# 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.)

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]

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