Let $A$ be a commutative ring and $f\in A[X]$ with $f \not\equiv 0$. Show that if $f|0$ in $A[X]$ then $\exists a \in A$ st $a \neq 0$ and $af = 0$

Asked Oct 08 '18 at 00:00

Active Oct 08 '18 at 00:14

Viewed 176 times

For example, let $f= \sum_{i=0}^n a_ix^i$. If every $a_i\not\in U(A)$, then $\exists b_i\in A$ with $b_i \neq 0$ for every $i$ such that $a_ib_i=0$, so I can take $a = \prod b_i$ reaches what I need.

I tried using $f|0$ saying that exists a $g$ such that $fg\equiv0$, but the only thing I've reached with this is this is that the independant term of $fg$ is $0$.

Another thing I noticed is that $Im(f) \in A-U(A)$

asked Oct 08 '18 at 00:00

Silkking

1

The notation $f \mid 0$ makes no sense here ($0$ is divisible by any element of a ring). Are you trying to say $f$ is a zero-divisor (i.e., there exists some nonzero $g \in A\left[X\right]$ such that $fg=0$) ? In that case, this is a known fact (see, e.g. https://math.stackexchange.com/questions/83121/zero-divisor-in-rx ). – darij grinberg Oct 08 '18 at 00:04
Thanks. I thought $f|0$ was the same, but I realize now that is not. I'll look at the original question – Silkking Oct 08 '18 at 00:21

Let $A$ be a commutative ring and $f\in A[X]$ with $f \not\equiv 0$. Show that if $f|0$ in $A[X]$ then $\exists a \in A$ st $a \neq 0$ and $af = 0$

0 Answers0