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I am studying elementary Field Theory. I have a problem that I have been wrestling with for a bit:

Let $p(x)$ be an irreducible polynomial over a field $F$. Suppose that $p(x)$ divides $f_1(x)....f_n(x)$ in $F[x]$. Prove that $p(x)$ divides $f_i(x)$ for some $i$ in {$1,...,n$}.

I'm assuming we can use induction to prove this but I'm not really sure how to go about it.

Thanks!

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2 Answers 2

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The base case of the induction is obvious.

Assume the inductive hypothesis $p\mid f_1f_2\cdots f_n$ implies $p\mid f_i$ for some $i\in\{1,\ldots,n\}$

Now, consider $p\mid f_1f_2\cdots f_nf_{n+1}$. If $p\mid f_1f_2\cdots f_n$, we're done by the inductive hypothesis. Suppose not, then $p\mid f_1f_2\cdots f_{n+1}$ and $p\not\mid f_1f_2\cdots f_n$ implies $p\mid f_{n+1}$ since $p$ is irreducible which completes the inductive step.

Addendum: This is basically a generalization of Euclid's lemma for the Euclidean domain of the ring of polynomials in $x$

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You have to prove that if $p$ divides $fg$, then $p$ divides $f$ or $p$ divides $g$, then you can proceed easily by induction. The argument is pretty much the same as for integers, you can look it up in any book.

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  • $\begingroup$ Any book? I'm 100% confident that the novel I am reading does not contain the argument Max needs! $\endgroup$
    – Rob Arthan
    Apr 10, 2018 at 21:07
  • $\begingroup$ Do you maybe see a post on here that might help? Thanks! $\endgroup$
    – Max
    Apr 10, 2018 at 21:12
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    $\begingroup$ Any book that deals with this material, of course. $\endgroup$
    – xyzzyz
    Apr 10, 2018 at 21:34
  • $\begingroup$ My answer helps exactly as much as the answer above mine. The answer above mine only does the easy induction I mentioned, and doesn't get to the crux of the issue that I explicitly point out: $p | fg \rightarrow p|f \lor p|g$. $\endgroup$
    – xyzzyz
    Apr 11, 2018 at 18:11
  • $\begingroup$ @xyzzyz: I agree that the answer that has been accepted is little better than yours. I do think it would be a kindness to give the OP more help with the interesting property of prime polynomials that you are appealing to. $\endgroup$
    – Rob Arthan
    Apr 11, 2018 at 18:37

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