The Krull dimension of a polynomial ring over a PID

abstract-algebracommutative-algebrakrull-dimensionmaximal-and-prime-ideals

Recently, I proved this statement:

Proposition: Let $R$ be a PID. The prime ideals of $R[y]$ are precisely the ideals of the following form:

  1. $(0)$,
  2. $(f(y))$ where $f$ is an irreducible polynomial in $R[Y]$
  3. $(p, f(y))$ where $p \in R$ is prime and $f(y)$ is irreducible in $R[y]$ and its image is irreducible in $(R/p)[y]$.

Note: Maximal ideals are of the third form and vice versa.

Using this statement I would like to show that the Krull dimension of $k[x,y]$ is 2 where $k$ is a field. However, I'm stuck at showing that there can't be a chain of prime ideals of the form $\{0\}\subsetneq (f)\subsetneq (p,g)$ i.e a chain involving every ideal of the types shown above.

Any ideas? In the case when R is a general PID would the Krull dimension be 2 or 3?

Best Answer

Every PID which is not a field has Krull dimension $1$. Also, every PID is Noetherian, and if $R$ is Noetherian whose Krull dimension is $n$, then $R[x]$ has Krull dimension $n+1$. Therefore the answer to your question is that that Krull dimension of $R[y]$ is $2$.

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