 PM

Originally Posted by topspin1617

Oh wow..... thank you. Now I feel like a complete moron lol, this is so obvious.

I do have a very good commutative algebra background. I just wasn't even thinking about Krull dimension. I was thinking more in the geometric direction, trying to figure out if there was something about the geometry here that forced that ideal to be maximal.

That still leaves my other question though... do we know that $B$ must be local? Again, it's just the way that the author mentions " $R$ dominates $B$ " which is only defined for local rings. Though I SUPPOSE that could be a typo/oversight/mistake.

i have no reason to believe that $B$ must be local itself. i suggest you read the rest of the proof. right after this the author talks about the localization of $B$ at $\mathfrak{n}$ , which would be $B$ itself if $\mathfrak{n}$ was the unique maximal ideal of $B$ . so i guess "domination" here was used only in this sense that the contraction of a maximal ideal of a overring of $B$ in $B$ is a maximal ideal of $B$ .

**topspin1617** · June 22nd 2011, 08:15 PM

Originally Posted by NonCommAlg

i have no reason to believe that $B$ must be local itself. i suggest you read the rest of the proof. right after this the author talks about the localization of $B$ at $\mathfrak{n}$ , which would be $B$ itself if $\mathfrak{n}$ was the unique maximal ideal of $B$ . so i guess "domination" here was used only in this sense that the contraction of a maximal ideal of a overring of $B$ in $B$ is a maximal ideal of $B$ .

You're right in that there doesn't seem to be any reason that $B$ should be local.

If that's how he meant domination, then he really shouldn't use it in this context. I'm just reading his own definition:

"If $A,B$ are local rings contained in a field $K$ , we say that $B$ dominates $A$ if $A\subseteq B$ and $\mathfrak{m}_B\cap A=\mathfrak{m}_A$ ."

**NonCommAlg** · June 22nd 2011, 08:26 PM

Originally Posted by topspin1617

You're right in that there doesn't seem to be any reason that $B$ should be local.

If that's how he meant domination, then he really shouldn't use it in this context. I'm just reading his own definition:

"If $A,B$ are local rings contained in a field $K$ , we say that $B$ dominates $A$ if $A\subseteq B$ and $\mathfrak{m}_B\cap A=\mathfrak{m}_A$ ."

yes, i saw the definition but that's not a big deal. the author is trying to prove that $B_{\mathfrak{n}}=R$ and this tells you that $B$ is not necessarily local.

**NonCommAlg** · June 22nd 2011, 08:52 PM

i just realized that, in my proof to your first question, we need to explain why $\mathfrak{n} \neq (0)$ . i think the author assumes from the beginning that $y \in \mathfrak{m}$ , which implies that $y \in \mathfrak{n}$ and so $\mathfrak{n} \neq (0)$ .
you might have another way to show this though.

**topspin1617** · June 22nd 2011, 09:43 PM

Originally Posted by NonCommAlg

yes, i saw the definition but that's not a big deal. the author is trying to prove that $B_{\mathfrak{n}}=R$ and this tells you that $B$ is not necessarily local.

I know it's not that big of a deal. I was just wondering if he purposely meant something there, in which case that would have mattered.

Originally Posted by NonCommAlg

i just realized that, in my proof to your first question, we need to explain why $\mathfrak{n} \neq (0)$ . i think the author assumes from the beginning that $y \in \mathfrak{m}$ , which implies that $y \in \mathfrak{n}$ and so $\mathfrak{n} \neq (0)$ .
you might have another way to show this though.

Well, he doesn't assume that from the beginning. He begins the following paragraph with "if, in addition, $y\in \mathfrak{m}_R....$ ". I'll have to think about why the intersection is nonzero in general.

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