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By Cohen structure theorem, a complete regular equicharacteristic Noetherian local ring is isomorphic to a power series. In particular, this should hold for finite extensions of power series $k[[t]][\alpha]$, where $\alpha$ is a root of $f(x) \in k[[t]][x]$, if it is regular.

How would this computation go in the simplest nontrivial case? Say I would like to exhibit the following isomorphism explicity: $$k[[t]][\alpha] \xrightarrow{\sim} k[[u]], \text{ where } \alpha^2 + a\alpha + b = 0.$$

And we assume that $\alpha$ is not algebraic over $k$. If $a=0, b=-t$, for example, then we can easily construct the isomorphism by $\alpha\mapsto u, t\mapsto u^2$. But how in general?

Thanks.

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    $\begingroup$ In characteristic not 2, you can complete squares and assume $a=0$. Then, we may clearly assume $b\neq 0$ and so write $b=t^mu$ where $u$ is a unit. Again, we may replace $\alpha$ by $\alpha/t^n$ where $n$ is the integral part of $m/2$. Thus, we may assume $m=0,1$. Can you now finish the argument? $\endgroup$
    – Mohan
    Commented Sep 1, 2018 at 16:39
  • $\begingroup$ This is related to Weierstrass preparation, correct? Yes, I understand the argument, but I would like to have more knowledge/control of the unit $u$ that can be factored off $b$. $\endgroup$
    – Somatic Custard
    Commented Sep 1, 2018 at 18:40
  • $\begingroup$ Any unit $u=u(0)v^2$ for some unit $v$ with $v(0)=1$ by Hensel's lemma. Thus, replacing $\alpha$ with $\alpha v^{-1}$, you can assume the equation is $\alpha^2+at^n=0$ with $0\neq a\in k$ and $n=0,1$. $\endgroup$
    – Mohan
    Commented Sep 1, 2018 at 22:03

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