**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.33**

Solution:

(1) Note first that if $a_0$ is a unit in $R$ and $a_i$ nilpotent in $R$ for $i > 0$, then $a_ix^i$ is nilpotent in $R[x]$ for $i > 0$. By Exercise 7.1.14, $p(x)$ is a unit in $R[x]$.

Now suppose $p(x)$ is a unit in $R[x]$, with $$q(x) = \sum_{i=0}^m b_ix^i and p(x)q(x) = 1.$$ We proceed by induction on $n$, the degree of $p$.

For the base case, suppose $p(x) = a_0$. Now $$p(x)q(x) = \sum_{i=0} a_0b_ix^i.$$ Comparing coefficients, we have $a_0b_0 = 1$ in $R$, so that $a_0$ is a unit in $R$. Now it is vacuously true that for $i > 0$, $a_i$ is nilpotent.

For the inductive step, suppose that for some $d$, if $r(x)$ is a polynomial of degree at most $d$ that is a unit, then the constant coefficient of $r(x)$ is a unit and all other coefficients are nilpotent. Suppose now that $n = d+1$. Comparing coefficients of $$p(x)q(x) = \sum_{k=0}^{n+m}(\sum_{i+j=k}a_ib_j)x^k = 1,$$ we see that $a_0b_0 = 1$. Thus $b_0$ is a unit in $R$.

We will now show by induction that for all $t$ with $$0 \leq t \leq m, a_n^{t+1}b_{m-t} = 0.$$ For the base case $t = 0$, we have $a_n^1b_{m-0} = 0$ by considering the $n+m$ coefficient of $p(x)q(x)$. For the inductive step, suppose that for some $t$, if $0 \leq s < t \leq m$, then $a_n^{s+1}b_{m-s} = 0$. Note that since $d=n > 1$, the $n+m-t$ coefficient of $p(x)q(x)$ is $0$ on one hand, and $\sum_{i+j=n+m-t} a_ib_j$ on the other. Multiplying by $a_n^t$, we see that $$\sum_{i+j=n+m-t} a_ia_n^tb_j = 0.$$ Note that if $j > m-t$, then $a_n^tb_j = 0$ by the induction hypothesis. If $j < m-t$, then $i> n$, a contradiction. Thus the only remaining term is the $(i,j) = (n,m-t)$ term, and we see that $a_n^{t+1}b_{m-t} = 0$. Thus the result holds by induction.

Letting $t = m$, we have $a_n^{m+1}b_0 = 0$. Since $b_0$ is a unit, we have $a_n^{m+1} = 0$, so that $a_n$ is nilpotent.

Now $p(x) – a_nx^n$ is a unit by Exercise 7.1.14, and has degree at most $d$. Thus, by the induction hypothesis, $a_0$ is a unit and $a_i$ nilpotent for $i > 0$.

(2) First, since $R$ and thus $R[x]$ is commutative, $\mathfrak{N}(R)$ is an ideal, and so if $a_i$ is nilpotent, then $p(x) = \sum_{i=0}^n a_ix^i$ is nilpotent.

Now suppose $p(x) = \sum_{i=0}^n a_ix^i$ is nilpotent. We proceed by induction on $n$, the degree of $p(x)$.

If $p(x) = a_0$ has degree $0$, then $(p(x))^m = a_0^m = 0$ for some $m$. Thus $a_0$ is nilpotent, and all of the coefficients of $p(x)$ are nilpotent.

Suppose now that for some $d \geq 0$, if $r(x)$ is a nilpotent polynomial of degree at most $d$, then all of the coefficients of $r(x)$ are nilpotent. Now suppose $p(x)$ has degree $n = d+1$ and that $(p(x))^m = 0$ for some $m \geq 1$.

We prove by induction that the $nm$ coefficient of $(p(x))^m$ is $a_n^m$. For the base case $m = 1$, the $n$ coefficient of $p(x)$ is indeed $a_n$. Suppose now that for some $m \geq 1$, the $nm$ coefficient of $p(x)$ is $a_n^m$. Now the $n(m+1)$ coefficient of $(p(x))^{m+1}$ is by definition $$\sum_{i+j = n(m+1)} a_ib_j,$$ where $b_j$ is the $j$ coefficient of $(p(x))^m$. If $i > n$, then $a_i$ does not exist. (Equivalently, is zero.) If $i < n$, then $j > nm$, so that $b_j$ does not exist. (Equivalently, is zero.) Thus the only remaining term is the $(i,j) = (n,nm)$ term, and we have $$\sum_{i+j = n(m+1)} a_ib_j = a_nb_{nm} = a_n^{m+1}.$$ Thus the result holds.

Now the nm coefficient of $(p(x))^m$ is $a_n^m$ on one hand and zero on the other. Thus $a_n^m = 0$, so that $a_n$ is nilpotent. Now $p(x) – a_nx^n$ is nilpotent by Exercise 7.1.14, and has degree at most $d$. Thus by the induction hypothesis all coefficients of $p(x) – a_nx^n$ are nilpotent, and so all coefficients of $p(x)$ are nilpotent.