Prime ideals of formal power series ring that are above the same prime ideal

MathOverflow Asked by A. C. Biller on January 5, 2022

Let $R$ denote a commutative ring with identity and let $R[[X]]$ denote the
ring of formal power series over $R$ in an indeterminate $X$. If $I$ is an ideal of $R$,
then $I[[X]]$, the set of power series of $R[[X]]$ with coefficients all in $I$. Now if $P$ is a prime ideal of $R$, then $P[[X]]$ is a prime ideal of $R[[X]]$ that is contained in the prime ideal $langle P, xrangle$ of $R[[X]]$ such that $langle P, X ranglecap R=P$. Now let $Q$ be a prime ideal of $R[[X]]$ such that $P[[X]]subseteq Q$ and $Qcap R=P$. Is there any characterization for such a $Q$? (I guess that $langle P, Xrangle$ is the only ideal with that property.)

One Answer

In general, $(P, X)$ is not the only prime containing $P[[X]]$ and contracting to $P$. I don't have anything to say about the problem of characterizing such primes, but in general it seems extremely hard. Let's focus on the case $P = 0$.

As a motivating example we can even use the integers. The ring $mathbb{Z}[[X]]$ is a UFD. For any prime $p$ and power series $F$, it is clear that $p +xF$ is irreducible in $mathbb{Z}[[X]]$ and hence prime. Moreover if we take $f in mathbb{Z}[X]$ to be such that the content $c(f)$ of $f$, that is the ideal generated by the coefficients of $f$ in $mathbb{Z}$, is coprime to $p mathbb{Z}$, then additionally $(p + Xf) cap mathbb{Z}= 0$. One way to show this would be to appeal to the Dedekind-Mertens content formula$^1$, which asserts that over any ring $R$, if $f$ is a polynomial of degree $n$, $G,H in R[[X]]$, with $fG=H$, then $c(f)c(G)^{n+1} = c(G)^{n} c(H)$. Here $c(F)$ denotes the content ideal of the power series $F$. From here, if we had $(p+Xf)G = p G_0 in mathbb{Z}$ then the D-M formula would imply $frac{1}{p} c(G)^k subseteq c(G)^k$ which would in turn imply $p$ is a unit (absurd). For every prime $p$, we have found infinitely many polynomials which are prime in $mathbb{Z}[[X]]$ and which lie over $0$ in $mathbb{Z}$. Moreover in this way we can be sure to find lots of distinct primes in $mathbb{Z}[[X]]$, which follows for example from this old post of mine on stackexchange.

I'm not sure to what extent this way of producing principal primes over $0$ generalizes to other rings. It does work verbatim for any Archimedean GCD domain $D$ for which $D[[X]]$ has its irreducible elements prime. The tough part is that last bit, which is a very delicate property. However, it is sufficient that $D[[X]]$ be a UFD, which is a well-studied problem. So for example this argument applies just as well to any regular UFD.

$^1$ See theorem 3.6 in the paper Zero divisors in power series rings by R. Gilmer, A. Grams, and T. Parker [Journal für die reine und angewandte Mathematik (1975), EUDML Link]

Answered by Badam Baplan on January 5, 2022

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