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Every polynomial $f(x) \in K[x]$ can be uniquely written as \begin{equation} f(x) = \alpha (p_1(x))^{a_1} \cdot (p_2(x))^{a_2} \cdot ... \cdot (p_k(x))^{a_k}, \end{equation} where $\alpha \in K, a_i \in \mathbb{N}$, with $p_i(x)$ irreducible for all $1 \leq i < k$.

Proof: I could prove the decomposition using induction on the degree of polynomial $f(x)$ but couldn't think of a way to prove uniqueness. Please help me.

Rob Arthan
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Pedro Souza
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  • $\frac{f(x)}{p_1(x)}$ has a unique decomposition by induction on the degree of $f$. – Robert Shore Aug 08 '19 at 22:20
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    You need to require the $p_i$ to be monic (to avoid identities like $2(x -1) = (2x - 2)$). – Rob Arthan Aug 08 '19 at 22:23
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    @EliotG: I don't think it's a duplicate: Gauss's lemma is a somewhat harder result than the the result that $K[x]$ has unique factorisation when $K$ is a field. – Rob Arthan Aug 08 '19 at 22:27
  • Euclidean domain $\implies$ PID $\implies$ UFD – hardmath Aug 08 '19 at 22:30
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    @hardmath: and the reverse implications are false and it turns out to be a bit easier to prove that $K[x]$ is a euclidean domain and hence a UFD when $K$ is known to be a field. – Rob Arthan Aug 08 '19 at 22:39
  • @RobA: I agree with you. The degree argument makes it easier to show irreducible elements are primes, which makes the factorization unique (up to unit factors). (The degree argument also helps to show units of $K[X]$ are just the nonzero constants.) – hardmath Aug 08 '19 at 22:45
  • @Rob Not clear what you mean, easier than what? – Bill Dubuque Aug 08 '19 at 23:14
  • @BillDubuque: I meant that it is easier to show that $R[x]$ is a UFD when $R$ is given to be a field rather than when $R$ is only given to be a UFD. – Rob Arthan Aug 08 '19 at 23:18

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