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:
- $\deg(a_n g) < \deg(g)$ since $fg = 0 \implies a_n b_m = 0$.
- Yet $(a_n g)f = a_n (fg) = 0$.
- Then since $\deg(a_n g) < \deg(g)$, we have that $\deg(a_n g) = 0$.
- But then we can evidently "repeat this argument" (as my textbook states) for all of the $a_i$.
- We obtain that $a_i g = 0$ for all of the $0 \le i \le n$.
- Then $a_i b_m = 0$ for all of the $0 \le i \le n$.
- 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?
- Suppose $f(x) \in R[x]$ is a zero-divisor.
- Let $0 \ne g$ be a polynomial of minimal degree in $R[x]$ s.t. $fg = 0$.
Suppose $f(x) = a_0 + a_1 x + \ldots + a_n x^n$ and $g(x) = b_0 + b_1 x + \ldots + b_m x^m$.
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. $fg = 0$.
Then $g = b_0 = b_m$.
- Then $fg = 0 = b_0 f$ and the proof is complete.