Assume throughout that \(R\) is a commutative ring with identity. Let \(N(R)\) be the set of nilpotent elements, \(ZD(R)\) the set of zero divisors, and \(\unitsof{R}\) the group of units.
Show that every nilpotent element is either zero or a zero divisor.
Show that if \(n\in N(R)\), then \(1+n\) is a unit. Deduce that \(N(R)+\unitsof{R}=\unitsof{R}\).
For \(f(x)=\sum_{k=0}^d a_kx^k\in R[x]\), show that \(f\in \unitsof{R[x]}\) iff \(a_0\in \unitsof{R}\) and \(a_k\in N(R)\) for every \(k\ge1\).
Show that \(f(x)\in N(R[x])\) iff \(a_k\in N(R)\) for every \(k\).
Show that \(f\in ZD(R[x])\) iff \(f\neq0\) and \(rf(x)=0\) for some \(0\neq r\in R\).
Solution
1. Every nilpotent element is either zero or a zero divisor.
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.
2. If \(n\in N(R)\), then \(1+n\) is a unit; consequently \(N(R)+\unitsof{R}=\unitsof{R}\).
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 \(0\in N(R)\).
3. A polynomial \(f(x)=\sum_{k=0}^d a_kx^k\) is a unit in \(R[x]\) if and only if \(a_0\) is a unit and every \(a_k\) for \(k\ge1\) is nilpotent.
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 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.
4. A polynomial is nilpotent if and only if all its coefficients are 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 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.
5. A nonzero polynomial \(f\in R[x]\) is a zero divisor if and only if some nonzero scalar \(r\in R\) annihilates it.
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 \(rf=0\).