Last class we were discussing the proof (using Zorn's lemma) that every commutative ring with unity has a maximal ideal. In the proof, we strongly use the fact that every proper ideal must not contain $1$, and so any union of these proper ideals must remain proper, after all this union must not contain $1$.
So my question is: is it possible to proof the same result without assuming the unity element? The argument with $1$ seems crucial to ensure the union is proper. Is there any ring (of course, without $1$) in which union of finite ideals is the whole ring?
Notice also that, if you have a rng $R$ without identity but which nevertheless is Noetherian, (meaning it satisfies the ACC on ideals) you can still conclude that the union of an ascending chain of proper ideals has a maximal element, so that its union is equal to the maximal element, and hence is proper. That leads us to consider only rngs which lack the ACC on ideals.
Take any non-Noetherian ring $R$ with identity and find a strictly increasing chain of ideals $I_0\subset I_1\subset I_2\subset\ldots$.
Then $I=\bigcup_{j=1}^\infty I_j$ is an ideal of $R$ which is not equal to any of the $I_j$.
Since $I_j\lhd R$, $I_j\lhd I$ as a ring without identity.
By construction, $I$ is a union of a chain of proper ideals of $I$.
Concretely, you could take the union of the chain $(x_1)\subset (x_1,x_2)\subset (x_1, x_2, x_3)\subset\ldots$ in $K[x_1, x_2, x_3, \ldots]$ (in countably many variables.)