Lemma 10.120.10. A polynomial ring over a UFD is a UFD. In particular, if $k$ is a field, then $k[x_1, \ldots , x_ n]$ is a UFD.

**Proof.**
Let $R$ be a UFD. Then $R$ satisfies the ascending chain condition for principal ideals (Lemma 10.120.8), hence $R[x]$ satisfies the ascending chain condition for principal ideals (Lemma 10.120.9), and hence every element of $R[x]$ has a factorization into irreducibles (Lemma 10.120.3). Let $S \subset R$ be the multiplicative subset generated by prime elements. Since every nonunit of $R$ is a product of prime elements we see that $K = S^{-1}R$ is the fraction field of $R$. Observe that every prime element of $R$ maps to a prime element of $R[x]$ and that $S^{-1}(R[x]) = S^{-1}R[x] = K[x]$ is a UFD (and even a PID). Thus we may apply Lemma 10.120.7 to conclude.
$\square$

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