Definitions problem - showing an ideal is not principal (and showing it is principal in only one case) Question is as follows:
Show the ideal $(2)+(x)$ of $\mathbb{Z}[x]$ is not principal. Then
Show that for a field $K$ the polynomial ring $K[x_1,...,x_n]$ is not a principal ideal for $n>1$
Other than not knowing where to start with the second one, I sense there might be some nice trick involving contrapositive... or something. Rather than induction starting at 2. 
What have I done so far you ask?
I've consulted Serge Lang's superb book "Undergraduate Algebra" and used the index to find "principal ideal" taking me to page 59. Here I learn what a principal left ideal generated by $a$ is. 
I've consulted R.B.J.T Allenby's book "Rings, Fields and Groups - An Introduction to Abstract Algebra" and again searched the index (second edition btw) which takes me to page 99 were it notes that an example is called a "principal ideal", I can glean something from this but I only like to play "guess the definition" when I've seen it but need my memory jogged not when I'm trying to find a definition.
Now I have fallen back onto searching for the definition and found some. I am just not sure how to apply it nor if I am looking at an alternate form of the convention. I'd really like a concrete definition and an example (so if you give a definition and a full example that isn't this question, I'd be very happy) While I search though I hope to tap into the wealth of experience here. I hate searching, I love reading.
 A: The first thing to note is that since $\mathbf{Z}$ is an integral domain, the well-known result that the degree of the product of two polynomials is the sum of their degrees holds in $\mathbf{Z}[x]$.
Now, suppose $(2)+(x)$ is principal. This means that there exists some $f \in \mathbf{Z}[x]$ such that any polynomial in $(2)+(x)$ is a multiple of $f$. In particular, $2$ is a multiple of $f$, so $f$ must be a constant polynomial. Moreover, $x$ must be a multiple of $f$, so we have $(f)(ax+b) = fax+fb = x$, so $fa = 1$, meaning $f$ is a unit. This implies that $(f) = \mathbf{Z}[x]$, which contradicts the assumption that $(f) = (2)+(x)$, because $1 \notin (2)+(x)$.
By a similar argument, the ideal $(x_1)+(x_2)$ is not principal in $K[x_1,\dots,x_n]$, $n > 1$.
A: $\!\begin{eqnarray}\rm {\bf Hint}\ \ \  Notice\ that\ the \ ideals \ \ (2) &\subsetneq&\rm (2,x)&\subsetneq&\rm (1)\ \ are\ distinct\ \ \ \,primes\,\ \ (or\,\ 1)\\ 
\rm because\ their\ residue\ rings\  \ \ \Bbb F_2[x] &\supsetneq&\ \ \, \Bbb F_2&\supsetneq&\rm  \Bbb (0)\ \ are\ distinct\ \ domains\ (or\,\ 0).\end{eqnarray}$
Therefore if $\rm\ (2,x) = (q)\,$ then $\rm\,q\mid 2\,$ properly, contra $\,2\,$ is prime so irreducible. 
