I'm trying to prove the converse of the following theorem. I think suggestion available at this website are mistaken or I didn't understand them correctly.

Theorem. Let $R$ be a commutative ring with $1$. Then $f(X)=a_0+a_1X+a_2X^2+ \cdots +a_nX^n$ is a unit in $R[X]$ if and only if $a_0$ is a unit in $R$ and $a_1,a_2, \ldots, a_n$ are all nilpotent in $R$.

Let $f,g$ be two polynomials of degree $n$ and $m$. Then the coefficients of their product will be given by the following relation: $c_k = \sum_{i=0}^k a_i b_{k-i}$. Here we don't get $a_n b_m = 0$ because it is not yet the coefficient of $x^{m+n}$. If I would prove that $a_n$ is nilpotent, the proof will be followed... Any comment where I'm confused?

Why this question isn't duplicate? This question is answered but all argue that coefficient of $x^{m+n}$ are $a_n b_m$ which I believe is wrong. $c_{m+n} = \sum_{i=0}^{m+n} a_i b_{m+n-i}$. If you expand this, it will be sum of product of those coefficients such that $X^iX^j$, where $i+j = m $

  • $\begingroup$ Here are some tips on formatting. $\endgroup$ – Richard D. James Nov 25 '14 at 5:34
  • $\begingroup$ Perhaps that would have been more accurate. But that was in a context of a problem. That is why I written down the problem as well. $\endgroup$ – madeel Nov 26 '14 at 11:23

$c_{m+n} = \sum_{i=0}^{m+n} a_i b_{m+n-i} = a_n b_m$ because for $i<n$ we have $m+n-i > m$, hence $b_{m+n-i}=0$, and for $i>n$ we have $a_i=0$.

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  • $\begingroup$ I got it. I should have get this simple point. Thanks for your help... $\endgroup$ – madeel Nov 26 '14 at 11:24

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