Let $A$ be a completion discrete valuation ring with quotient field $K$, $L/K$ finite and separable, and $B$ the integral closure of $A$ in $L$. Let $P, \mathfrak P$ be the unique maximal ideals of $A, B$.

I know that if $B/ \mathfrak P$ is a separable extension of $A/P$, then $B = A[x]$ for some $x \in B$. Is there an example of an inseparable extension of residue fields where $B$ is not a simple extension of $A$?

  • $\begingroup$ See the Wikipedia page for the Primitive Element Theorem, it gives counterexamples for the non-separable extensions. $\endgroup$ – Mastrel Jul 7 '14 at 3:15
  • $\begingroup$ I'm sorry but I don't see how the article gives an example of what I'm looking for. Can you explain in more detail? $\endgroup$ – D_S Jul 7 '14 at 3:30
  • $\begingroup$ Your right, it doesn't. Sorry, when I looked up the Primitive Element Theorem I forgot that you supposed you started with a DVR. $\endgroup$ – Mastrel Jul 7 '14 at 3:37

If $B=A[x]$, then $B/Q=A/P[x+Q]$. (I use $Q$ for the maximal ideal of $B$ - simpler to type.) So you need to find an example with a residue field extension that possesses no primitive element.

An example of a field extension $k\subset l$ without primitive element is $\mathbb{F}_p(s^p,t^p)\subset \mathbb{F}_p(s,t)$, where $s,t$ are algebraically independent elements over the finite field $\mathbb{F}_p$ with $p$ elements. The extension is of degree $p^2$.

Let $k\subset l$ be this extension and let $A:=k[[z]]$ be the power series in $z$ with coefficients in $k$. Consider the polynomials $X^p+zX-s^p,X^p+zX-t^p\in A[X]$.

Both polynomials are irreducible over $K:=k((z))$ , because they are irreducible modulo $P=zA$.

Both polynomials are separable: their derivative is $z$.

Let $x$ be a root of the first, and $y$ of the second polynomial. Let $B$ be the integral closure of $A$ in $L:=K(x,y)$.

The extension $K\subset L$ is separable of degree $\leq p^2$.

The elements $x,y$ are integral over $A$, hence $A[x,y]\subseteq B$ and therefore

$B/Q\supseteq k[x+Q,y+Q]=k[s,t]=l$,

which in turn implies $[B/Q:A/P]\geq [l:k]=p^2$. By the fundamental inequality of valuation theory

$[B/Q:A/P]\leq [L:K]\leq p^2$,

hence $B/Q=l$, which is what we wanted.


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