# Commutative Algebra/Jacobson rings and Jacobson spaces

## Definition and elementary characterisations

Definition 14.1:

A Jacobson ring is a ring such that every prime ideal is the intersection of some maximal ideals.

Before we strive for a characterisation of Jacobson rings, we shall prove a lemma first which will be of great use in one of the proofs in that characterisation.

Lemma 14.2:

Let ${\displaystyle R}$ be a Jacobson ring and let ${\displaystyle I\leq R}$ be an ideal. Then ${\displaystyle R/I}$ is a Jacobson ring.

Proof:

Let ${\displaystyle p\leq R/I}$ be prime. Then ${\displaystyle P:=\pi _{I}^{-1}(p)}$ is prime. Hence, according to the hypothesis, we may write

${\displaystyle P=\bigcap _{\alpha \in A}m_{\alpha }}$,

where the ${\displaystyle m_{\alpha }}$ are all maximal. As ${\displaystyle \pi _{I}}$ is surjective, we have ${\displaystyle \pi _{I}(P)=\pi _{I}(\pi _{I}^{-1}(p))=p}$. Hence, we have

${\displaystyle p=\pi _{I}\left(\bigcap _{\alpha \in A}m_{\alpha }\right)=\bigcap _{\alpha \in A}\pi _{I}(m_{\alpha })}$,

where the latter equality follows from ${\displaystyle \forall \alpha \in A:y+I\in \pi _{I}(m_{\alpha })}$ implying that for all ${\displaystyle \alpha }$, ${\displaystyle y=x_{\alpha }+i_{\alpha }}$, where ${\displaystyle x_{\alpha }\in m_{\alpha }}$ and ${\displaystyle i_{\alpha }\in I\subseteq m_{\alpha }}$ and thus ${\displaystyle y\in m_{\alpha }}$. Since the ideals ${\displaystyle \pi _{I}(m_{\alpha })}$ are maximal, the claim follows.${\displaystyle \Box }$

Theorem 14.3:

Let ${\displaystyle R}$ be a ring. The following are equivalent:

1. ${\displaystyle R}$ is a Jacobson ring.
2. Every radical ideal (see def. 13.1) is an intersection of maximal ideals.
3. For every ${\displaystyle p\leq R}$ prime the Jacobson radical of ${\displaystyle R/p}$ equals the zero ideal.
4. For every ideal ${\displaystyle I\leq R}$, the Jacobson radical of ${\displaystyle R/I}$ is equal to the nilradical of ${\displaystyle R/I}$.

Proof 1: We prove 1. ${\displaystyle \Rightarrow }$ 2. ${\displaystyle \Rightarrow }$ 3. ${\displaystyle \Rightarrow }$ 4. ${\displaystyle \Rightarrow }$ 1.

1. ${\displaystyle \Rightarrow }$ 2.: Let ${\displaystyle I}$ be a radical ideal. Due to theorem 13.3,

${\displaystyle I=\bigcap _{I\subseteq p \atop p{\text{ prime}}}p}$.

Now we may write each prime ideal ${\displaystyle p}$ containing ${\displaystyle I}$ as the intersection of maximal ideals (we are in a Jacobson ring) and hence obtain 1. ${\displaystyle \Rightarrow }$ 2.

2. ${\displaystyle \Rightarrow }$ 3.: Let ${\displaystyle p\leq R}$ be prime. In particular, ${\displaystyle p}$ is radical. Hence, we may write

${\displaystyle p=\bigcap _{i\in I}m_{i}}$,

where the ${\displaystyle m_{i}}$ are maximal. Now suppose that ${\displaystyle x+p}$ is contained within the Jacobson radical of ${\displaystyle R/p}$. According to theorem 13.7, ${\displaystyle (1-xy)+p}$ is a unit within ${\displaystyle R/p}$, where ${\displaystyle y\in R}$ is arbitrary. We want to prove ${\displaystyle x\in p}$. Let thus ${\displaystyle k\in I}$ be such that ${\displaystyle x\notin m_{k}}$. Then ${\displaystyle \langle x\rangle +m_{k}=R}$ and thus ${\displaystyle 1=xy+s}$ with ${\displaystyle y\in R}$ and ${\displaystyle s\in m_{k}}$, that is ${\displaystyle s=1-xy}$. Let ${\displaystyle a+p}$ be the inverse of ${\displaystyle s+p}$, that is ${\displaystyle as-1\in p}$. This means ${\displaystyle as-1\in m_{i}}$ for all ${\displaystyle i\in I}$, and in particular, ${\displaystyle as-1\in m_{k}}$. Hence ${\displaystyle 1\in m_{k}}$, contradiction.

3. ${\displaystyle \Rightarrow }$ 4.: Let ${\displaystyle I\leq R}$. Assume there exists ${\displaystyle x+I\in R/I}$ and a prime ideal ${\displaystyle q\leq R/I}$ such that ${\displaystyle x\notin q}$, but ${\displaystyle x\in m}$ for all maximal ${\displaystyle m\leq R/I}$. Let ${\displaystyle \pi _{I}:R\to R/I}$ be the canonical projection. Since preimages of prime ideals under homomorphism are prime, ${\displaystyle p:=\pi _{I}^{-1}(q)}$ is prime.

Let ${\displaystyle m'}$ be a maximal ideal within ${\displaystyle R/p}$. Assume ${\displaystyle x+p\notin m'}$. Let ${\displaystyle \pi _{p}:R\to R/p}$ be the canonical projection. As in the first proof of theorem 12.2, ${\displaystyle J:=\pi _{p}^{-1}(m')}$ is maximal.

We claim that ${\displaystyle K:=\pi _{I}(J)}$ is maximal. Assume ${\displaystyle 1+I\in K}$, that is ${\displaystyle i-1\in J}$ for a suitable ${\displaystyle i\in I}$. Since ${\displaystyle I\subseteq p\subseteq J}$, ${\displaystyle 1\in J}$, contradiction. Assume ${\displaystyle K}$ is strictly contained within ${\displaystyle L\leq R/I}$. Let ${\displaystyle x+I\in L\setminus K}$. Then ${\displaystyle x\in \pi _{I}^{-1}(L)}$. If ${\displaystyle x\in \pi _{I}^{-1}(K)}$, then ${\displaystyle x+I\in K}$, contradiction. Hence ${\displaystyle \pi _{I}^{-1}(L)\supsetneq \pi _{I}^{-1}(K)=J}$ and thus ${\displaystyle 1\in \pi _{I}^{-1}(L)}$, that is ${\displaystyle 1+I\in L}$.

Furthermore, if ${\displaystyle x+I\in K}$, then ${\displaystyle x\in \pi _{I}^{-1}(\pi _{I}(J))}$. Now ${\displaystyle \pi _{I}^{-1}(\pi _{I}(J))=I+J=J}$ since ${\displaystyle I\subseteq J}$. Hence, ${\displaystyle x\in J}$, that is, ${\displaystyle \pi _{p}(x)\in M'}$, a contradiction to ${\displaystyle x+p\notin m'}$.

Thus, ${\displaystyle x}$ is contained within the Jacobson radical of ${\displaystyle R/p}$.

4. ${\displaystyle \Rightarrow }$ 1.: Assume ${\displaystyle q\leq R}$ is prime not the intersection of maximal ideals. Then

${\displaystyle q\subsetneq \bigcap _{q\subseteq m\leq R \atop m{\text{ maximal}}}m}$.

Hence, there exists an ${\displaystyle x\in R}$ such that ${\displaystyle q\subseteq m\Rightarrow x\in m\setminus q}$ for every maximal ideal ${\displaystyle m}$ of ${\displaystyle R}$.

The set ${\displaystyle \{(x+q)^{n}|n\in \mathbb {N} _{0}\}}$ is multiplicatively closed. Thus, theorem 12.3 gives us a prime ideal ${\displaystyle p\leq R/q}$ such that ${\displaystyle x\notin p}$.

Let ${\displaystyle m}$ be a maximal ideal of ${\displaystyle R/q}$ that does not contain ${\displaystyle x}$. Let ${\displaystyle \pi :R\to R/q}$ be the canonical projection. We claim that ${\displaystyle \pi ^{-1}(m)}$ is a maximal ideal containing ${\displaystyle p}$. Indeed, the proof runs as in the first proof of theorem 12.2. Furthermore, ${\displaystyle \pi ^{-1}(m)}$ does not contain ${\displaystyle x}$, for if it did, then ${\displaystyle \pi (x)=x+p\in m}$. Thus we obtained a contradiction, which is why every maximal ideal of ${\displaystyle R/q}$ contains ${\displaystyle x}$.

Since within ${\displaystyle R/q}$, the Jacobson radical equals the Nilradical, ${\displaystyle x}$ is also contained within all prime ideals of ${\displaystyle R/q}$, in particular within ${\displaystyle p}$. Thus we have obtained a contradiction.${\displaystyle \Box }$

Proof 2: We prove 1. ${\displaystyle \Rightarrow }$ 4. ${\displaystyle \Rightarrow }$ 3. ${\displaystyle \Rightarrow }$ 2. ${\displaystyle \Rightarrow }$ 1.

1. ${\displaystyle \Rightarrow }$ 4.: Due to lemma 3.10, ${\displaystyle R/I}$ is a Jacobson ring. Hence, it follows from the representations of theorem 13.3 and def. 13.6, that Nilradical and Jacobson radical of ${\displaystyle R/I}$ are equal.

4. ${\displaystyle \Rightarrow }$ 3.: Since ${\displaystyle p}$ is a radical ideal (since it is even a prime ideal), ${\displaystyle R/p}$ has no nilpotent elements and thus it's nilradical vanishes. Since the Jacobson radical of that ring equals the Nilradical due to the hypothesis, we obtain that the Jacobson radical vanishes as well.

3. ${\displaystyle \Rightarrow }$ 2.: I found no shorter path than to combine 3. ${\displaystyle \Rightarrow }$ 1. with 1. ${\displaystyle \Rightarrow }$ 2.

2. ${\displaystyle \Rightarrow }$ 1.: Every prime ideal is radical.${\displaystyle \Box }$

Remaining arrows:

1. ${\displaystyle \Rightarrow }$ 3.: Let ${\displaystyle p}$ be a prime ideal of ${\displaystyle R}$. Now suppose that ${\displaystyle x+p}$ is contained within the Jacobson radical of ${\displaystyle R/p}$. According to theorem 13.7, ${\displaystyle (1-xy)+p}$ is a unit within ${\displaystyle R/p}$, where ${\displaystyle y\in R}$ is arbitrary. Write

${\displaystyle p=\bigcap _{i\in I}m_{i}}$,

where the ${\displaystyle m_{i}}$ are maximal. We want to prove ${\displaystyle x\in p}$. Let thus ${\displaystyle k\in I}$ be such that ${\displaystyle x\notin m_{k}}$. Then ${\displaystyle \langle x\rangle +m_{k}=R}$ and thus ${\displaystyle 1=xy+s}$ with ${\displaystyle y\in R}$ and ${\displaystyle s\in m_{k}}$, that is ${\displaystyle s=1-xy}$. Let ${\displaystyle a+p}$ be the inverse of ${\displaystyle s+p}$, that is ${\displaystyle as-1\in p}$. This means ${\displaystyle as-1\in m_{i}}$ for all ${\displaystyle i\in I}$, and in particular, ${\displaystyle as-1\in m_{k}}$. Hence ${\displaystyle 1\in m_{k}}$, contradiction.

3. ${\displaystyle \Rightarrow }$ 1.: Let ${\displaystyle p\leq R}$ be prime. If ${\displaystyle p}$ is maximal, there is nothing to show. If ${\displaystyle p}$ is not maximal, ${\displaystyle R/p}$ is not a field. In this case, there exists a non-unit within ${\displaystyle R/p}$, and hence, by theorem 12.1 or 12.2 (applied to ${\displaystyle I=(a)}$ where ${\displaystyle a}$ is a non-unit), ${\displaystyle R/p}$ contains at least one maximal ideal. Furthermore, the Jacobson radical of ${\displaystyle R/p}$ is trivial, which is why there are some maximal ideals ${\displaystyle m_{i},i\in I}$ of ${\displaystyle R/p}$ such that

${\displaystyle \bigcap _{i\in I}m_{i}=\emptyset }$.

As in the first proof of theorem 12.2, ${\displaystyle K_{i}:=\pi ^{-1}(m_{i})}$ are maximal ideals of ${\displaystyle R}$. Furthermore,

${\displaystyle p=\bigcap _{i\in I}K_{i}}$.

2. ${\displaystyle \Rightarrow }$ 4.: Let ${\displaystyle {\mathcal {N}}_{I}}$ be the nilradical of ${\displaystyle R/I}$. We claim that

${\displaystyle K:=\pi _{I}^{-1}({\mathcal {N}}_{I})=r(I)}$.

Let first ${\displaystyle k\in K}$, that is, ${\displaystyle k+I\in {\mathcal {N}}_{I}}$. Then ${\displaystyle k^{n}+I=0+I}$, that is ${\displaystyle k^{n}\in I}$ and ${\displaystyle k\in r(I)}$. The other inclusion follows similarly, only the order is in reverse (in fact, we just did equivalences).

Due to the assumption, we may write

${\displaystyle r(I)=\bigcap _{\alpha \in A}m_{\alpha }}$,

where the ${\displaystyle m_{\alpha }}$ are maximal ideals of ${\displaystyle R}$.

Since ${\displaystyle \pi _{I}}$ is surjective, ${\displaystyle \pi _{I}(\pi _{I}^{-1}({\mathcal {N}}_{I}))={\mathcal {N}}_{I}}$. Hence,

${\displaystyle {\mathcal {N}}_{I}=\pi _{I}(r(I))=\pi _{I}\left(\bigcap _{\alpha \in A}m_{\alpha }\right)=\bigcap _{\alpha \in A}\pi _{I}(m_{\alpha })}$,

where the last equality follows from ${\displaystyle \forall \alpha \in A:y+I\in \pi _{I}(m_{\alpha })}$ implying that ${\displaystyle y=x_{\alpha }+i_{\alpha }}$ for ${\displaystyle i_{\alpha }\in I\subseteq m_{\alpha }}$ and ${\displaystyle x_{\alpha }\in m_{\alpha }}$ and hence ${\displaystyle y\in m_{\alpha }}$ for all ${\displaystyle \alpha }$. Furthermore, the ${\displaystyle \pi _{I}(m_{\alpha })}$ are either maximal or equal to ${\displaystyle R/I}$, since any ideal ${\displaystyle J}$ of ${\displaystyle R/I}$ properly containing ${\displaystyle \pi _{I}(m_{\alpha })}$ contains one element ${\displaystyle y+I}$ not contained within ${\displaystyle \pi _{I}(m_{\alpha })}$, which is why ${\displaystyle y\notin \pi _{I}^{-1}(\pi _{I}(m_{\alpha }))=m_{\alpha }}$, hence ${\displaystyle \pi _{I}^{-1}(J)=R}$ and thus ${\displaystyle J=\pi _{I}(\pi _{I}^{-1}(J))=R/I}$.

Thus, ${\displaystyle {\mathcal {N}}_{I}}$ is the intersection of some maximal ideals of ${\displaystyle R/I}$, and thus the Jacobson radical of ${\displaystyle R/I}$ is contained within it. Since the other inclusion holds in general, we are done.

4. ${\displaystyle \Rightarrow }$ 2.: As before, we have

${\displaystyle \pi _{I}^{-1}({\mathcal {N}}_{I})=r(I)}$.

Let now ${\displaystyle {\mathcal {J}}_{I}}$ be the Jacobson radical of ${\displaystyle R/I}$, that is,

${\displaystyle {\mathcal {J}}_{I}=\bigcap _{\alpha \in A}m_{\alpha }}$,

where the ${\displaystyle m_{\alpha }}$ are the maximal ideals of ${\displaystyle R/I}$. Then we have by the assumption:

${\displaystyle \bigcap _{\alpha \in A}\pi _{I}^{-1}(m_{\alpha })=\pi _{I}^{-1}\left({\mathcal {J}}_{I}\right)=\pi _{I}^{-1}\left({\mathcal {N}}_{I}\right)=r(I)}$.

Furthermore, as in the first proof of theorem 12.2, ${\displaystyle \pi _{I}^{-1}(m_{\alpha })}$ are maximal.

## Goldman's criteria

Now we shall prove two more characterisations of being a Jacobson ring. These were established by Oscar Goldman.

Theorem 14.4 (Goldman's first criterion):

Let ${\displaystyle R}$ be a ring. ${\displaystyle R}$ is Jacobson if and only if ${\displaystyle R[x]}$ is.

This is the hard one, and we do it right away so that we have it done.

Proof:

One direction (${\displaystyle \Leftarrow }$) isn't too horrible. Let ${\displaystyle R[x]}$ be a Jacobson ring, and let ${\displaystyle p_{0}\leq R}$ be a prime ideal of ${\displaystyle R}$. (We shall denote ideals of ${\displaystyle R}$ with a small zero as opposed to ideals of ${\displaystyle R[x]}$ to avoid confusion.)

We now define

${\displaystyle p:=p_{0}R[x]+xR[x]}$.

This ideal contains exactly the polynomials whose constant term is in ${\displaystyle p_{0}}$. It is prime since

${\displaystyle fg\in p\Rightarrow f\in p\vee g\in p}$

as can be seen by comparing the constant coefficients. Since ${\displaystyle R[x]}$ is Jacobson, for a given ${\displaystyle a}$ that is not contained within ${\displaystyle p_{0}}$, and hence not in ${\displaystyle p}$, there exists a maximal ideal ${\displaystyle m}$ containing ${\displaystyle p}$, but not containing ${\displaystyle a}$. Set ${\displaystyle m_{0}:=m\cap R}$. We claim that ${\displaystyle m_{0}}$ is maximal. Indeed, we have an isomorphism

${\displaystyle R[x]/m\cong R/m_{0}}$

via

${\displaystyle a_{n}x^{n}+\cdots +a_{1}x+a_{0}+m\mapsto a_{0}+m_{0}}$.

Therefore, ${\displaystyle R[x]/m}$ is a field if and only if ${\displaystyle R/m_{0}}$ is. Hence, ${\displaystyle m_{0}}$ is maximal, and it does not contain ${\displaystyle a}$. Since thus every element outside ${\displaystyle p_{0}}$ can be separated from ${\displaystyle p_{0}}$ by a maximal ideal, ${\displaystyle R}$ is a Jacobson ring.

The other direction ${\displaystyle \Rightarrow }$ is a bit longer.

We have given ${\displaystyle R}$ a Jacobson ring and want to prove ${\displaystyle R[x]}$ Jacobson. Hence, let ${\displaystyle p\leq R[x]}$ be a prime ideal, and we want to show it to be the intersection of maximal ideals.

We first treat the case where ${\displaystyle p\cap R=\{0\}}$ and ${\displaystyle R}$ is an integral domain.

Assume first that ${\displaystyle p}$ does contain a nonzero element (i.e. is not equal the zero ideal).

Assume ${\displaystyle g\in R[x]}$ is contained within all maximal ideals containing ${\displaystyle p}$, but not within ${\displaystyle p}$. Let ${\displaystyle f\in p}$ such that ${\displaystyle f}$ is of lowest degree among all nonzero polynomials in ${\displaystyle p}$. Since ${\displaystyle p\cap R=\{0\}}$, ${\displaystyle \deg f\geq 1}$. Since ${\displaystyle R}$ is an integral domain, we can form the quotient field ${\displaystyle K=\operatorname {Quot} R}$. Then ${\displaystyle R[x]\subseteq K[x]}$.

Assume that ${\displaystyle f}$ is not irreducible in ${\displaystyle K[x]}$. Then ${\displaystyle f=f_{1}f_{2}}$, ${\displaystyle f_{1},f_{2}\in K[x]}$, where ${\displaystyle f_{1}}$, ${\displaystyle f_{2}}$ are not associated to ${\displaystyle f}$. Let ${\displaystyle \alpha ,\beta ,\gamma }$ such that ${\displaystyle \alpha f,\beta f_{1},\gamma f_{2}\in R[x]}$. Then ${\displaystyle \alpha \beta \gamma f=\alpha (\beta f_{1})(\gamma f_{2})}$. As ${\displaystyle p}$ is prime, wlog. ${\displaystyle \alpha \beta f_{1}\in p}$. Hence ${\displaystyle \deg f_{1}=\deg f}$. Thus, ${\displaystyle f}$ and ${\displaystyle f_{1}}$ are associated, contradiction.

${\displaystyle K[x]}$ is Euclidean with the degree as absolute value. Uniqueness of prime factorisation gives a definition of the greatest common divisor. Since ${\displaystyle f}$ is irreducible in ${\displaystyle K[x]}$ and ${\displaystyle g\notin p}$, ${\displaystyle \gcd(f,g)=1}$. Applying the Euclidean algorithm, ${\displaystyle 1=fh_{1}+gh_{2}}$, ${\displaystyle h_{1},h_{2}\in K[x]}$. Multiplication by an appropriate constant ${\displaystyle b}$ yields ${\displaystyle b=fbh_{1}+gbh_{2}}$, ${\displaystyle bh_{1},bh_{2}\in R[x]}$. Thus, ${\displaystyle b\in p+gR[x]}$. Hence, ${\displaystyle b}$ is contained within every maximal ideal containing ${\displaystyle p}$. Further, ${\displaystyle p\cap R=\{0\}\Rightarrow b\notin p}$.

Let ${\displaystyle m_{0}\leq R}$ be any maximal ideal of ${\displaystyle R}$ not containing ${\displaystyle a}$. Set

${\displaystyle I:=m_{0}R[x]+p}$.

Assume ${\displaystyle I=R[x]}$. Then ${\displaystyle 1=u(x)+v(x)}$, ${\displaystyle u\in m_{0}R[x],v\in p}$. We divide ${\displaystyle v}$ by ${\displaystyle f}$ by applying a polynomial long division algorithm working for elements of a general polynomial ring: We successively eliminate the first coefficient of ${\displaystyle v}$ by subtracting an appropriate multiple of ${\displaystyle f}$. Should that not be possible, we multiply ${\displaystyle v}$ by the leading coefficient of ${\displaystyle f}$, that shall be denoted by ${\displaystyle a}$. Then we cannot eliminate the desired coefficient of ${\displaystyle v}$, but we can eliminate the desired coefficient of ${\displaystyle av}$. Repeating this process gives us

${\displaystyle a^{n}v(x)=f(x)h(x)+i(x)}$, ${\displaystyle \deg i<\deg f}$

for ${\displaystyle h,i\in R[x]}$. Furthermore, since this equation implies ${\displaystyle i\in p}$, we must have ${\displaystyle i=0}$ since the degree of ${\displaystyle f}$ was minimal among polynomials in ${\displaystyle p}$. Then

{\displaystyle {\begin{aligned}a^{n}&=a^{n}v(x)+a^{n}u(x)\\&=f(x)h(x)+a^{n}u(x)\\&=f(x)h(x)+r(x)\end{aligned}}}

with ${\displaystyle r(x):=a^{n}u(x)\in m_{0}R[x]}$. By moving such coefficients to ${\displaystyle r(x)}$, we may assume that no coefficient of ${\displaystyle h}$ is in ${\displaystyle m_{0}}$. Further, ${\displaystyle h}$ is nonzero since otherwise ${\displaystyle a^{n}\in m_{0}\Rightarrow a\in m_{0}}$. Denote the highest coefficient of ${\displaystyle h}$ by ${\displaystyle \delta }$, and the highest coefficient of ${\displaystyle r}$ by ${\displaystyle \epsilon }$. Since the highest coefficients of ${\displaystyle fh}$ and ${\displaystyle r}$ must cancel out (as ${\displaystyle \deg f\geq 1}$),

${\displaystyle a\delta =-\epsilon }$.

Thus, ${\displaystyle a\notin m_{0}}$ and ${\displaystyle \delta \notin m_{0}}$, but ${\displaystyle -\epsilon \in m_{0}}$, which is absurd as every maximal ideal is prime. Hence, ${\displaystyle I\subsetneq R[x]}$.

According to theorem 12.2, there exists a maximal ideal ${\displaystyle m\leq R[x]}$ containing ${\displaystyle I}$. Now ${\displaystyle m\cap R}$ does not equal all of ${\displaystyle R}$, since otherwise ${\displaystyle m=R[x]}$. Hence, ${\displaystyle m_{0}\subseteq m\cap R}$ and the maximality of ${\displaystyle m_{0}}$ imply ${\displaystyle m\cap R=m_{0}}$. Further, ${\displaystyle m}$ is a maximal ideal containing ${\displaystyle p}$ and thus contains ${\displaystyle b}$. Hence, ${\displaystyle b\in m_{0}}$.

Thus, every maximal ideal ${\displaystyle m_{0}}$ that does not contain ${\displaystyle a}$ contains ${\displaystyle b}$; that is, ${\displaystyle ab\in m_{0}}$ for all maximal ideals ${\displaystyle m_{0}}$ of ${\displaystyle R}$. But according to theorem 12.3, we may choose a prime ideal ${\displaystyle p_{0}}$ of ${\displaystyle R}$ not intersecting the (multiplicatively closed) set ${\displaystyle \{(ab)^{n}|n\in \mathbb {N} \}}$, and since ${\displaystyle R}$ is a Jacobson ring, there exists a maximal ideal ${\displaystyle m_{0}}$ containing ${\displaystyle p_{0}}$ and not containing ${\displaystyle ab}$. This is a contradiction.

Let now ${\displaystyle p\leq R[x]}$ be the zero ideal (which is prime within an integral domain). Assume that there are only finitely many elements in ${\displaystyle R[x]}$ which are irreducible in ${\displaystyle K[x]}$, and call them ${\displaystyle f_{1},\ldots ,f_{n}}$. The element

${\displaystyle f_{1}(x)\cdots f_{n}(x)+1}$

factors into irreducible elements, but at the same time is not divisible by any of ${\displaystyle f_{1},\ldots ,f_{n}}$, since otherwise wlog.

${\displaystyle f_{1}(x)\cdots f_{n}(x)+1=f_{1}(x)\cdot s(x)\Leftrightarrow 1=f_{1}(x)(s(x)-f_{2}(x)\cdots f_{n}(x))}$,

which is absurd. Thus, there exists at least one further irreducible element not listed in ${\displaystyle f_{1},\ldots ,f_{n}}$, and multiplying this by an appropriate constant yields a further element of ${\displaystyle R[x]}$ irreducible in ${\displaystyle K[x]}$.

Let ${\displaystyle f\in R[x]}$ be irreducible in ${\displaystyle K[x]}$. We form the ideal ${\displaystyle \langle f\rangle \leq K[x]}$ and define ${\displaystyle I_{f}:=R[x]\cap \langle f\rangle }$. We claim that ${\displaystyle I_{f}}$ is prime. Indeed, if ${\displaystyle a(x)b(x)\in \langle f\rangle }$, then ${\displaystyle a}$ and ${\displaystyle b}$ factor in ${\displaystyle K[x]}$ into irreducible components. Since ${\displaystyle K[x]}$ is a unique factorisation domain, ${\displaystyle f}$ occurs in at least one of those two factorisations.

Assume there is a nonzero element ${\displaystyle w(x)}$ contained within all the ${\displaystyle I_{f}}$, where ${\displaystyle f}$ is irreducible over ${\displaystyle K[x]}$. ${\displaystyle w}$ factors in ${\displaystyle K[x]}$ uniquely into finitely many irreducible components, leading to a contradiction to the infinitude of irreducible elements of ${\displaystyle K[x]}$. Hence,

${\displaystyle \bigcap _{f\in K[x] \atop f{\text{ irreducible}}}I_{f}=\{0\}}$,

where each ${\displaystyle I_{f}}$ is prime and ${\displaystyle I_{f}\cap R=\{0\}}$. Hence, by the previous case, each ${\displaystyle I_{f}}$ can be written as the intersection of maximal elements, and thus, so can ${\displaystyle p=\{0\}}$.

Now for the general case where ${\displaystyle R}$ is an arbitrary Jacobson ring and ${\displaystyle p\leq R[x]}$ is a general prime ideal of ${\displaystyle R[x]}$. Set ${\displaystyle p_{0}:=p\cap R}$. ${\displaystyle p_{0}}$ is a prime ideal, since if ${\displaystyle ab\in p_{0}}$, where ${\displaystyle a,b\in R}$, then ${\displaystyle a\in p}$ or ${\displaystyle b\in p}$, and hence ${\displaystyle a\in p_{0}}$ or ${\displaystyle b\in p_{0}}$. We further set ${\displaystyle q:=p_{0}R[x]}$. Then we have

${\displaystyle R[x]/q\cong (R/p_{0})[x]}$

via the isomorphism

${\displaystyle \varphi :a_{n}x^{n}+\cdots +a_{1}x+a_{0}+q\mapsto (a_{n}+p_{0})x^{n}+\cdots +(a_{1}+p_{0})x+(a_{0}+p_{0})}$.

Set

${\displaystyle R':=R/p_{0}}$ and ${\displaystyle p':=\varphi (\pi _{q}(p))}$.

Then ${\displaystyle R'}$ is an integral domain and a Jacobson ring (lemma 14.2), and ${\displaystyle p'}$ is a prime ideal of ${\displaystyle R'[x]}$ with the property that ${\displaystyle p'\cap R'=\{0\}}$. Hence, by the previous case,

${\displaystyle p'=\bigcap _{p'\subseteq m'\leq R' \atop m'{\text{ max.}}}m'}$.

Thus, since ${\displaystyle q\subseteq p}$,

${\displaystyle p=\pi _{q}^{-1}(\varphi ^{-1}(p'))=(\varphi \circ \pi _{q})^{-1}\left(\bigcap _{p'\subseteq m'\leq R' \atop m'{\text{ max.}}}m'\right)=\bigcap _{p'\subseteq m'\leq R' \atop m'{\text{ max.}}}(\varphi \circ \pi _{q})^{-1}(m')}$,

which is an intersection of maximal ideals due to lemma 12.4 and since isomorphisms preserve maximal ideals.${\displaystyle \Box }$

Theorem 14.5 (Goldman's second criterion):

A ring ${\displaystyle R}$ is Jacobson if and only if for every maximal ideal ${\displaystyle m\in R[x]}$, ${\displaystyle m_{0}:=m\cap R}$ is maximal in ${\displaystyle R}$.

Proof:

The reverse direction ${\displaystyle \Leftarrow }$ is once again easier.

Let ${\displaystyle p_{0}\leq R}$ be a prime ideal within ${\displaystyle R}$, and let ${\displaystyle a\notin p_{0}}$. Set

${\displaystyle I:=p_{0}R[x]+(ax-1)R[x]}$.

Assume ${\displaystyle I=R[x]}$. Then there exist ${\displaystyle f\in p_{0}R[x]}$, ${\displaystyle g\in R[x]}$ such that

${\displaystyle 1=f(x)+(ax-1)g(x)}$.

By shifting parts of ${\displaystyle g}$ to ${\displaystyle f}$, one may assume that ${\displaystyle g}$ does not have any coefficients contained within ${\displaystyle p_{0}}$. Furthermore, if ${\displaystyle g=0}$ follows ${\displaystyle 1\in p_{0}R[x]}$. Further, ${\displaystyle p_{0}R[x]\cap R=p_{0}}$, since if ${\displaystyle ch\in R}$, ${\displaystyle c\in p_{0}}$, ${\displaystyle h\in R[x]}$, then ${\displaystyle c}$ annihilates all higher coefficients of ${\displaystyle h}$, which is why ${\displaystyle ch}$ equals the constant term of ${\displaystyle h}$ times ${\displaystyle c}$ and thus ${\displaystyle ch\in p_{0}}$. Hence ${\displaystyle g\neq 0}$ and let ${\displaystyle b}$ be the leading coefficient of ${\displaystyle g}$. Since the nontrivial coefficients of the polynomial ${\displaystyle f(x)+(ax-1)g(x)}$ must be zero for it being constantly one, ${\displaystyle ab\in p_{0}}$, contradicting the primality of ${\displaystyle p_{0}}$.

Thus, let ${\displaystyle m\leq R[x]}$ be maximal containing ${\displaystyle I}$. Assume ${\displaystyle m}$ contains ${\displaystyle a}$. Then ${\displaystyle ax-(ax-1)=1\in m}$ and thus ${\displaystyle m=R[x]}$. ${\displaystyle m}$ contracts to a maximal ideal ${\displaystyle m_{0}}$ of ${\displaystyle R}$, which does not contain ${\displaystyle a}$, but does contain ${\displaystyle p_{0}}$. Hence the claim.

The other direction is more tricky, but not as bad as in the previous theorem.

Let thus ${\displaystyle R}$ be a Jacobson ring. Assume there exists a maximal ideal ${\displaystyle m\leq R[x]}$ such that ${\displaystyle R\cap m}$ is not maximal within ${\displaystyle R}$. Define

${\displaystyle p_{0}:=m\cap R}$ and ${\displaystyle p:=p_{0}R[x]}$. ${\displaystyle p_{0}}$ is a prime ideal, since if ${\displaystyle a,b\in R}$ such that ${\displaystyle ab\in R}$, ${\displaystyle a\in m}$ or ${\displaystyle b\in m}$ and hence ${\displaystyle a\in p_{0}}$ or ${\displaystyle b\in p_{0}}$. Further
${\displaystyle R[x]/p\cong (R/p_{0})[x]}$

via the isomorphism

${\displaystyle \varphi :a_{n}x^{n}+\cdots +a_{1}x+a_{0}+p\mapsto (a_{n}+p_{0})x^{n}+\cdots +(a_{1}+p_{0})x+(a_{0}+p_{0})}$.

According to lemma 12.5, ${\displaystyle \pi _{p}(m)}$ is a maximal ideal within ${\displaystyle R[x]/p}$. We set

${\displaystyle R':=R/p_{0}}$ and ${\displaystyle m':=\varphi (\pi _{p}(m))}$.

Then ${\displaystyle R'}$ is a Jacobson ring that is not a field, ${\displaystyle m'}$ is a maximal ideal within ${\displaystyle R'}$ (isomorphisms preserve maximal ideals) and ${\displaystyle m'\cap R'=\{0\}}$, since if ${\displaystyle w\in R[x]}$ is any element of ${\displaystyle m}$ which is not mapped to zero by ${\displaystyle \pi _{p}}$, then at least one of ${\displaystyle a_{n}+p_{0},\ldots ,a_{1}+p_{0}}$ must be nonzero, for, if only ${\displaystyle a_{0}\notin p_{0}}$, then ${\displaystyle a_{0}\in (m\cap R)\setminus p_{0}}$, which is absurd.

Replacing ${\displaystyle R}$ by ${\displaystyle R'}$ and ${\displaystyle m}$ by ${\displaystyle m'}$, we lead the assumption to a contradiction where ${\displaystyle R}$ is an integral domain but not a field and ${\displaystyle m\cap R=\{0\}}$.

${\displaystyle m}$ is nonzero, because else ${\displaystyle R[x]}$ would be a field. Let ${\displaystyle f\neq 0}$ have minimal degree among the nonzero polynomials of ${\displaystyle m}$, and let ${\displaystyle a\in R}$ be the leading coefficient of ${\displaystyle f}$.

Let ${\displaystyle n_{0}\leq R}$ be an arbitrary maximal ideal of ${\displaystyle R}$. ${\displaystyle n_{0}}$ can not be the zero ideal, for otherwise ${\displaystyle R}$ would be a field. Hence, let ${\displaystyle b\in n_{0}}$ be nonzero. Since ${\displaystyle m\cap R=\{0\}}$, ${\displaystyle b\notin m}$. Since ${\displaystyle m}$ is maximal, ${\displaystyle m+\langle b\rangle =R[x]}$. Hence, ${\displaystyle 1=g(x)+bh(x)}$, where ${\displaystyle g\in m}$ and ${\displaystyle h\in R[x]}$. Applying the general division algorithm that was described above in order to divide ${\displaystyle g}$ by ${\displaystyle f}$ and obtain

${\displaystyle a^{n}h(x)=s(x)f(x)+r(x)}$

for suitable ${\displaystyle n\in \mathbb {N} }$ and ${\displaystyle r,s\in R[x]}$ such that ${\displaystyle \deg r<\deg f}$. From the equality holding for ${\displaystyle h}$ we get

${\displaystyle a^{n}bh(x)=a^{n}(1-g(x))=bs(x)f(x)+br(x)\Leftrightarrow a^{n}-br(x)=bs(x)f(x)+a^{n}g(x)}$.

Hence, ${\displaystyle a^{n}-br(x)\in m}$, and since the degree of ${\displaystyle f}$ was minimal in ${\displaystyle m}$, ${\displaystyle a^{n}-br(x)=0}$. Since all coefficients of ${\displaystyle br(x)}$ are contained within ${\displaystyle n_{0}}$ (since they are multiplied by ${\displaystyle b}$), ${\displaystyle a^{n}\in n_{0}}$. Thus ${\displaystyle a\in n_{0}}$ (maximal ideals are prime).

Hence, ${\displaystyle a}$ is contained in all maximal ideals of ${\displaystyle R}$. But since ${\displaystyle R}$ was assumed to be an integral domain, this is impossible in view of lemma 12.3 applied to the set ${\displaystyle S=\{a^{n}|n\in \mathbb {N} _{0}\}}$, yielding a prime ideal ${\displaystyle p_{0}\leq R}$ which is separated from ${\displaystyle a}$ by a maximal ideal since ${\displaystyle R}$ is a Jacobson ring. Hence, we have obtained a contradiction.${\displaystyle \Box }$