Unique Factorization of Monic Polynomials

Asked Nov 24 '20 at 02:09

Active Nov 24 '20 at 02:34

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Let $N(x)$ be any monic polynomial over a field $\mathbb{F}$.

Prove that there exists irreducible polynomials $P_{1}(x),P_{2}(x),...,P_{h}(x)$ and positive integers $r_{1},r_{2},...,r_{h}$ such that $N(x)=P^{r_{1}}_{1}(x),P^{r_{2}}_{2}(x),...,P^{r_{h}}_{h}(x)$

and if $R_{1}(x),R_{2}(x),...,R_{k}(x)$ is a separate set of irreducible polynomials with distinct integers $s_{1},s_{2},...,s_{k}$ such that $N(x)=R^{s_{1}}_{1}(x),R^{s_{2}}_{2}(x),...,R^{s_{k}}_{k}(x)$ then $h=k$.

Could I invoke the Fundamental Theorem of Algebra to show this is true? Any hints would be helpful. Thanks.

edited Nov 24 '20 at 02:34

asked Nov 24 '20 at 02:09

Chairman Meow

1

Just for notation: $\mathbb{Z}$ is not a field. And FToA is not helpful because the irreducible factors might not be linear factors, i.e. the polynomial could have complex roots. For instance, $x^2+x+1$ is irreducible. – Ethan Dlugie Nov 24 '20 at 02:17
The $\mathbb{Z}$ was a typo. Thanks for the feedback. – Chairman Meow Nov 24 '20 at 02:34
Then the fundamental theorem of algebra is especially not relevant because that theorem is about polynomials specifically over $\mathbb{C}$. – Ethan Dlugie Nov 24 '20 at 02:38
You seem to be looking for UFD (unique factorization domain) property. Every field is UFD, and if $R$ is UFD, then the polynomial ring $R[x]$ is UFD as well. – Sil Nov 24 '20 at 09:51
$\Bbb F[x]$ is Euclidean, so a PID, so a UFD. That irreducbles are prime follows immediately from Euclidean $\Rightarrow$ all gcds exist, as in $\Bbb Z.,$ Is there something that is not clear about these well-known results (already proved here many times) – Bill Dubuque Nov 24 '20 at 10:18
Great, thanks! That is really helpful. I will try to come up with a solution and post that so I can get more feedback. Thanks again for the help all. – Chairman Meow Nov 24 '20 at 20:28

Unique Factorization of Monic Polynomials

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