Show that $\langle 2,x \rangle$ is not a principal ideal in $\mathbb Z [x]$ Hi
I don't know how to show that $\langle 2,x \rangle$ is not principal and the definition of a principal ideal is unclear to me. I need help on this, please. 
The ring that I am talking about is $\mathbb{Z}[x]$ so $\langle 2,x \rangle$ refers to $2g(x) + xf(x)$ where $g(x)$, $f(x)$ belongs to $\mathbb{Z}[x]$.
 A: 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.
A: 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 $\langle f(x) \rangle$.
A: We have $\langle 2, x \rangle = \{ a_n x^n + \cdots a_1 x + a_0 \mid a_0 \in 2 \mathbb{Z}, a_1, \ldots, a_n \in \mathbb{Z} \}$. 
Suppose that $\langle 2, x \rangle$ is principal. Then there is some polynomial $f(x)$ in $\langle 2, x \rangle$ such that $\langle f(x) \rangle = \langle 2, x \rangle$. Therefore $x \in \langle 2, x \rangle = \langle f(x) \rangle$ and $2 \in \langle 2, x \rangle = \langle f(x) \rangle$. 
It follows that $2 = f(x) g(x)$, $g(x) \in \mathbb{Z}[x]$. Therefore $f(x)=c \in \mathbb{Z}$. 
Since $x \in \langle f(x) \rangle$, $c$ must be $1$ or $-1$ (for example, if $c=2$, then $\langle f(x) \rangle = \langle c \rangle = \{a_n x^n + \cdots a_1 x + a_0 \mid a_0, \ldots, a_n \text{ are even}\} \neq \langle 2, x \rangle$ which is a contradiction to our assumption). 
But the ideal of $\mathbb{Z}[x]$ generated by $1$ or $-1$ is $\mathbb{Z}[x]$. Since $\langle f(x) \rangle \neq \mathbb{Z}[x]$, we obtain a contradiction again.
A: 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.
A: 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.
A: An ideal $\langle a_1, \dots, a_k \rangle$ is the smallest ideal containing these elements, explicitly the set of all linear combinations $r_1a_1+\dots + r_ka_k$ where the $r_i$ are arbitrary elements from the ring.
A principal ideal is an ideal that can be generated by a single element.
So first of all, you have to say which ring you are looking at to have a definite question.
Now, you could comment on whether you understand this definition of principal ideal.
If the ideal $\langle 2,x \rangle$ were principal, the generator would have to divide 2. What are the integer polynomial divisors of 2?
A: 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. 
A: We prove by contradiction.
Suppose that $(2,X)$ is a principal ideal. Then, we have $(2,X)=(f(X))$ for some polynomial $f(X) \in \mathbb{Z}[X]$.
Therefore, we have
\begin{equation}
f(X)\cdot g(X) = 2
\end{equation}
for some polynomial $g(X) \in \mathbb{Z}[X]$.
By inspecting the degree, we have
\begin{equation}
\text{deg}(f(X)\cdot g(X)) = \text{deg}(f(X)) + \text{deg}(g(X)) = \text{deg}(2) = 0.
\end{equation}
Hence, we must have
\begin{equation}
\text{deg}(f(X)) = 0 = \text{deg}(g(X)),
\end{equation}
since the degree of polynomials cannot be negative.
This gives us that both $f(X)$ and $g(X)$ are constant polynomials (integers). Therefore, we have either $f(X)=\pm 1$ or $f(X)=\pm 2$.
Suppose that $f(X)=\pm 1$, then we have
\begin{equation}
\pm 1 = 2\alpha(X) + X\cdot\beta(X)
\end{equation}
for some polynomials $\alpha(X),\beta(X)\in\mathbb{Z}[X]$.
Let the coefficient of the constant term in $\alpha(X)$ be an integer $x\in\mathbb{Z}$, then observe that the coefficient of the constant term of $2\alpha(X)$ must be an even number. Hence, we obtain $1=2x$. But, there is no integer $x$ that satisfies this equation. Therefore, $f(X)\neq\pm 1$.
We are left with $f(X)=\pm 2$. Similarly, observe that the coefficient of the term $X$ of $2\alpha(X)$ must be an even number while the coefficient of the term $X\cdot \beta(X)$ can be either odd or even. Hence, $f(X)\neq \pm 2$ also.
We have ruled out all possibilities for $f(X)$. Therefore, we conclude that $(2,X)$ can never be a principal ideal of $\mathbb{Z}[X]$.
