Is $(XY - 1)$ a maximal ideal in $k[[X]][Y]$, and if so, how can I see it?

It is at least prime because the generator is irreducible, and by the same argument it is maximal among all principal ideals. But I haven't gotten further than that - Finding units in the quotient ring didn't turn out well, either.

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    $\begingroup$ Is $k[[x]][y]/(xy-1) \cong k[[x]][x^{-1}]$ a field? $\endgroup$
    – Bill Cook
    Jan 15, 2012 at 19:40
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    $\begingroup$ Thanks, this isomorphy is very helpful. I immediately thought that it is obvious that this is a field, but as it turns out, I'm not sure yet $\endgroup$
    – argon
    Jan 15, 2012 at 19:53
  • $\begingroup$ Maybe this will help: what are the units in $k[[X]]$? $\endgroup$ Jan 15, 2012 at 19:57
  • $\begingroup$ Ah yes, I vaguely remember having proven that all elements of $K[[X]]$ except for X are invertible, is that true? In that case, I'd be done. $\endgroup$
    – argon
    Jan 15, 2012 at 20:01
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    $\begingroup$ @argon That's not quite true. The units are the formal power series having non-zero constant terms. So all non-zero elements look like $X^n \times (\text{a unit})$. $\endgroup$ Jan 15, 2012 at 20:05

1 Answer 1


If $A$ is a commutative ring and $s\in A$ any element , the ring $ B=A[Y]/(sY-1)$ is isomorphic to the localized ring $S^{-1}A$, where $S$ is the multiplicative set $S=\lbrace 1,s,s^2,s^3,... \rbrace \subset A$.
The prime ideals of the localized ring $B$ are in bijection with the prime ideals of $A$ disjoint from $S$: this is perhaps the fundamental fact about localization.

In your case $A=k[[X]]$, $s=X$ and the prime ideals of the DVR $A=k[[X]$ are $(0)$ and $M=(X)$.
The only surviving prime ideal in $B=S^{-1}A$ is the zero ideal, since obviously $M$ is killed by localization, and thus $B$ is a field (called the field of Laurent series k((X)) over $k$).
If you go back to the other definition of $B$ as $B=k[[X]][Y]/(XY-1)$, you see that your ideal $(XY-1)$ was indeed maximal.

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    $\begingroup$ Dear @user26857: you are free to suppose $s$ not nilpotent, of course. However my answer is correct even in that case: the ring $S^{-1} A=0$ is then the zero ring and its prime spectrum, the empty set, is in perfect bijection with the set of prime ideals of $A$ disjoint from $S$, which also happens to be the empty set. And the icing on the cake is that for $s$ nilpotent we also have $(sY-1)=(1)\subset A[Y]$ so that $B=A[Y]/(sY-1)=0$ too :-) $\endgroup$ Mar 29, 2014 at 21:39

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