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

Proof:

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

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

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

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

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

Theorem 14.3:

Let $R$ be a ring. The following are equivalent:

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

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

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

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

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

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

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

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

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

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

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

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

Thus, $x$ is contained within the Jacobson radical of $R/p$.

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

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

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

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

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

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

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

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

4. $\Rightarrow$ 3.: Since $p$ is a radical ideal (since it is even a prime ideal), $R/p$ 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. $\Rightarrow$ 2.: I found no shorter path than to combine 3. $\Rightarrow$ 1. with 1. $\Rightarrow$ 2.

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

Remaining arrows:

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

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

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

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

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

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

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

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

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

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

Due to the assumption, we may write

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

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

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

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

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

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

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

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

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

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

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

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

Furthermore, as in the first proof of theorem 12.2, $\pi _{I}^{-1}(m_{\alpha })$ 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 $R$ be a ring. $R$ is Jacobson if and only if $R[x]$ is.

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

Proof:

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

We now define

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

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

$fg\in p\Rightarrow f\in p\vee g\in p$

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

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

via

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

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

The other direction $\Rightarrow$ is a bit longer.

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

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

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

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

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

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

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

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

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

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

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

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

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

$a\delta =-\epsilon$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

via the isomorphism

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

Set

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

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

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

Thus, since $q\subseteq p$,

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

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

Theorem 14.5 (Goldman's second criterion):

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

Proof:

The reverse direction $\Leftarrow$ is once again easier.

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

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

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

$1=f(x)+(ax-1)g(x)$.

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

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

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

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

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

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

via the isomorphism

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

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

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

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

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

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

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

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

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

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

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

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