# units in discrete valuation rings

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 $A[x] = B$.

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I presume you mean "let $x$ be a lift of $y$ to $B$"? – 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.

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