TITLE: Prove $(2, x)$ is not a free $R$-module.
QUESTION [8 upvotes]: Let $R = \mathbb Z[x]$ and let $M = (2, x)$ be the ideal generated by $2$ and $x$, considered as an $R$-module.   

Show that $\{2, x\}$ is not a basis for $M$.
Show that any 2 elements of $M$ are linearly dependent.
Show that $M$ is not a free $R$-module.


My approach so far: Suppose by way of contradiction $\{2, x\}$ is in fact a basis for $M$, then it would follow that $2$ and $x$ are linearly independent, that is for $\alpha(x), \beta(x) \in \mathbb Z[x]$, if $2\alpha(x) + \beta(x)x = 0$ then it must follow that $\alpha(x) = \beta(x) = 0$, but observe that $\alpha(x) = x$ and $\beta(x) = -2$ satisfies this equation and they are both non-zero. Conclude that $\{2, x\}$ is a linearly dependent set, a contradiction to our hypothesis that $\{2, x\}$ is a basis of $M$. Conclude that $\{2, x\}$ is not a basis for $M$.
Now I am stuck on the second part. I choose two arbitrary elements of $M = (2, x) = \{2f(x) + g(x)x : f, g \in \mathbb Z[x]\}$. Say $a, b \in M$ then by construction $a(x) = 2f_1(x) + g_1(x)x$ and $b(x) = 2f_2(x) + g_2(x)x$. If $\alpha, \beta \in \mathbb Z[x]$, that is, $a, b$ scalar and if $\alpha(x)a(x) + \beta(x)b(x) = 0$ we want to show that $\alpha(x)$, $\beta(x)$ are not both necessarily both zero. I started trying to choose values of $\alpha$ and $\beta$ nonzero such that the equality holds, but it never works out. Any help here would be appreciated.
For the final part, I'm assuming the second part will help. I'm guessing if we show the dimension of $M$ over $\mathbb Z[x]$ must be 2, and then since we know any two elements in $M$ must be linearly dependent, no basis could exist. But my reasoning is a guess, I don't know why mathematically the dimension must be 2, or even if this is the best approach to this part of the problem.
Finally I searched online before asking the question here, I found two proofs using the concept of 'rank', but I'd prefer not to use this in my proof since I am unfamiliar with the concept.

REPLY [7 votes]: Your first part is correct.
As for the second part, think bigger: show that any two elements of an ideal $I$ in a commutative ring are linearly dependent.  (Hint: this is just as easy as the first part!)
As for the third part: in view of the second part, in any commutative ring the ideals which are free as $R$-modules are necessarily principal ideals.  (In a domain the converse is also true: any ideal of the form $\langle a \rangle$ is isomorphic to $R$ as an $R$-module.  In a general ring this is true iff $a$ is not a zero-divisor: if $a$ is a zero-divisor, $\langle a \rangle$ has a nonzero annihilator ideal $I$ and $\langle a \rangle \cong R/I$ is not a free module.)  So you want to show that $I = \langle 2,x \rangle$ is not principal, which is fairly straightforward (and in fact has been asked on this site before).  One possible starting point: notice that $I$ is a prime ideal...