$\mathbb{Z}_5[x]$ is a UFD

Question

Prove that $\mathbb{Z}_5[x]$ is a unique factorization domain.

My approach is to prove that $\mathbb{Z}_5[x]$ is a PID, which implies that it is a UFD.

Proof:

Suppose there exists an ideal $I$ in $\mathbb{Z}_5[x]$ such that it is generated by two or more elements of $\mathbb{Z}_5[x]$. That is, $I = \langle g_1(x), g_2(x), ..., g_n(x)\rangle$. Then $I=\{a_1(x)g_1(x)+a_2(x)g_2(x)+...+a_n(x)g_n(x):a_i(x)\in \mathbb{Z}_5[x] \}$. Consider $\max\{a_i(x)g_i(x)\}=\deg_{max} (I)$. Then, since $\mathbb{Z}_5$ is a PID, $\langle a_i(x)g_i(x)\rangle = \langle g_1(x), g_2(x), ..., g_n(x)\rangle$. Hence, $\mathbb{Z}_5[x]$ is a PID. This implies that $\mathbb{Z}_5[x]$ is a UFD.

It would be interesting to know one's opinion on my proof.

Answer 1

Hint:

You have to use that, in a polynomial ring over a field, you can perform Euclidean divisions, and consider a non-zero polynomial of least degree in the ideal.

Answer 2

By $\mathbb{Z}_5$ I assume you mean $\mathbb{Z}/5\mathbb{Z}$. This is a field, since 5 is prime. Any field is a PID. See Why any field is a principal ideal domain?

EDIT: After comments, I realized that while the OP said, "My approach is to prove that $\mathbb{Z}_5$ is a PID," what the OP meant to say was "My approach is to prove that $\mathbb{Z}_5[x]$ is a PID." My answer, while a good answer to the first, is not a good answer to the second.