# Zerodivisors in polynomial rings

Let $R$ be a ring and $R[x]$ its polynomial ring. Let $a_0 + a_1x + \cdots + a_n x^n = f$ be a zero-divisor in $R[x]$. Let $b_0 + b_1 x + \cdots + b_m x^m = g \in R[x]$ be the minimal non-zero polynomial s.t. $fg = 0$.

I am trying to understand a proof that $b_m f = 0$. The proof goes like this:

1. $\deg(a_n g) < \deg(g)$ since $fg = 0 \implies a_n b_m = 0$.
2. Yet $(a_n g)f = a_n (fg) = 0$.
3. Then since $\deg(a_n g) < \deg(g)$, we have that $\deg(a_n g) = 0$.
4. But then we can evidently "repeat this argument" (as my textbook states) for all of the $a_i$.
5. We obtain that $a_i g = 0$ for all of the $0 \le i \le n$.
6. Then $a_i b_m = 0$ for all of the $0 \le i \le n$.
7. Then $b_m f = 0$ as desired.

Original Question: How does (4) work? While it is clear to me that $(a_ig)f = 0$, I don't understand why $\deg(a_i g) < \deg(g)$. It seems we can no longer appeal to $(1)$ for this.

Modified Question: Isn't (4)-(7) completely superflous? That is, can't we argue more concisely that $b_m f = 0$ as follows?

1. Suppose $f(x) \in R[x]$ is a zero-divisor.
2. Let $0 \ne g$ be a polynomial of minimal degree in $R[x]$ s.t. $fg = 0$.
3. Suppose $f(x) = a_0 + a_1 x + \ldots + a_n x^n$ and $g(x) = b_0 + b_1 x + \ldots + b_m x^m$.

4. Consider $a_n g = a_n (b_0 + b_1 x + \ldots + b_m x^m) = a_n b_0$ since if $m > 0$, then $deg(a_n g) < deg(g)$ and $a_n g f = 0$ would contradict our choice of $g$ as a polynomial of minimal degree s.t. $fg = 0$.

5. Then $g = b_0 = b_m$.

6. Then $fg = 0 = b_0 f$ and the proof is complete.

Since $a_ng=0$ you get $a_nb_{m-1}=0$. Now look at the coefficient of $x^{m+n-1}$ in the product $fg$ and find that $a_{n-1}b_m=0$. Now repeat the argument for $a_{n-1}g$, and so on.
• I just now understood what's going on in this proof. I wasn't understanding the full implication of $a_n g = 0$. And no, your comment was very helpful. – Techn1cal Dec 8 '13 at 14:09