I wanted to prove the following:

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

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> 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 $R$.
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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:

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

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!