Let $p(x)$ be irreducible in $K[x]$ & $p(x)|(r_1(x)...r_n(x))$ where $r_i(x) \in K[x] \forall 1 \leq i \leq n$. Then $p(x)|r_i(x)$ ...

Question

Theorem. Let $p(x)$ be irreducible in $K[x]$ and $p(x)|(r_1(x)...r_n(x))$ where $r_i(x) \in K[x]$ for all $1 \leq i \leq n$. Then $p(x)|r_i(x)$ for at least one $1 \leq i \leq n $.

Proof. I read that the proof should be by induction on $n$, but I don't understand how to do this, what's the given formula that you usually substitute values of $n$ into? I just don't know where to start.

Dietrich Burde · Accepted Answer · 2020-04-21T13:21:17.823

1

Since any irreducible element in $K[X]$ is also prime, we have that $p\mid r_1\cdots r_n$ implies that $p\mid r_i$ for some $i$. Formally, we could do induction on $n$, the case $n=1$ being trivial.

edited Apr 21 '20 at 13:21

answered Apr 21 '20 at 13:14

Dietrich Burde

130,978

How do we know that it's a PID? – MathGeek1998 Apr 21 '20 at 13:16
How do we know that? It is well-known :) But we don't need it. – Dietrich Burde Apr 21 '20 at 13:18

Let $p(x)$ be irreducible in $K[x]$ & $p(x)|(r_1(x)...r_n(x))$ where $r_i(x) \in K[x] \forall 1 \leq i \leq n$. Then $p(x)|r_i(x)$ ...

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