# Proof verification: $\langle 2, x \rangle$ is a prime, not principal ideal

I want to prove that $\langle 2, x \rangle$ is a prime, but not principal, ideal of $\mathbb{Z}[X]$.

$\langle 2, x \rangle$ is prime: Suppose $f(x) = f_n x^n + \ldots + f_1 x + f_0$ and $g(x) = g_m x^m + \ldots + g_1 x + g_0$ are polynomials in $\mathbb{Z}[X]$ with $f(x)g(x) \in \langle 2, x \rangle$. Then $f(x)g(x) = xp(x) + 2q(x)$ for $p(x), q(x)\in \langle 2, x \rangle$. The polynomial on the right side has an even constant term. Thus, $f(x)g(x)$ must have an even constant term implies at least one of $f_0$ or $g_0$ must be even. Without loss of generality, assume $f_0$ is even, so $f_0 = 2k$. Then $f(x) = x(f_n x^{n - 1} + \ldots + f_1) + 2k \in \langle 2, x \rangle$. Hence, $\langle 2, x \rangle$ is prime.

$\langle 2, x \rangle$ is not principal: Suppose it is principal. Then $\langle 2, x \rangle = \langle \alpha \rangle$ for $\alpha\in \mathbb{Z}[X]$. So $\alpha \mid 2$ and $\alpha \mid x$. $\alpha \mid 2 \implies \alpha = 1$ or $2$. If $\alpha = 2$, then $\alpha$ can't divide $x$. So $\alpha = 1$. So $\langle 2, x \rangle = \langle 1 \rangle$ and $1 = 2p(x) + xq(x)$ where $p(x), q(x) \in \mathbb{Z}[X]$. But this can't happen since the constant term on the right side is even. Hence, $\langle 2, x \rangle$ is not principal.

Does this look right?

• this is related link as to show it is not pricipal Commented Dec 4, 2015 at 18:07
• On a purely cosmetic level, it didn't look quite right. $<2, x>$ is too "pointy" and $(2, x)$ is too "round". In my opinion, $\langle 2, x \rangle$ is the perfect middleground. Commented Dec 5, 2015 at 18:58
• Please don't ask multiple questions in a single post. Commented Dec 5, 2015 at 19:01
• The proof looks fine. For the primality of the ideal you can also see what is the quotient $\frac{\mathbb{Z} [x]}{(2,x)}$ that is isomorphic to $\mathbb{Z}_2$ which is a field, so you can conclude that $(2,x)$ is maximal and so prime. Commented Dec 5, 2015 at 19:05