## The ring of polynomials over a field with no linear term is not a UFD

Let $F$ be a field. Prove that the subset $R \subseteq F[x]$ consisting of all polynomials whose linear coefficient is zero is a subring. Prove also that $R$ is not a unique factorization domain.

To show that $R$ is a subring, we need to see that it is closed under subtraction and multiplication. To that end, let $\alpha = a_0 + x^2a(x)$ and $\beta = b_0 + x^2 b(x)$ be in $R$. Then $\alpha - \beta = (a_0 - b_0) + x^2(a(x) - b(x)) \in R$ and $\alpha\beta = a_0b_0 + x^2(b_0a(x)x^2 + a_0b(x)x^2 + x^4a(x)b(x)) \in R$. So $R$ is a subring.

Next, we claim that $x^2$ and $x^3$ are irreducible in $R$. To see this, note that no element of $R$ has degree 1. If $x^2 = p(x)q(x)$ factors in $R$, then computing the degree of both sides we have $\mathsf{deg}(p(x)) + \mathsf{deg}(q(x)) = 2$. Since neither of these degrees is 1, one must be 0, so that without loss of generality, $p(x)$ is constant and thus a unit. So $x^2$ is irreducible. Similarly, $x^3$ is irreducible because if two nonnegative integers sum to 3 then one of them must be 0 or 1. So $x^2$ and $x^3$ are irreducible in $R$.

Now $x^6 = x^2x^2x^2 = x^3x^3$, so that $x^6$ has at least two distinct irreducible factorizations in $R$. Thus $R$ is not a unique factorization domain.