0
$\begingroup$

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 $

$\endgroup$
  • $\begingroup$ Here are some tips on formatting. $\endgroup$ – André 3000 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
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.