2
$\begingroup$

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 $A$?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.