Field of polynomials mod n? I have a few questions and i am looking for some clarification.
1) Is it correct that one can define a field $(Z_n, +, X)$ of integers mod $n$, where all the elements are integers $a$ such that $gcd(a, n)$ = $1$? In other words, this can be a field even when $n$ is not prime, just so long as we delete some elements? Van anyone tell me if this partiular field (if it is a field) has a name?
2) If we call the field above $F$, then is $F(x)$ a field? So the only factors of polynomials are either quadratics (with complex roots over $C{(x)}$) or linear factors of the form $x-a$ for $a \in F$?

About (1), as KCd says it's pretty easy to show that no such field can exist (although you can make a perfectly good group this way).  
For example suppose you had some set $S \subset Z / nZ$ with addition and multiplication modulo $n$.  Let $\overline{m}$ be the multiplicative identity, ie $\overline{m}\overline{x} = \overline{x}$ for all $\overline{x} \in S$.  Then letting $x \in Z$ be any representative of $\overline{x}$ we have that $\overline{m} + \overline{m} + ... + \overline{m}$ ($x$ times) equals $\overline{x}$.  So $\overline{m}$ generates $S$, and every representative of $S$ is a multiple of $m$.  Note that the set of equivalence classes is the same then if we changed to modulo $lcm(m,n)$ (since $\overline{x} = \overline{0}$ iff $n \mid x$ which implies $lcm(n,m)$ divides $x$ anyways).  Thus $S = mZ/lcm(n,m)Z = mZ/kmZ$ for some $k$.  And then $mZ/mkZ \simeq Z/kZ$ implying $k$ must be a prime, otherwise we have zero divisors.  
