Hi Pete, this sounds like a lot of fun! I wish I could be there (-:
Here is a concrete and useful property of flatness, you can explain it without using Tor. Suppose $R\to S$ is a flat extension.
Then if $I$ is an ideal of $R$, tensoring the exact sequence:
$$ 0 \to I \to R \to R/I \to 0$$
with $S$ gives that $I\otimes_RS = IS$. The left hand side is somewhat abstract object, but the right hand side is very concrete.
There are very natural extensions which are flat but not projective. For example, if $R$ is Noetherian and $\dim R>0$, then $S=R[[X]]$ is flat but never projective over $R$.
Let me give an alternative proof of the result in Ron Maimon's answer, that the canonical map $V\to V^{**}$ can consistently, in the absence of the axiom of choice, be surjective. The only technical advantage of my proof is that it uses only the consistency of ZF plus countable choice plus "all sets (in Polish spaces) have the Baire property". That consistency was proved (by Shelah) relative to just ZF, whereas the consistency of "all sets are Lebesgue measurable" needs an inaccessible cardinal. I think my proof is also a bit simpler than the one using measure. (Also, I don't need to mention ultrafilters, which some people might consider an advantage.)
Let $f:V^*\to\mathbb R$ be a linear map, and let me identify $V^*$ with the space of infinite sequences of reals. Topologize $V^*$ with the product topology, where each factor $\mathbb R$ has the usual topology of the reals. That makes $V^*$ a Polish space, so I can use the assumption about Baire category. In particular, if I partition $\mathbb R$ into intervals of length 1, then the inverse images of these intervals under $f$ have the Baire property, and they can't all be meager, by the Baire category theorem. So at least one of them, call it $f^{-1}(I)$, differs by a meager set from a nonempty open set. Inside that nonempty open set, I can find a basic open set of the form `B=$\prod_iU_i$ where, for some $n$, the first $n$ of the factors $U_i$ are intervals of some length $\delta$ and the later factors $U_i$ are $\mathbb R$. For the first $n$ indices $i$, let $U'_i$ be the interval with the same midpoint as $U_i$ but only half the length, and let $B'$ be the product that is like $B$ except using the $U'_i$ instead of the $U_i$ for the first $n$ factors. Consider an arbitrary $z\in V^*$ whose first $n$ components are smaller than $\delta/2$ in absolute value. Then translation by $z$ in $V^*$ maps $B'$ homeomorphically to another subset of $B$. So the two sets $f^{-1}(I)\cap B'$ and $\{x\in B':z+x\in f^{-1}(I)\}$ are both comeager in $B'$ and therefore must intersect. Let $x$ be in their intersection. Both $f(x)$ and $f(z+x)$ are in the interval $I$ of length 1. Subtracting (and remembering that $f$ is linear), we get that $|f(z)|\leq 2$.
Summarizing, we have that $|f(z)|\leq 2$ for all $z\in V^*$ whose first $n$ components are smaller than $\delta/2$ in absolute value. By linearity, if the first $n$ components of $z$ are smaller than $\alpha\delta/2$ in absolute value, for some positive $\alpha$, then $|f(z)|\leq2\alpha$. In particular, if the first $n$ components of $z$ are zero, then so is $f(z)$. That is, the kernel of $f$ includes the subspace $N$ consisting of those $z\in V^*$ whose first $n$ components vanish. So $f$ factors through the quotient $V^*/N$, which is finite-dimensional (in fact, $n$-dimensional). Knowing what linear functionals on finite-dimensional spaces look like, we immediately conclude that $f$ is given by inner product with a member of $V$ (having non-zero components in at most the first $n$ positions).
Best Answer
What carries over?
As Peter pointed out, a submodule of a free $\mathbb{Z}$-module though free need not have a complement. Indeed each submodule of a free $\mathbb{Z}$-module is free, but a quotient module need not be, for instance $\mathbb{Z}/2\mathbb{Z}$. Also a $\mathbb{Z}$-module is free if and oly if it is projective; this entails that a kernel of a map of free modules does have a complement.
The set $\mathrm{Hom}(F,G)$ for free $\mathbb{Z}$-modules need not be free. If $F$ is free of countably infinite rank and $G=\mathbb{Z}$, then $\mathrm{Hom}(F,G)\cong\prod_{j=1}^\infty\mathbb{Z}$ which remarkably is not free over $\mathbb{Z}$. But $F\otimes G$ is free for free $F$ and $G$.