Last edited by JaneFairfax (2008-08-04 08:53:19)

Ricky wrote:

A ring R is a field if and only if its only ideals are 1 and R.

Last edited by JaneFairfax (2008-08-04 20:06:26)

Last edited by JaneFairfax (2008-08-04 13:00:20)

The proof of the theorem in post #4 is a lot more straightforward if you use a bit more of the machinery.

Just as in group theory, the lattice isomorphism theory holds for rings modulo ideals.  Specifically, there is an inclusion preserving bijection between subrings of R that contain I and the set of subrings of R/I.  Furthermore, A is an ideal containing I in R if and only if A/I is an ideal in R/I.  In other words, the lattice composed of subrings that contain I and the lattice of R/I "look" exactly the same.  The proof of this is rather tedious, but I'm told not difficult.

Now, armed with this fact and my theorem in post #3, the proof follows right away.  M is maximal if and only if there aren't any nontrivial ideals in R containing M, excluding M itself.  There aren't any nontrivial ideals in R containing M excluding M itself if and only if there aren't any nontrivial ideals in R/M.  Finally, there aren't any nontrivial ideals in R/M if and only if R/M is a field.