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Let $R$ be a commutative ring. Let $p(x) = a_n x^n + a_{n-1} x^{n-1} +...+a_1 x +a_0 \in R[x]$.

Prove that, $p(x)$ is a unit in $R[x]$ iff $a_0$ is a unit and $a_1 , a_2 ,... , a_n $ are nilpotent in $R$


I could prove that $a_0$ is a unit $R$

and that $a_i $ is a zero divisor in $R$

but I couldn't show that it's a nilpotent !

any hints ?!

This is problem $\#33.(a)$ page 250 from Dummit and Foote's text , 3rd ed .

On a website, they said that the proof can be done by induction but I couldn't follow the proof.

any nice hints to help me solve this problem ?

no lemon no melon
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Fawzy Hegab
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  • Related, but not the same: http://math.stackexchange.com/questions/30380/finding-invertible-polynomials-in-polynomial-ring-mathbbz-nx?lq=1 – rschwieb May 03 '13 at 19:52

2 Answers2

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The following result is very useful:

Lemma. The sum of a nilpotent and a unit is a unit.

$(\Leftarrow)$ If $a_0\in R^\times$ and $a_1,\dots,a_n$ are nilpotent, then clearly $a_1x+\cdots+a_nx^n$ is nilpotent (just raise it to a high enough power and every term will contain $a_i^N=0$). Then $f(x)$ is the sum of a nilpotent and a unit, hence a unit.

$(\Rightarrow)$ Comparing the constant and the $x^{m+n}$ terms in $fg=1$ gives $a_0b_0=1$, so $a_0$ is a unit, and $a_nb_m=0$, which starts the induction. Comparing the coefficients of $x^{m+n-r}$ yields: $$ a_{n-r}b_m+a_{n-r+1}b_{m-1}+\cdots+a_nb_{m-r}=0. $$Can you go from here? You want to try and show that $a_n$ is nilpotent, since then $f(x)-a_nx^n$ is the sum of a nilpotent and a unit.

Warren Moore
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Hint: WOLOG, let $a_0=1$, $p_n(x)=a_nx^n+a_{n-1}x+\dots+1=a_n x^n+p_{n-1}(x)$. Suppose $q_m(x)=b_mx^m+b_{m-1}x^{m-1}+\dots+1=b_mx^m+q_{m-1}$ is the inverse of $p_n$. Show $p_{n-1}q_{m-1}=1$, and then $a_n$ is nilpotent. Now you know how the induction works.

Ma Ming
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