1

Is there a simple classification of the prime ideals in $k[[x]][y]$ for $k$ an algebraically closed field? This is a two-dimensional ring so we can divide all prime ideals according to their heights, possible values being $0$, $1$, $2$.

Since the ring is an integral domain, there is only one prime ideal of height $0$, the zero ideal. Not sure about the rest.

1 Answers1

2

The ring $A = k[\![X]\!]$ is a discrete valuation ring and as such

  • it is a local domain, and
  • it is a principal ideal domain, hence
  • it is a unique factorisation domain.

However, Gauß classifies all prime elements in polynomial rings over unique factorisation domains, see wiki/Gauß’s Lemma. So we know all principal prime ideals. We use the structure of $A$ as a local domain to get the rest.

So now let $\mathfrak p ⊆ A[Y]$ be a nontrivial prime ideal.

  • If $X ∈ \mathfrak p$, then it corresponds to a prime ideal in $k[Y] = A/(X)[Y]$, so $\mathfrak p = (X,f)$ for some polynomial $f ∈ A[Y]$ that is irreducible in $k[Y]$, so linear in our case.
  • If $X \notin \mathfrak p$, then it corresponds to a prime in $Q[Y] = A_X[Y]$, where $Q = A_X = \operatorname{Quot} A$ is the quotient field of $A$, which is the same as $A$ localized far $X$. Hence $\mathfrak p = (f)$ for some polynomial $f ∈ A[Y]$ which is irreducible in $Q[Y]$ and has content $1$.

All in all, the primes in $k[\![X]\!][Y]$ are

  • $(0)$ and $(X)$,
  • $(f)$ for polynomials $f ∈ A[Y]$ irreducible in $Q[Y]$ of content $1$, and
  • $(X, f)$ for polynomials $f ∈ A[Y]$ irreducible in $k[Y]$.

You don’t need $k$ to be algebraically closed, but it helps with finding the irreducible polynomials in $k[Y]$ and probably in $Q[Y]$, too.

user26857
  • 52,094
k.stm
  • 18,539