Nowadays we are interested in such number systems. Indeed, in any symbolic mathematics system that handles calculus (e.g. Macsyma, Maple, Mathematica), one is forced to be interested in general "elementary number" systems closed under "elementary" operations, e.g. see Chow's monthly article What is a closed form number? But such transcendental number theory is much more difficult than algebraic number theory. There are many open problems with no resolution in sight, e.g. Schanuel's Conjecture. There has been much interesting model-theoretic work done on such topics in the past few decades, e.g. see work by Daniel Richardson and Lou van den Dries to get an entry point into this literature, and see the literature on symbolic mathematical computation for algorithms and heuristics.
You know that if there is an isomorphism $h:A\to B$ from an algebraic structure (monoid, group, ring, etc.) $A$ onto another algebraic structure $B$ of the same kind, then $B$ is essentially just $A$ ‘in disguise’: the two structures are essentially the same structure. In other words, Isomorphisms are the maps that preserve the structure exactly.
Homomorphisms preserve some of the structure. (Here some may be all, since every isomorphism is a homomorphism. That is, it’s some in the sense of $\subseteq$, not $\subsetneqq$.) They preserve the operations, but they may allow elements that ‘look enough alike’ to be collapsed to a single element. For instance, the usual group homomorphism from $\Bbb Z$ to $\Bbb Z/2\Bbb Z$ (for which you use the notation $\Bbb Z_2$) ‘says’ that all even integers are essentially the same and collapses them all to the $0$ of $\Bbb Z/2\Bbb Z$. Similarly, it ‘says’ that all odd integers are essentially the same and collapses them all to the $1$ of $\Bbb Z/2\Bbb Z$. It wipes out any finer detail than odd versus even. When you learn in grade school that even $+$ even $=$ even, odd $+$ even $=$ odd, and so on, you’re essentially doing the same thing.
The kernel of the homomorphism is a measure of how much detail is wiped out: the bigger the kernel, the more detail is lost. In the example of the last paragraph, the kernel is the entire set of even integers: the fact that all even integers are in the kernel says that they’re all being seen as somehow ‘the same’, and even more specifically, ‘the same’ as $0$. An isomorphism has a trivial kernel: the only thing that it sees as looking like $0$ is $0$ itself, and no detail is lost.
Another way to put it is that a homomorphic image of an algebraic structure is a kind of approximation to that structure. If the homomorphism is an isomorphism, it’s a perfect approximation; otherwise, it’s more or less crude approximation. As the kernel of the homomorphism gets bigger, the crudeness of the approximation increases. In the case of groups, if the kernel is the whole group, then the homomorphic image is the trivial group, and all detail is lost: all that’s left is the fact that we started with a group.
Best Answer
Thanks to both J.W. Tanner and Jyrki Lahtonen for their help with this question - here's my attempt at making the solution rigorous.
Suppose the field $\mathbb{C}(x)$ of rational functions over $\mathbb{C}$ is algebraically closed. Take the polynomial $P(y) = y^2 - x$; then since $\mathbb{C}(x)$ is algebraically closed, $P$ must have a root in $\mathbb{C}(x)$. But $P(y)=0$ if and only if $y^2 = x$, for some $y \in \mathbb{C}(x)$. Suppose that $y = p(x)/q(x)$. Then we have that: $$ \frac{p^2(x)}{q^2(x)} = x \implies p^2(x) = xq^2(x) $$ Hence considering degrees, we see that we must have $2\text{deg}(p) = 1+2\text{deg}(q)$, and so $2(\text{deg}(p) - \text{deg}(q))=1 \implies \text{deg}(p) - \text{deg}(q) = 1/2$, a contradiction as $p,q$ are polynomials. Thus no such $y$ exists, i.e. $P$ has no root in $\mathbb{C}(x)$ and hence $\mathbb{C}(x)$ is not algebraically closed.