Exercise 1.2

Question

Let $A$ be a ring and let $A[x]$ be the ring of polynomials in an
  indeterminate $x$, with coefficients in $A$.
  Let $f = a_0 + a_1x + \cdots + a_nx^n \in A[x]$.
  Prove that
  \begin{enumerate}[label=oman*)]
    \item $f$ is a unit in $A[x] \Leftrightarrow a_0$ is a unit in $A$
      and $a_1,\ldots,a_n$ are nilpotent.
    \item $f$ is nilpotent $\Leftrightarrow a_0,a_1,\ldots,a_n$ are nilpotent.
    \item $f$ is a zero-divisor $\Leftrightarrow$ there exists $a 
e 0$
      in $A$ such that $af = 0$.
    \item $f$ is said to be \emph{primitive} if $(a_0,a_1,\ldots,a_n) = (1)$.
      Prove that if $f,g \in A[x]$, then $fg$ is primitive
      $\Leftrightarrow f$ and $g$ are primitive.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $i)$] $\Rightarrow$. We induce on the degree $n$ of $f$. If $n=0$, then $f = a_0 \in A^ imes$. So suppose $n > 0$. Let $b_0 + b_1x + \cdots + b_mx^m$ be the inverse of $f$. By matching degrees in $x$ in the equation $fg = 1$, we have the system of equations \begin{equation}\label{eq:1.2} \begin{aligned} a_nb_m &= 0\ a_{n-1}b_m + a_nb_{m-1} &= 0\ a_{n-2}b_m + a_{n-1}b_{m-1} + a_nb_{m-2} &= 0\ &\ \,\vdots\ a_0b_0 &= 1. \end{aligned} \end{equation} The last row gives that $a_0,b_0$ are units. We now claim that $a_n^{r+1}b_{m-r} = 0$ for $0 \le r \le m$. The case when $r=0$ is the first row in \eqref{eq:1.2}. Suppose the claim holds for all $0 \le r < k$. We then have \begin{multline*} a_n^k(a_{n-k}b_m + a_{n-k+1}b_{m-1} + \cdots + a_{n-1}b_{m-k+1} + a_nb_{m-k})\ = a_n^ka_{n-k}b_m + a_n^ka_{n-k+1}b_{m-1} + \cdots + a_n^ka_{n-1}b_{m-k+1} + a_n^{k+1}b_{m-k} = 0 \end{multline*} by multiplying the $(k+1)$th line from \eqref{eq:1.2} by $a_n^k$. All but the last term vanish by inductive hypothesis, and so $a_n^{k+1}b_{m-k} = 0$. In particular, for $r=m$ we have $a_n^{m+1}b_0 = 0$, which implies $a_n^{m+1} = 0$ since $b_0$ is a unit, i.e., $a_n$ is nilpotent. Now $f - a_nx^n$ is a unit by Exercise 1, hence by inductive hypothesis $a_1,\ldots,a_{n-1}$ are units as well. \par $\Leftarrow$. $g = \sum_{i=1}^n a_ix^i$ is nilpotent since $\mathfrak{N}$ is an ideal by Prop.~1.7, so $f = a_0 + g$ is a unit by Exercise 1. \end{proof} \begin{proof}[Proof of $ii)$] We have: $\Rightarrow$. $1 + f$ is a unit by Exercise 1, and so $a_1,\ldots,a_n$ are nilpotent by $i)$. $a_0$ is nilpotent as well for otherwise $f^m$ has constant term $a_0^m e 0$ for any $m$. \par $\Leftarrow$. If $a_0,a_1,\ldots,a_n \in \mathfrak{N}$, $f \in \mathfrak{N}$ since $\mathfrak{N}$ is an ideal by Prop.~1.7. \end{proof} \begin{proof}[Proof of $iii)$] $\Leftarrow$. Trivial since $a \in A \subset A[x]$. \par $\Rightarrow$. Suppose not, and let $g = b_0 + b_1x + \cdots + b_mx^m$ be a polynomial of least degree $m > 0$ such that $fg = 0$. We claim $a_{n-r}g = 0$ for all $0 \le r \le n$. First, $a_nb_m=0$, hence $a_ng = 0$ since it has degree less than $m$ but annihilates $f$. Now suppose that $a_{n-r}g = 0$ for all $0 \le r < r_0$. Then, we have that \begin{equation*} fg = \left(a_0 + a_1x + \cdots + a_{n-r_0}x^{n-r_0} ight)g = 0 \end{equation*} by inductive hypothesis. Then, $a_{n-r_0}b_m = 0$, hence $a_{n-r_0}g = 0$ since it has degree less than $m$ but annihilates $f$. By induction, $a_{n-r}g = 0$ for all $0 \le r \le n$, hence letting $a = b_m$ gives $af = 0$, a contradiction. \end{proof} \begin{proof}[Proof of $iv)$] $\Rightarrow$. Suppose $fg$ is primitive but $f$ is not. Then, $\mathfrak{a} = (a_0,a_1,\ldots,a_n) \subset \mathfrak{m} \subsetneq A$ by Cor.~$1.4$, where $\mathfrak{m}$ is a maximal ideal. Now consider the natural homomorphism $\varphi\colon A[x] o A/\mathfrak{m}[x]$ induced by the map $A o A/\mathfrak{m}$ on coefficients. $\varphi(f) = 0$ implies that $\varphi(fg) = 0$, and so the coefficients of $fg$ are in $\mathfrak{m}$, a contradiction. \par $\Leftarrow$. Suppose $f,g$ are primitive but $fg$ is not. Then, the coefficients of $fg$ generate an ideal $\mathfrak{a}$, and $\mathfrak{a} \subset \mathfrak{m} \subsetneq A$ by Cor.~$1.4$, where $\mathfrak{m}$ is again a maximal ideal. The same map $\varphi$ defined above has $\varphi(fg) = 0$, and so $\varphi(f)\varphi(g) = 0$, i.e., either $\varphi(f) = 0$ or $\varphi(g) = 0$, for if either were a zero divisor, then there exists $a \in A/\mathfrak{m}$ such that $a\varphi(f) = 0$ by $iii)$, contradicting that $A/\mathfrak{m}$ is a field. Thus, the coefficients of either $f$ or $g$ are contained in $\mathfrak{m}$, a contradiction. \end{proof}

Exercise 1.2

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Exercise 1.2

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