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Let $\mathbb{Z}[[X]]$ be the formal power series ring over $\mathbb{Z}$.

I want to understand

  • the set of prime ideals $\rm{Spec}(\mathbb{Z}[[X]])$
  • maximal ideals $\rm{Spm}(\mathbb{Z}[[X]])$ and
  • the Jacobson radical $J(\mathbb{Z}[[X]])$.

How can one write these sets explicitly?

JSquared
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神宮寺春姫
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  • Not sure for the classification of characteristic $0$ prime ideals. For example $\Bbb{Z}[[X]]/(X-p)$ is isomorphic to the $p$-adic integers. – reuns Apr 25 '23 at 08:24

1 Answers1

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Maximal ideals are of the form $(p,X)$ for a prime $p$ (see Wiki) and the Jacobson radical is the intersection of the maximal ideals, so it equals $(X)$. Alternatively there is a proof at Jacobson radical of formal power series over an integral domain.

[Edit: a previous version of this answer contained a wrong description of height 1 primes in the formal power series ring.]

In case of the polynomial ring $\text{Spec}\,\mathbb Z[X]$ all prime ideals may be described as being of the form $(p,f)$ for $p$ prime or zero and $f$ irreducible mod $p$ or zero (see Spectrum of $\mathbb{Z}[x]$).

A similar characterization for $\mathbb Z[[X]]$ is that every nonzero prime ideal is either:

  • a maximal ideal of height 2 of the form $(p,X)$, or
  • a height 1 prime ideal of the form $(f)$ for $f\in\mathbb Z[[X]]$ irreducible

This follows since it is also a UFD, (hence height 1 primes are principal) and its Krull dimension is also 2 (hence height 2 primes are maximal).

It should be noted that a polynomial may change its irreducibility when considered as a formal power series. See https://www.jstor.org/stable/27642533 for examples and discussion of the irreducibility problem for integer power series. In terms of the map of schemes $$\text{Spec}\, \mathbb Z[[X]] \to \text{Spec}\,\mathbb Z[X]$$ this means that

  • there is some ramification over irreducible polynomials which split as a power series
  • besides power series, the generic fiber also includes some polynomials (which are reducible as a polynomial, but all but one of the factors has constant term $\pm1$, thus is a unit when considered as a power series).
Ben
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    The second to last sentence is wrong. For example, the fiber over $(X)$ to $\mathbb{Q}[X]$ has more things, for example the ideal $(X - p)$ for a prime $p$. – JSquared Apr 10 '23 at 18:56
  • @JSquared Sorry, I can’t follow your comment. The second to last sentence in this answer is about DVRs, but your comment mentions a polynomial ring. $(X-p)$ is not a proper ideal in the formal power series ring, as $1/(X-p)$ is expressible by geometric series. – Ben Apr 12 '23 at 00:29
  • Let $f = \sum_{n=1}^\infty a_nX^n$ be some power series such that $(X-p)(f) = 1$. Then, by the multiplication law for formal power series, $a_0 * p = 1$, (colloquially, $a = 1/p$. But $p$ does not have an inverse in $\mathbb{Z}$.

    If you want to write it as a geometric series, you'd have to write it as $1/(1-(1+p-x))$. Then if you expand the resulting geometric series and collect in terms of x, component will be an infinite series of with binomial coefficients, which won't converge in $\mathbb{Z}_p$--but if it converged to an integer it would have had to in $\mathbb{Z}_p$.

     – JSquared Apr 18 '23 at 20:50
  • @JSquared I don’t see your point. The second to last sentence is about formal power series over a field, not a general ring. It then applies that fact only to $\mathbb F_p$ and $\mathbb Q$. It seems like you are considering the case when $k=\mathbb Z$, which is not a field, and the statement would no longer hold. – Ben Apr 20 '23 at 00:23
  • Perhaps I've not communicated my objection well. I'm objecting to the claim that "the prime ideals are all of the form $(X,p),(X),(p)$ or $(0)$ for a prime $p$", for example $(X-p)$ is a proper ideal of $\mathbb{Z}[[X]]$ which is not of this form, the proof is the first two sentences of my previous comment. – JSquared Apr 24 '23 at 19:36
  • @JSquared Thanks I see what you mean now. The first bullet point in the answer is already wrong then. – Ben Apr 25 '23 at 03:16
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    @JSquared since the question is closed i thought it best to edit the answer than delete it. Please let me know if you find something wrong – Ben Apr 25 '23 at 05:54