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I am trying to deduce that $\mathbb{Z}[x]$ is a UFD given the fact that the product of two primitive polynomials $fg$, given $f,g\in{\mathbb{Z}[x]}$, is primitive (I have managed to prove this myself). I am also aware (and have proven) of Gauss' Lemma which states that, given $f\in\mathbb{Z}[x]$ primitive, if $f$ is reducible over $\mathbb{Q}$, then it follows that $f$ is reducible over $\mathbb{Z}$. I am also aware (and have proven) the fact that $\mathbb{Q}[x]$ is a UFD. How does this therefore all culminate to prove that $\mathbb{Z}[x]$ is a UFD? (feel like I'm missing something rather straightforward here...). Thank you.

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Any $f(x)$ in $\mathbb{Z}[x]$ can be written uniquely as $f(x) = c(f)p(x)$ where $p$ is primitive. As $\mathbb{Z}$ is a UFD then $c(f)$ can be uniquely written as a product of prime (integers). As $\mathbb{Q}[x]$ is a UFD then $p(x)$ can be written as a product of irreducible rational polynomials. Any reduction of $p(x)$ into rational polynomials can be replaced with a reduction of $p(x)$ into integer polynomials, again in a unique fashion (up to the units of $\mathbb{Z}$).

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  • $\begingroup$ So how does this show that $\mathbb{Z}[x]$ is a UFD... please could you elaborate $\endgroup$ – P-S.D Aug 16 '18 at 11:24
  • $\begingroup$ Do you know the definition of a UFD? $\endgroup$ – asdf Aug 16 '18 at 13:37
  • $\begingroup$ To show that $\mathbb{Z}[x]$ is a UFD I must show that every non-zero elements has a unique factorisation into a product of (not necessarily distinct) primes and a unit (in this case $1$ or $-1$). How does your answer show this? $\endgroup$ – P-S.D Aug 16 '18 at 14:20

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