# Quotient of Jacobson ring is Jacobson as in Eisenbud

I wanted to prove the following:

Let $R$ be a Jacobson ring, $\mathfrak{p}<R$ a (prime) ideal. Then $R/\mathfrak{p}$ is Jacobson.

The statement has been taken from Eisenbud's Commutative Algebra with a View Towards Algebraic Geometry, p. 132 in the proof of lemma 4.20.

I want to prove it with what I have available there, hence:

A ring $R$ is Jacobson if $$\mathfrak{p}=\bigcap_{\substack{\mathfrak{m}\in\max(R)\\ \mathfrak{p}\subseteq\mathfrak{m}}}\mathfrak{m}\quad\forall \mathfrak{p}\in \operatorname{prime}(R)$$ where $\max(R)$ is the collection of maximal ideals and $\operatorname{prime}(R)$ is the collection of prime ideals in $R$.

We have to show that $\mathfrak{q}\in\operatorname{prime}(R/\mathfrak{p})$ implies $\mathfrak{q}=\bigcap_{\mathfrak{m}\in\max(R/\mathfrak{p}), \mathfrak{q}\subseteq \mathfrak{m}}\mathfrak{m}$. Let $\mathfrak{r}:=\bigcap_{\mathfrak{m}\in\max(R),\pi^{-1}(\mathfrak{q})\subseteq\mathfrak{m}}\pi(\mathfrak{m})$. In what follows I denote by $\pi:R\to R/\mathfrak{p}$ the canonical projection. I used the following claim:

Let $R$ be a ring, $\mathfrak{p}<R$ an ideal (not necessarily prime), $\mathfrak{m}$ a maximal ideal containing $\mathfrak{p}$, then $\pi(\mathfrak{m})$ is a maximal ideal and $\pi$ yields a bijection between maximal ideals containing $\mathfrak{p}$ in $R$ and ideals in $R/\mathfrak{p}$.

\begin{align*}\pi^{-1}(\mathfrak{r})=&\bigcap_{\substack{\mathfrak{m}\in\max(R)\\\pi^{-1}(\mathfrak{q})\subseteq\mathfrak{m}}}\pi^{-1}(\pi(\mathfrak{m}))\\ =&\bigcap_{\substack{\mathfrak{m}\in\max(R)\\\pi^{-1}(\mathfrak{q})\subseteq\mathfrak{m}}}\mathfrak{m}\qquad\text{(as of the 1-1 correspondence)}\\ =&\pi^{-1}(\mathfrak{q})\qquad\text{(as \pi^{-1}(\mathfrak{q})< R is prime and R is Jacobson)}\end{align*}

Because $\pi(\mathfrak{m})$ is an ideal for every $\mathfrak{m}\in\max(R)$ such that $\mathfrak{p}\subseteq\mathfrak{m}$ by the 1-1 correspondence, so is $\mathfrak{r}$ and as of the 1-1 correspondence between maximal ideals containing $\mathfrak{q}$ and maximal ideals containing $\pi^{-1}(\mathfrak{q})$, $\mathfrak{r}=\bigcap_{\mathfrak{m}\in\max(R/\mathfrak{p}),\mathfrak{q}\subseteq\mathfrak{m}}\mathfrak{m}$.

Thus the 1-1 correspondence implies $\mathfrak{r}=\mathfrak{q}$.

Question: does this make sense? Is this correct reasoning? Thank you!

## 2 Answers

What you have written doesn't quite make sense (though these may just be typos): if $\mathfrak{r}:=\bigcap_{\mathfrak{m}\in\max(R),\pi^{-1}(\mathfrak{q})\subseteq\mathfrak{m}}\mathfrak{m}$, then $\pi^{-1}(\mathfrak{r})$ is not $\bigcap_{\substack{\mathfrak{m}\in\max(R)\\\pi^{-1}(\mathfrak{q})\subseteq\mathfrak{m}}}\pi^{-1}(\pi(\mathfrak{m}))$. Also, note that for any maximal ideal $\mathfrak{m}$ of $R$, $\pi^{-1}(\pi(\mathfrak{m})) = \mathfrak{m}$ always holds. As an alternative way to view the Jacobson condition, consider the following: $\DeclareMathOperator{nil}{nil} \DeclareMathOperator{rad}{rad}$

Proposition: Let $\nil(R)$ be the nilradical of $R$ (i.e. intersection of all primes), $\rad(R)$ the Jacobson radical of $R$ (i.e. intersection of all maximal ideals). The following are equivalent:
i) $R$ is Jacobson
ii) $\nil(R/I) = \rad(R/I)$ for every ideal $I$
iii) $\nil(R/p) = \rad(R/p)$ for every prime $p$

The implications (i) $\implies$ (ii) and (iii) $\implies$ (i) follow immediately from the correspondence of primes and maximal ideals in the quotient (try proving this carefully!). Once this characterization is shown though, the property of being Jacobson is seen to pass to any quotient, since any quotient of $R/p$ is still a quotient of $R$.

• Thank you. $\mathfrak{r}$ should indeed be the intersection of the images. I will check your proposed statement. I wanted to do it in the "self-contained" way mentioned above though. – M. Luethi Jan 24 '14 at 21:26

there is some nice reference on this topic you may want to have a look: https://en.wikibooks.org/wiki/Commutative_Algebra/Jacobson_rings_and_Jacobson_spaces

For a solution in the "self-contained" way as you want: Let $$p$$ be a prime in $$R/I$$, and it can be pulled back to the unique prime $$P$$ in $$R$$. Denote $$\pi$$ the canonical projection from $$R$$ to $$R/I$$. We also assume that

$$$$P = \cap M_{\alpha} \text{ ,where M_{\alpha} is maximal.}$$$$ Then we have the following:

$$$$p=\pi(\pi^{-1}(p)) =\pi(P)=\pi(\cap M_{\alpha})=\cap \pi (M_{\alpha})$$$$

It's worth noticing that the last equality doesn't hold in general. In this case, the nontrivial direction holds since $$$$\forall y \in\pi(\cap M_{\alpha}),\forall \alpha, (y+ I)\subset M_{\alpha}+I$$$$ This leads to $$y=x_{\alpha}+i_{\alpha}$$ ,where $$x_{\alpha} \in M_{\alpha}$$ and $$i_{\alpha} \in I$$

So $$\forall \alpha , y\in M_{\alpha}$$

• I don't think there is some big idea behind so I prefer to post the little trick in detail. – Ivan Jan 10 '19 at 21:28
• Also, the reference gives another nice characterization of Jacobson rings. – Ivan Jan 10 '19 at 21:33