If you find any mistakes, please make a comment! Thank you.

Characterize the units and nilpotent elements of a polynomial ring


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.


Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
Close Menu