Atyiah Macdonald Ex 1.2(iii)

Question

$f\in A[x]$ is a zero divisor if and only if there is $a\neq 0\in A$ with $a\cdot f=0$

One direction is trivial. The other direction I find the hint confusing. It says we should choose a zerodivisor $g$ of $f$ with a minimal degree. We write $f=\sum^n a_ix^i$ and $g=\sum^m b_j x^j$. Then all the coefficients of $fg$ are zero. In particular the highest degree coeeficent is zero, $a_nb_m=0$. By assumption then $a_ng$ must be zero since it has a degree less than $g$ and it annihilates $g$.

Then the hint says to use induction on $a_{n-r}$. Are we inducting on $r$? It is also not clear to me how to result follows from this?

This question is not a duplicate of Zero divisor in $R[x]$ as I am specifically asking how to use the hint and do this problem with the induction.

Answer 1

You computed the coefficient of $x^{n+m}$ in $fg$ and got that $a_ng=0$. Now let's compute the coefficient of $x^{n+m-1}$ in $fg$. We get:

$a_{n-1}b_m+a_nb_{m-1}=0$

Since $a_ng=0$ we know that in particular $a_nb_{m-1}=0$, and thus it follows that $a_{n-1}b_m=0$. So again, $a_{n-1}g$ is a polynomial of degree less than $m$ which annihilates $f$, and so $a_{n-1}g=0$.

Continue like this by induction and show that $a_{n-r}g=0$ for $r=0,1,...,n$. In particular this implies:

$a_0b_m=a_1b_m=...=a_nb_m=0$

And so $b_mf=0$.