Proving Isomorphism without Explicit Functions in Abstract Algebra

[quasar_4]

I am having a very hard time with a general concept of proving something. If I have some arbitrary function mapping one ring, let's say R, to another ring, S, and want to prove that R is isomorphic to S, then I need to show that there exists a bijective homomorphism between R and S. But how do I do this if I don't know explicitly what f is? In general, how does one show that two things are isomorphic without a function?

It seems a lot easier in linear algebra with vector spaces since one can often make some kind of dimension argument, but I don't know what to do in the abstract algebra case.

 

[matt grime]

Deciding whether two groups (i.e. just the additive parts) are isomorphic is a HARD problem. I mean that, roughly, in the proper sense of the word. There is no way to do it in general. If the ring is finite, then the work is expected to be exponenetial in the number of elements of the ring. In general it is easier to decide when things are not isomorphic. Of course, in any particular case, there may be (frequently are) short cuts, and things to exploit. The nearest things I can think of to structure properties are

1) Algebras decomposing into matrix sub algebras

2) Morita equuivalence i.e. when two rings have the same module theory (but are very much not required to be isomorphic) e.g. If F is any field, then F and M_n(F) are morita equivalent, but not isomorphic for n>1.

 

[quasar_4]

hm, so I can't just define some bijective function between them that is also a homomorphism?

 

[matt grime]

Well, if they are isomorphic, such a thing will exist, but there is in general no way to find any such homomorphism.

 

[mathwonk]

without a mapping, it is usually not possible to show things are isomorphic. i.e. to show two things are isomorphic, first look for a natural map between them. then try to show it has an inverse, or at least to comopute its kernel. then try to show that it is surjective.

example, if we have nested increasing modules N<M<P over the same ring, to show the

P/M is isomorphic to [P/N]/[M/N], find a map. One can easily map P to [P/N]/[M/N], by sending an element of P to the equivalence class mod M/N of its equivalence class mod N.

But a map in the other direction is also at hand. Take an element of P and send it to its equivalence class mod M. Then everything in N goes to zero, so we get a map alsp from P/N to P/M. But here again everything in P/N which is represented by an element of P which is actually in M goes to zero. So we get an induced map[P/N]/[M/N]-->P/M, which is inverse to the previous one since composing takes the class of element {x} to itself.

the composition being the identity is trivial, it was the welldefiendness that was needed. confusing huh?