A subring of polynomial ring with coefficients in a DVR that is not noetherian

Question

Let $R$ be a discrete valuation ring, $K$ its field of fractions and $A=\{f\in K[T],f(0)\in R\}$. Let $\mathfrak{m}$ be the maximal ideal of $R$, $\mathfrak{m'}={\mathfrak{m}+KT+KT^2+\cdots}$.

1) How to show $A$ is not a Noetherian ring?

I cannot construct the infinite ascending chain of ideals.

2) How to show $KT+KT^2+\cdots$ is the only prime ideal between $0$ and $\mathfrak{m'}$ ?

(It is in Gortz and Wedhorn, Algebraic Geometry I, p. 279, Ex. 10.9.)

For the second part, you can notice that $A\to R$ defined by $f\mapsto f(0)$ is a surjective ring map, with kernel $\mathfrak{m}'$. From this you know that $A/\mathfrak{m}'\cong R$. You should be able to conclude then. — Alex Youcis, Oct 06 '13 at 07:34
@AlexYoucis The kernel of your map is not $m'$. Even supposing that you consider $m'=TK[T]$, is not clear (to me) how this helps. — , Oct 06 '13 at 08:43

score 4 · Accepted Answer · answered Oct 06 '13 at 09:55

4

Hints.

1) Assume that the ideal $TK[T]$ is finitely generated and conclude that $K$ is a finitely generated $R$-module, a contradiction.

2) Let $P$ be a prime ideal of $A$, $(0)\subset P\subset\mathfrak m'$. Then $P\cap R=\mathfrak m$ or $P\cap R=(0)$. In the first case $P=\mathfrak m'$. In the later case note that $S^{-1}A=K[T]$, where $S=R-\{0\}$. Therefore $S^{-1}P$ is generated by a polynomial $f\in P$ which is irreducible in $K[T]$. What can you deduce from this?

answered Oct 06 '13 at 09:55

Thanks, here is my reasoning according to your hint, I don't know if it is a bit tedious. (1)If $TK[T]$ is finite generated, then $K$ is integral over $A$, so it will be integral over R, which is contradiction since $R$ is integrally closed. (2)Then I can deduce from the correspondence that $P$ has form $(f)^{c}$, so $f$ has constant term zero and $f$ irreducible, so $f=T$. – Oct 06 '13 at 10:31
@mqx Actually I think 2) is not true, but it's true that the height of $m'$ is 2, that is, there are no primes $p\subset p'$ strictly contained between $(0)$ and $m'$. – Oct 06 '13 at 11:08
Let me see, by localization we have $R+KT+..\rightarrow K[T]$, if $P$ is a prime ideal contained in $\mathfrak{m}$ and $P\cap R=0$, then $S^{-1}P$ corresponds to a unique $(f)$ in $K[T]$, whose contraction is $P$, the contraction of $f\cap R=0$ so $f$ must be T since it is irreducible, is this right? And how to prove the assertion on height directly? – Oct 06 '13 at 11:30
@mqx Not only $T$ has this property! If $\mathfrak m=(t)$ then $f(T)=t+T$ is also irreducible and $(f)\cap R=(0)$. (I don't know what you mean by proving the assertion about height directly. It's obvious by passing to $K[T]$.) – Oct 06 '13 at 13:38
I see, it is because injective integral extension preserves dimension. Sorry for my slow mind.. – Oct 06 '13 at 13:47

A subring of polynomial ring with coefficients in a DVR that is not noetherian

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