# Maximal ideal question, + invertibility

1. Feb 24, 2008

### sihag

I'm unable to think of an example of a maximal ideal that is not prime in a finite commutative ring without unity.

Of course the context is, the result stating that in a commutative ring with unity, an ideal is maximal iff it is prime.

Somebody help me out with this ?

Another question is for the ring of 2 x 2 matrices over reals (lets call it M), its subring of matrices with all entries equal (lets call it R) is a ring with a different unity. (wrt usual matrix operations of addition and multiplication)

the unity is 1/2 in each entry.
Of course all non-zero elements of R are invertible, the inverse being (1/4)x (where x is the 'equal entry' for the element of R).
My question is why does this happen, though such matrices belonging to R are all singular.

My own reasoning is that since the determinant is 0 for all matrices belonging to R, we are really not defining anything by the determinant calculation. Also the adjoint of any element of R doesn't belong to R in this case, since the entries will not be equal for adjoint (negatives for the (2,1)th and (1,2)th entry).
But, I'm really not convinced by my own reasons. I know it's tantalizing. Does it have to with the necessary and sufficient criterion relating the determinant with the fact of invertibility of the matrix?

Help me out with this too, someone?

Thank you.

2. Feb 24, 2008

### Hurkyl

Staff Emeritus
Well, obviously, you have to find where the proof for the unital case fails. Have you tried reviewing the proof with a fine-toothed comb to find where the failure happens? Here, I think the only problem is (hint in white) that if you have an ascending chain of proper ideals, you aren't guaranteed their union is proper. (In particular, this means you're going to have to look at rngs that aren't Noetherian) (highlight to read)

Well, I'm not really sure what you mean by "why". But my best guess is that you should look at algebraic properties -- can you think of any unusual equations that the identity of the subrng must satisfy? In particular (additional hint colored white) you might be interested in the idempotents of your rng (Highlight to read, but only after you've reflected on what I've already said!)

Last edited: Feb 24, 2008
3. Feb 24, 2008

### morphism

This is not true. For example, in Z[x], a commutative ring with unity, <x> is prime but not maximal. But your statement is true for Artinian rings (and for PIDs).

Last edited: Feb 24, 2008
4. Feb 24, 2008

### Hurkyl

Staff Emeritus
Hrm... I think I was answering a question different than the one you meant to ask with this section.

5. Feb 24, 2008

### morphism

Yeah - it looks like you were considering the question of whether or not a proper ideal sits inside a maximal ideal.

As for the original question, my gut reaction was to look at non-Artinian rings, and examine an infinite descending chain of ideals (since ideals are in particular non-unital rings). This actually did produce an example: in Z, we have the infinite descending chain 2Z > 4Z > 8Z > ..., and 4Z is a maximal but not prime ideal of the rng 2Z.

But then I realized that the OP was asking for an example in a finite commutative ring without unity. At first I thought this would be impossible to find, since finite rings are trivially Artinian, but I think I did manage to construct an example. Let S=Z/8Z and let R=2S (the ideal of S generated by 2). R is a finite commutative ring without unity (it has no nonzero idempotents). Now if we let I={0,4}, then this is an ideal of R that is maximal but not prime.

In conclusion: non-unital rings suck.

6. Feb 24, 2008

### sihag

mistake, i meant in a finite commutative ring with unity, an ideal is maximal iff it is prime.
yes it should work, if we use {0,2,4,6} addition and multiplication modulo 8
{0,4} is maximal but not prime. (2*6=4,under mult. modulo 8, but neither 2 nor 6 belong to the ideal). got it.

What i meant was, it feels counter-intuitive. I've got to think about it more.