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Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ be a polynomial over a commutative ring $R$. Prove that

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

(b) $p$ is nilpotent in $R[x]$ iff $a_0,a_1,\ldots,a_n$ are nilpotent in $R$.

I have got no lead in part (a) and in part (b) I have proved the if part. for the only if part can we use the fact that a polynomial is vanish everywhere iff all its coefficients are $0$?

Community
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Grobber
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  • By definition, the zero polynomial is the one with all its coefficients equal to $0$. – egreg Dec 14 '13 at 17:08
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    About your last question: no, we can't over a general commutative ring. For example, take $;x(x-1)\in\left(\Bbb Z/2\Bbb Z\right)[x];$ . We must be careful distinguishing between polynomials and polynomial functions – DonAntonio Dec 14 '13 at 17:08
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    First prove (b); then for (a) prove the more general fact that $u + n$ for a unit $u$ and a nilpotent element $n$ is a unit again. – Magdiragdag Dec 14 '13 at 17:09
  • Also, don't think about the vanishing of the polynomial ($p(a) = 0$ for all $a \in R$); instead think about when the polynomial is $0$ (all its coefficients are $0$). The latter implies the former, but not conversely. For this problem, the vanishing of a polynomial plays no role. – Magdiragdag Dec 14 '13 at 17:11
  • DonAntonio, can you elaborate on the differences between polynomial and polynomial functions? – Grobber Dec 14 '13 at 17:17
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    http://math.stackexchange.com/questions/83885/proving-a-basic-property-of-polynomial-rings –  Dec 14 '13 at 17:22
  • http://math.stackexchange.com/questions/263896/nilradical-of-polynomial-ring –  Dec 14 '13 at 17:22
  • I think the example I gave is better than 1000 words, @Grobber : the function $;x^2-x;$ is exactly the same as the zero function on the field with two elements, whereas the polynomial $;x^2-x;$ is a quadratic, very non-zero polynomial in the ring of polynomials over the field with two elements. – DonAntonio Dec 14 '13 at 17:22

2 Answers2

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For (a) suppose that $q(x)=\sum_{i=0}^m b_i x^i$ is the multiplicative inverse of $p(x).$ Now compute $1=p(x) q(x).$ The constant term is $a_0 b_0,$ which is equal to $1,$ so $a_0$ is a unit. Now, try to see what identities the other coefficients satisfy...

Igor Rivin
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For the "only-if"-part of b) consider the first coefficent and show that it is nilpotent. Then use the fact that the nilpotent elements form an ideal to derive b) inductively.

benh
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