TITLE: Interpretation of ideals as unitless subsets
QUESTION [5 upvotes]: One way to prove that a field $K$ has no ideals except the entire field and the trivial ideal is to note the fact that every element $x$ has an inverse. By the definition of an ideal, if $x$ is in the ideal then $x^{-1}x$ is because $x^{-1} \in K$. But now we have that 1 is in the ideal, and so again by the definition of an ideal we have that every element is in the ideal. Therefore it is either the entire field or trivial.
However, this works for any would-be ideal that has a unit; hence my question. I don't see how this coheres particularly with the idea that ideals are generalizations of things like "multiple of $n$", or that we use them to form quotient rings.
Can someone please explain whether this has a deeper meaning or if it's not really important? I think it might have something to do with what is written in the "motivation" section in the Wikipedia article for ideals but I'm not really sure.
Edit: I do realize that not all subsets without a unit are ideals. Sorry for the confusion.

REPLY [5 votes]: One way to think of an ideal as being the set of multiples of the (possibly non-existent) g.c.d of all the elements that contain it.  (I am thinking here of the case of a commutative ring with $1$, so that distinctions between left, right, and two-sided don't matter.)
In the integers, for example, any set of elements has a g.c.d. with all the  reasonable properties that you could want, and furthermore if $\{a_i\}_{i \in I}$ is a set of integers,
then the ideal generated by the $a_i$ is in fact principal, and its generator is a
g.c.d. of this collection.  In a more general ring, g.c.d.s don't necessarily exist, or even if they do, they don't have all the properties that they do in the integers.  So rather than trying to work with g.c.d.s, we can introduce ideals, which generally have better properties (or, rather, do many of the same jobs in more general rings that g.c.d.s and their set of multiples do in the context of the integers).
With this in mind, one sees (a) why ideals are the natural kernels of quotient maps: think about the case of the integers, working {\em modulo} $n$ means setting all multiples of $n$ equal to zero; (b) why an ideal with a unit will be the trivial ideal (i.e. the whole ring): because the g.c.d. of any set containing a unit will have to be $1$.