Ideals and quotients

Define \[ \phi:R/I\longrightarrow R/J, \qquad r+I\longmapsto r+J. \] Because \(I\subseteq J\), this map is well-defined and surjective. Its kernel is exactly \(J/I\). The first isomorphism theorem therefore gives the displayed isomorphism.

An ideal is prime exactly when its quotient ring is an integral domain. By <1>1, \[ R/J\cong (R/I)/(J/I). \] Hence one quotient is a domain exactly when the other is.

An ideal is maximal exactly when its quotient ring is a field. Again <1>1 identifies the two quotient rings, so the two maximality conditions are equivalent.

Suppose first that \(J\) is radical. If \[ (r+I)^n\in J/I, \] then \(r^n+I\in J/I\), which means \(r^n\in J\). Radicality of \(J\) gives \(r\in J\), hence \(r+I\in J/I\). Thus \(J/I\) is radical.

Conversely, suppose \(J/I\) is radical and \(r^n\in J\). Then \[ (r+I)^n=r^n+I\in J/I. \] Hence \(r+I\in J/I\), so \(r\in J\). Therefore \(J\) is radical.

Combine <1>2–<1>4 with the ordinary ideal correspondence for the quotient map \(R\to R/I\).

Let \(a\in R\) be nilpotent and nonzero. Choose the least \(m\ge1\) such that \(a^m=0\). Since \(a\neq0\), one has \(m\ge2\), and minimality gives \(a^{m-1}\neq0\). Thus \[ a\,a^{m-1}=0 \] with both factors nonzero, so \(a\) is a zero divisor.

Choose \(N\) with \(n^N=0\). Then \[ (1+n)\sum_{j=0}^{N-1}(-n)^j=1, \] so \(1+n\) is a unit.

Now let \(u\in \unitsof{R}\) and \(n\in N(R)\). Since \(R\) is commutative, \(u^{-1}n\) is nilpotent, and \[ u+n=u(1+u^{-1}n) \] is a product of units. Thus \(N(R)+\unitsof{R}\subseteq \unitsof{R}\). The reverse inclusion follows from \(0\in N(R)\).

Suppose first that \(a_0\in \unitsof{R}\) and every \(a_k\) for \(k\ge1\) is nilpotent. Put \[ h(x)=\sum_{k=1}^d a_kx^k. \] Each summand \(a_kx^k\) is nilpotent. A finite sum of nilpotent elements in a commutative ring is nilpotent: if \(t_i^{e_i}=0\), then every monomial in \((t_1+\cdots+t_r)^{e_1+\cdots+e_r}\) contains some \(t_i^{e_i}\) as a factor. Hence \(h\) is nilpotent, and so is \(a_0^{-1}h\). By <1>2, \[ f=a_0(1+a_0^{-1}h) \] is a unit.

Conversely, suppose \(fg=1\) in \(R[x]\). Comparing constant terms gives \(a_0b_0=1\), so \(a_0\) is a unit. Let \(\mathfrak p\) be any prime ideal of \(R\). Reducing modulo \(\mathfrak p\) gives a unit \(\overline f\) in the domain polynomial ring \((R/\mathfrak p)[x]\). Its only units are nonzero constants, so \(a_k\in\mathfrak p\) for every \(k\ge1\). Since this holds for every prime ideal, each such \(a_k\) lies in the nilradical and is nilpotent.

If every coefficient \(a_k\) is nilpotent, then each term \(a_kx^k\) is nilpotent, so their finite sum \(f\) is nilpotent by the argument in <1>3.

Conversely, suppose \(f^N=0\). For every prime ideal \(\mathfrak p\subset R\), reduction modulo \(\mathfrak p\) gives \[ \overline f^{\,N}=0 \] in the domain \((R/\mathfrak p)[x]\). Hence \(\overline f=0\), so every coefficient \(a_k\) lies in every prime ideal. Thus every \(a_k\) lies in the nilradical and is nilpotent.

If \(0\neq r\in R\) and \(rf=0\), then the nonzero constant polynomial \(r\) annihilates the nonzero polynomial \(f\), so \(f\) is a zero divisor.

Conversely, suppose \(f\) is a zero divisor. Choose a nonzero polynomial \[ g=b_0+b_1x+\cdots+b_mx^m \] of minimal degree among all nonzero \(g\) satisfying \(fg=0\). Write \[ f=a_0+a_1x+\cdots+a_nx^n. \] The coefficient of \(x^{n+m}\) in \(fg\) is \(a_nb_m\), so \(a_nb_m=0\). Hence either \(a_ng=0\) or \(\deg(a_ng)<m\). But \[ f(a_ng)=a_n(fg)=0. \] Minimality of \(m\) therefore forces \(a_ng=0\).

Replacing \(f\) by \(f-a_nx^n\) and repeating the same argument gives, by descending induction, \[ a_i g=0 \qquad\text{for every }i. \] Choose any nonzero coefficient \(b_j\) of \(g\). Then \(b_ja_i=0\) for every \(i\), so with \(r=b_j\neq0\) one has \(rf=0\).

If \(r^n=0\) and \(x\in R\), then \((xr)^n=x^nr^n=0\).

Let \(r^n=s^m=0\) and \(N=n+m-1\). In \((r+s)^N=\sum_{k=0}^N\binom Nk r^ks^{N-k}\), each term has \(k\ge n\) or \(N-k\ge m\), so each term is \(0\). Together with \(0\in\nilrad{R}\) and \(-r\in\nilrad{R}\), steps <1>1 and <1>2 show that \(\nilrad{R}\) is an ideal.

Let \(\bar r=r+\nilrad{R}\) with \(\bar r^n=0\). Then \(r^n\in\nilrad{R}\), so \((r^n)^m=0\) for some \(m\), hence \(r^{nm}=0\) and \(r\in\nilrad{R}\), that is, \(\bar r=0\).

If the left ideal \(R(1-j)\) were proper, it would lie in a maximal left ideal \(M\). Since \(j\in J(R)\subseteq M\), we would have \[ 1=(1-j)+j\in M, \] a contradiction. Hence \(R(1-j)=R\), so some \(u\in R\) satisfies \[ u(1-j)=1. \] Then \[ 1-u=-uj\in J(R) \] because \(J(R)\) is a two-sided ideal. Applying the same argument to \(1-u\in J(R)\) shows that \(u=1-(1-u)\) has a left inverse \(v\), so \(vu=1\). If \(a=1-j\), then \(ua=1\) and \[ v=v(ua)=(vu)a=a. \] Thus \(au=1\) as well, and \(a=1-j\) is a unit.

Since \(J(R)\) is a two-sided ideal, \(xr\in J(R)\). Apply <1>1 to \(j=xr\).

We first use the elementary identity: if \(1-ab\) is a unit, then so is \(1-ba\), with inverse \[ (1-ba)^{-1}=1+b(1-ab)^{-1}a. \] Hence the hypothesis also gives that \(1-rx\) is a unit for every \(r\in R\).

Let \(M\) be a maximal left ideal. If \(x\notin M\), then maximality gives \[ M+Rx=R. \] Thus \(1=m+rx\) for some \(m\in M\) and \(r\in R\), so \[ m=1-rx. \] But \(1-rx\) is a unit, impossible for an element of the proper left ideal \(M\). Therefore \(x\in M\) for every maximal left ideal \(M\), hence \(x\in J(R)\).

Multiplication of cosets gives \((a+\operatorname{Nil}(R))^n=a^n+\operatorname{Nil}(R)\) for every \(n\ge1\). Since \(a+\operatorname{Nil}(R)\) is nilpotent, \(a^n+\operatorname{Nil}(R)=\operatorname{Nil}(R)\) for some \(n\ge1\), that is, \(a^n\in\operatorname{Nil}(R)\).

By step <1>1, \(a^n\) is nilpotent, so \((a^n)^m=0\) for some \(m\ge1\), and \(a^{nm}=(a^n)^m\).

By step <1>2, \(a\) is nilpotent, so \(a\in\operatorname{Nil}(R)\). Equivalently, \(R/\operatorname{Nil}(R)\) has no nonzero nilpotent elements.

Let \(x^n=0\) with \(n\ge1\), and let \(\mathfrak p\) be prime. Then \(x^n=0\in\mathfrak p\). If \(x^k\in\mathfrak p\) with \(k\ge2\), then \(x\cdot x^{k-1}\in\mathfrak p\) gives \(x\in\mathfrak p\) or \(x^{k-1}\in\mathfrak p\); induction on \(k\) gives \(x\in\mathfrak p\).

Put \(S=\{f^n:n\ge0\}\); since \(f\) is not nilpotent, \(0\notin S\), so \((0)\in\mathcal F\). The union of a nonempty chain in \(\mathcal F\) is an ideal disjoint from \(S\), hence an upper bound in \(\mathcal F\). Zorn’s lemma gives a maximal element.

Since \(1=f^0\in S\), \(\mathfrak p\neq R\), and \(f=f^1\in S\) gives \(f\notin\mathfrak p\). Let \(a,b\notin\mathfrak p\). By maximality, \(\mathfrak p+(a)\) and \(\mathfrak p+(b)\) meet \(S\): \(f^m\in\mathfrak p+(a)\) and \(f^n\in\mathfrak p+(b)\) for some \(m,n\ge0\). Then \(f^{m+n}\in(\mathfrak p+(a))(\mathfrak p+(b))\subseteq\mathfrak p+(ab)\). If \(ab\in\mathfrak p\), then \(f^{m+n}\in\mathfrak p\cap S=\varnothing\), a contradiction; so \(ab\notin\mathfrak p\).

Step <1>1 gives \(\operatorname{Nil}(R)\subseteq P\). Steps <1>2 and <1>3 show that every non-nilpotent \(f\) lies outside some prime ideal, so \(P\subseteq\operatorname{Nil}(R)\).