Show that $\langle 2,x \rangle$ is not a principal ideal in $\mathbb Z [x]$

I think it's relatively easy to see that $I=\langle 2,x \rangle = \{a_nx^n+\dots+a_1x+a_0; a_0\text{ is even}\}$.

Now, suppose that $I=\langle f(x) \rangle$ for some $f(x)\in I$.

If $f(x)$ is a constant polynomial, then $\langle f(x) \rangle$ contains only polynomials with even coefficients, and we do not get $x$.

If $f(x)$ is of degree at least $1$, then non-zero polynomials in $\langle f(x) \rangle$ have degree at least $1$, and we do not get $2$.

So $I$ is not of the form $\langle f(x) \rangle$.

I want to record a somewhat less elementary, but perhaps more conceptual answer.

Note first that $\langle 2 \rangle$, $\langle x \rangle$ and $\langle 2, x \rangle$ are all prime ideals of $\mathbb{Z}[x]$. Indeed, the quotients by these ideals are isomorphic, respectively, to $(\mathbb{Z}/2\mathbb{Z})[x]$, $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$, which are all integral domains.

So in particular we have a proper inclusion of nonzero prime ideals

$0 \subsetneq \langle x \rangle \subsetneq \langle x, 2 \rangle$

in which the smaller ideal is principal. Now let $R$ be any integral domain and let $I \subset J$ be a proper inclusion of nonzero prime ideals, with $I$ a principal ideal. Then $J$ cannot be principal. Indeed, suppose $I = \langle x \rangle$ with $x$ a prime element. Suppose also $J = \langle y \rangle$. Then $x \in J$, so that there exists $a \in R$ with $x = ay$. Since $ay = x \in I$ and $I$ is prime, we have either $a \in I$ or $y \in I$. If $a \in I$, then $a = bx$, so $x = byx$ or $x(1-by) = 0$ in the domain $R$; since $x \neq 0$ we conclude $by = 1$, i.e., $y$ is a unit and therefore $J = R$, contradiction. Similarly if $y \in I$, then $y = bx$, so $x = abx$ and we conclude that $a$ is a unit and thus $I = J$, contradiction.

Added: A variant on the above argument is: if $0 \subsetneq I = \langle a \rangle \subsetneq \langle b \rangle = J$ with both $I$ and $J$ prime, then $a$ and $b$ are both irreducible elements and $b$ properly divides $a$, contradiction. This is technically a stronger fact because in an arbitrary domain a generator of a principal prime ideal is necessarily an irreducible element but the converse generally does not hold. However, the easiest way to show that an element $a \in R$ is irreducible is to show that $\langle a \rangle$ is prime, or equivalently that $R/\langle a \rangle$ is a domain. To show that a nonprime element $x$ is irreducible is more delicate.

Remark: If $R$ is a commutative Noetherian ring, then if $J$ is any nonzero principal prime ideal, there cannot be any nonzero prime ideal $I$ -- principal or otherwise -- with $0 \subsetneq I \subsetneq J$. This is a special case of Krull's Principal Ideal Theorem.

Below is a complete, rigorous elementary proof - easily comprehensible to a high-school student.

We show $\rm\,(2,x) = (f)\, $ in $\rm\,\mathbb Z[x]\,$ yields a parity contradiction, by simply evaluating polynomials.

$\rm\ \ f\, \in\, (2,x)\, \Rightarrow\, f\, =\, 2\, G + x\, H.\: $ Eval at $\rm\, x\! =\! 0\ \Rightarrow\ \color{#0a0}{f(0)} = 2\,G(0) = \color{#c00}{2n}\,$ for some $\rm\: n\in \mathbb Z$

$\rm\ \ 2\, \in\, (f)\ \Rightarrow\ 2\, =\, f\, g\:\ \Rightarrow\ deg(f) = 0\ \ \Rightarrow\ \ \color{#c00}f\ =\ \color{#0a0}{f(0)}\ =\ \color{#c00}{2n}$

$\rm\ \ x\, \in\, (f)\,\ \Rightarrow\,\ x\ =\ \color{#c00}f\, h\ =\ \color{#c00}{2n}h.\,\ $ Eval at $\rm\ x\! =\! 1\ \Rightarrow\ 1\, =\ 2n\,h(1)\ \Rightarrow\ 1\,$ is even $\, \Rightarrow\!\Leftarrow$

Remark $\ $ The above proof works over any domain where $\,2\ne 0\,$ and $\rm\,2\,$ is not a unit. $ $ i.e. $\rm\:2\nmid 1.\:$ In particular, it works over any domain with a nontrivial sense of parity, i.e. having $\rm\:\mathbb Z/2\:$ as ring image, e.g. the Gaussian integers, or the rationals writable with odd denominator - see this post. Conversely, the result is false if $\rm\,2 = 0\,$ or a unit since then $\rm\,(2,x) = (x)\,$ or $\,(1)\,$ is principal.

Further, the proof still works if we replace $\,2\,$ by any element $\,c\,$ of the coefficient domain $\,D,\,$ yielding: $\ (c,x)\,$ is principal in $\,D[x]\iff c=0\,$ or $\,c\,$ is a unit. Therefore we deduce

Theorem $ $ If $\,D\,$ is a domain then $\,D[x]\,$ is a PID $\iff D\,$ is a field.

since the direction $(\Leftarrow)$ is well-known via the Euclidean algorithm.

See here for generalizations to coeff rings from domains to rings.

One way to see that $\langle 2,x \rangle$ is not principal is to note that $\mathbb{Z}[x]$ is a UFD(See example 3 in the wiki page), and both $2$ and $x$ are primes. So if the ideal is principal, then $2$ and $x$ will share a common divisor. Contradiction. It is not as down to earth as Martin's solution, but it is a way to look at the problem.

Suppose to the contrary that $(2, x) = \{2p(x) + xq(x) : p(x), q(x) \in \mathbb Z\}$ is a principal ideal where $(2, x) = (a(x))$ for some $a(x) \in \mathbb Z[x]$. Observe that $(2, x)$ is proper since the constant term must be even. Moreover it is immediate that $2 \in (a(x))$ and by definition there exists $p(x) \in \mathbb Z[x]$ such that $2 = p(x)a(x)$. But observe that $0 =\deg p(x)a(x) = \deg p(x) + \deg a(x)$ which implies that $\deg p(x) = \deg a(x) = 0$. Since 2 is prime, we it must follow that $a(x), p(x) \in \{\pm1, \pm2\}$. But if $a(x) = \pm1$ then $(a(x)) = R$ which is a contradiction to $(a(x))$ being a proper ideal. Hence it must follow that $a(x) = \pm2$. So $(a(x)) = (2) = (-2)$. But by construction it must also follow that $x \in (a(x)) = (2)$ so there must exist $q(x) \in \mathbb Z[x]$ such that $x = 2q(x)$. The only way this can happen is if $q(x) = \frac12 x$ which is impossible since $q(x)$ can only have integer coefficients. Hence we have arrived at a contradiction to our hypothesis that $(2, x)$ is a principal ideal.

Note: This immediately implies that $\mathbb Z[x]$ is not a PID.