Solving for Simple: Proving M_2(Z_2) is a Simple Ring

[ehrenfest]

[SOLVED] show a matrix ring is simple

Homework Statement

Show that the matrix ring M_2(Z_2) is a simple ring; that is, M_2(Z_2) has no proper nontrivial ideals.

In my book M_2(Z_2) is the ring of 2 by 2 matrices whose elements are in Z_2.

Homework Equations

The Attempt at a Solution

I wrote down all of the elements of the ring. I stared at them for awhile and I couldn't think of anything to do. We want to show that if H is a nontrivial additive subgroup of the ring, then M_2(Z_2) H must equal M_2(Z_2). But how...

 

[morphism]

Hint: Think about the "standard bases matrices e_ij" of M_2(Z_2); here e_ij is the matrix with a '1' in the (i,j)th position and zeros elsewhere. An ideal is closed under left and right multiplication from the ring, so in particular if a is in your ideal, then so are e_ij*a and a*e_ij.

In general, the ideals of R correspond to the ideals of M_n(R) via the order-preserving bijection I <-> M_n(I). Consequently, M_n(R) is simple iff R is.

 

[ehrenfest]

Is an order-preserving bijection the same thing as an isomorphism?

 

[morphism]

No. Well, it depends on what context the word "isomorphism" is being used in. I'm just setting up a bijection between the ideals of two rings; order-preserving here means if I is a subset of J [as ideals of R], then M_n(I) is a subset of M_n(J) [as ideals of M_n(R)]. This does NOT mean that R and M_n(R) are isomorphic. The fact that M_n(R) shares simplicity with R is a by-product of their Morita equivalence -- this means that these two rings are very similar but not necessarily isomorphic. But this is neither here nor there.

Another irrelevant remark is that order-preserving bijections can be considered "isomorphisms" if we're working in the 'category' (I'm using this term loosely) of ordered sets, where the underlying structure is the order on the sets. This is consistent with thinking about "isomorphisms" as structure-preserving maps.

 

[ehrenfest]

Hint: Think about the "standard bases matrices e_ij" of M_2(Z_2); here e_ij is the matrix with a '1' in the (i,j)th position and zeros elsewhere. An ideal is closed under left and right multiplication from the ring, so in particular if a is in your ideal, then so are e_ij*a and a*e_ij.

In general, the ideals of R correspond to the ideals of M_n(R) via the order-preserving bijection I <-> M_n(I). Consequently, M_n(R) is simple iff R is.

Hmmm. So let I be a nontrivial ideal of M_2(Z_2). We know that there are at least two elements in I and there is one that is not 0, call it a. a must have some component that is not 0. Say that component is (i.j). Obviously you cannot get the whole ring just by applying the basis matrices on the right and on the left since that only gives 8 matrices. But you can apply the elementary matrices in succession and apparently that will give you the whole ring. But how to prove that? Again, I wrote out the 8 products that result from applying the basis matrices on the right and I didn't see the proof :(

 

[morphism]

Well, suppose we have a nonzero element a in the ideal, whose (1,2)th entry is nonzero. Then e₁₁*a*e₂₁ is the matrix with the (1,2)th entry of a in the (1,1)th position, and e₂₁*a*e₂₂ is the matrix with the (1,2)th entry of a in the (2,2)th position. Both of these are in the ideal, and so is their sum, which is the identity matrix. But if an ideal contains a unit (i.e. an invertible element), it must be the entire ring.

Try to generalize this. You'll have to get your hands dirty to see how the matrices e_ij behave.

 

[Hurkyl]

I wrote down all of the elements of the ring. I stared at them for awhile and I couldn't think of anything to do. We want to show that if H is a nontrivial additive subgroup of the ring, then M_2(Z_2) H must equal M_2(Z_2). But how...

Well, there aren't many nontrivial additive subgroups, are there? And you only need to consider the minimal ones anyways...

 

[ehrenfest]

Well, there aren't many nontrivial additive subgroups, are there? And you only need to consider the minimal ones anyways...

There are tons of additive subgroups; every nonzero element of the ring has additive order 2.