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I am studying Jacobson rings, using this file by Matthew Emerton and this source.

I am trying to understand the proof of the following:

Theorem. if $A$ is a Jacobson ring (that is, $\mathfrak p= \operatorname{Jac}(\mathfrak p)$ for all primes $\mathfrak p$ of $A$), then so is the ring of polynomials $A[X]$.

It appears in page 4 of Emerton's notes. He uses two lemmata:

Lemma 1. If $A\subseteq B$ are domains such that $A$ is Jacobson, and for some $0\neq a \in A$, the induced morphism $A_a \to B_a$ is integral, then $\operatorname{Jac}(B)=0$.

Lemma 2. If $\operatorname{Jac}( A ) = 0$, then $\operatorname{Jac} (A[X] ) = 0$ too.

(For clarity, following the notation of Bosch, $\operatorname{Jac}(B)$ is the intersection of all maximal ideals of $B$, and $\operatorname{Jac}(\mathfrak a) = \bigcap_{\mathfrak {a \subseteq m}} \mathfrak m$ is the intersection of all maximal ideals containing $\mathfrak a$). The proof Emerton gives goes as follows:

Proof. Let $\varphi:A[X]\to B$ be an epimorphism onto an integral domain $B$, and denote $A'=\varphi (A)$. It is sufficient to show that $\operatorname{Jac}(B)=0$. If $A'[X]\to B$ is an isomorphism, the result follows from lemma 2. So assume $0\neq \mathfrak p := \operatorname{Ker}(A'[X]\to B)$. Then we may find a non-zero element $a\in A'$ s.t. after inverting $a$, the ideal $\mathfrak p$ is generated by a monic polynomial $f\in A_a ' [X]$. Thus $B_a=A_a'[X]/f$ is finite over $A'_a$, and so lemma 1 shows again that $\operatorname{Jac}(B)=0$.

My questions:

  1. Why can we find a non-zero element $a$ of $A'$ as stated? More generally, the end of the proof is very unclear to me. Why is the extension $A'_a \to B_a$ finite? It would be very helpful if someone can explain it slower, in a bit more details.
  2. Atiyah-MacDonald problem 1.4 (see here) shows that $\operatorname{nil}(A[X])=\operatorname{Jac}(A[X])$. Is there a way to give a (perhaps shorter) proof using it?
Robert
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1 Answers1

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In $A'[x]$, there is a well-defined notion of degree. Let $\mathfrak p$ be the kernel of the map $A'[x]\to B$ and let $f\in\mathfrak p-0$ be an element of minimal degree $n$. By construction, $n>0$. After localising at the leading coefficient of $f$, if necessary, we may suppose that $f$ is monic. (I will not clutter the notation any further and just assume that $f$ is monic. Make sure you realise that and why the localisation cannot introduce a "new" element of smaller degree, which might happen in the presence of zero-divisors.) It is indeed the case that $f$ generates $\mathfrak p$, then (see below), but that's actually immaterial to the proof that $B/A'$ is finite, for if the kernel of $\varphi\colon A'[x]\to B$ contains a monic polynomial $f=x^n-\sum_{i=0}^{n-1}f_ix^i$, $f_i\in A'$, then $\varphi(x^n)=\sum_{i=0}^{n-1}f_i\varphi(x^i)$ is a linear combination of the $\varphi(x_i)$, $i<n$, and by a straightforward inductive argument, so is each $\varphi(x^m)$, $m>n$. Therefore, $B$ is generated by the finite set $1,\varphi(x),\dots,\varphi(x^{n-1})$ (whether or not $f$ actually generates the kernel).

But since I've realised that only after I have written down the proof that $\mathfrak p$ is actually generated by $f$, I might as well post it: Since we can do polynomial division by a monic(!) polynomial in any polynomial ring, we get that for every $g\in \mathfrak p$, there exists a decomposition $g=hf-r$ where $\deg r<\deg f$. But, $r=hf-g\in\mathfrak p$ cannot be of smaller degree as $f$ unless $r=0$; thus, $g\in (f)$. In conclusion, $\mathfrak p=(f)$, as claimed.

Ben
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