Let $B/A$ be an extension of discrete valuation rings such that the purely inseparable extension of residue fields $l/k$ has a primitive element, i.e., $l=k(y)$ for some element $y$ in $l$. I want to know if one can plug in another discrete valuation ring as follows.

Let $x$ be a lift of $y$ to $B$. Is $A[x]$ a dvr? What are necessary and sufficient conditions?

Of course, the answer is yes if $e=1$. Then $A[x] = B$.

  • $\begingroup$ I presume you mean "let $x$ be a lift of $y$ to $B$"? $\endgroup$ – David Loeffler Jan 28 '12 at 9:42

As $k\subseteq l$ is a purely inseparable extension, characteristic of $k$ should be positive, say $p$. The only fields I know with positive characteristic are finite fields and $F(T_1,\cdots,T_n)$ (where $F$ is a field with positive characteristic like finite fields). I started to make an example.

Let $K=F_{p^n}[T]$ and $L=F_{p^m}[T]$, $n<m$, we already know their discrete valuation rings. $B$ can not be the trivial valuation because then $k=l=\{0\}$ and some don't accept fields have one element and for others who accept, this isn't an inseparable extension .If $A$ won't be trivial then $A=(F_{p^n})_{p(T)}$ and $A=(F_{p^m})_{q(T)}$, $p(T)\in F_{p^n}[T],g(T)\in F_{p^m}[T]$ indecomposable polynomials and we have $k=F_{p^n},l=F_{p^m}$ which is a purely extension. Then As $A\subseteq B$ and $B\neq L$ we have to have $q(T)=u p(T)$ for a $u\in l$. The only discrete valuation ring of $L$ which can be written as $A[x]$ is $B$ and the necessary and sufficient condition on $x$ is $y\in A[x]$ for example $x=y$ or $x=T+y$ or $x=y^i$ which $y^i$ is another possible primitive element of $l$. (And absolutely it can't be anything such as $1$ or any other element of $K$ so I can't get what the questioner was to say by $e=1$ (which I think he it is a typo e $->$ x) then $A[x]=B$).

Now if $A$ be trivial then since $T,T^{-1}\in A\subseteq B$, $B$ should be trivial which we said it is not the case.

Can we guess this result is global, I mean the answer of the main question should be "The necessary and sufficient condition is $y\in A[x]$"?

If we change the question in this form that instead of the residue fields, $K\subseteq L$ is a purely extension then except the above example we have another class of examples with the same result. We can get $K=F_{p^n},L=F_{p^m}$, $n<m$. And the only valuation rings so discrete valuation rings of finite fields are trivial ones (because their multiplicative groups are cyclic). Then again we have to have $A=K,B=L$. And $A[x]=L$ means $K[x]=L$ and so $y\in A[x]$.

Again can we say answer in the new case is "The necessary and sufficient condition is $y\in A[x]$"?

Here at least I gave an answer in a class of examples.


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.