Simple Modules, A Maximal Left Ideal

Building the Ring

This is an example of a ring mod a maximal left ideal, leading to a simple module, which then pulls back to a different maximal left ideal. It is rather technical, and not critical for an understanding of simple rings and modules. You can skip ahead if you like.

Let F be a field of characteristic 0, and let x and y be indeterminants that do not commute. Hence the polynomials F[x,y] form a noncommutative ring.

Consider the ideal generated by yx-y-1, then mod out by this ideal. In other words, R consists of cosets of the ideal generated by yx-y-1.

For each coset, find a canonical coset representative in F[x,y]. Given a polynomial in F[x,y], find any instance of yx and replace it with y+1. A term of degree n has been replaced with two terms of degree n-1. Find another instance of yx and do the same. Repeat until there are no such factors. The canonical cosrep is a polynomial where each term is an element of F times a power of x times a power of y.

Building a Maximal Left Ideal

Let H be the left ideal generated by y. If a polynomial belongs to H, every term ends in a positive power of y. Take a moment and verify that H is a left ideal. Multiply by anything on the left, and before we replace instances of yx, every term still ends in y. When an instance of yx is replaced, the two daughter terms still end in y. The product lies in H, and H is a left ideal.

We can see that H is not a two sided ideal, since y times x-1 = 1. Is H maximal?

Suppose we adjoin xⁿ to H. Multiply by y on the left and replace yx with y+1, giving yx^n-1+x^n-1. Apply the same substitution, again and again, until you have y+x^n-1+x^n-2+…+x+1. We can ignore y, since that is already in H. Multiply the remaining polynomial by y again, on the left, and simplify, and find (n-1)y plus a polynomial that starts with x^n-2 and ends with n-2. Repeat this process, multiplying by y on the left, until you have 1. Thus xⁿ cannot be folded into H without spanning all of R.

We could have started with any polynomial in F[x]. If the lead coefficient is c, multiply by y on the left, as often as necessary, and discard multiples of y along the way, to produce c in F. Therefore H is a maximal left ideal.

A Simple R Module

The cosets of H in R are represented by the polynomials in F[x]. Thus our simple R module M is, essentially, F[x], although the action of R is a bit unusual. Premultiplication by x does the usual thing, but premultiplication by y spins off a new polynomial with lower degree, e.g. y*x² = x+1.

Selecting a Generator

Now let's reverse the process. Let 1 act as generator and ask which elements of R map 1 into the 0 coset of H. The answer is H, and we get the same left ideal back again. But what if we select another generator, such as x. Note that y² drives x into H, and so does xy-1. Let these span a left ideal J. Is J a maximal ideal?

There is no room for y, since x*y combines with xy-1 to produce 1.

Try to bring in some other polynomial p. If a term of p contains xy, replace that factor with 1. Thus each term is a power of x or a power of y. And higher powers of y are already part of H, thanks to y², so each term of p is y, or a power of x, or a constant.

Multiply p by x, then replace xy with 1. Now p is a polynomial in x.

Multiply by y on the left an reduce the degree of p, while introducing y. Multiply again and the degree on x decreases, while y becomes y². Repeat until you are left with powers of y plus a constant c, which was the lead coefficient on p. (We did this earlier.) The higher powers of y drop away, leaving a polynomial that is linear in y. Multiply by y and get rid of y², leaving cy. Multiply by c inverse to get y. We already said y was forbidden, hence there is no room for our polynomial p. The left ideal J is maximal, and different from H.

Not Artinian

Notice that the left ideal generated by y² lies properly in H. Verify this for each yⁿ, and build a descending chain of left ideals, hence R is not left artinian.

Minimal Ideal

Let p be a nonconstant polynomial and consider the two sided ideal generated by p. Multiply by y on the left to reduce the degree on x. Continue doing this until the powers of x go away, leaving a polynomial in y.

If p is our polynomial in y, multiply on the right by x-1. This carries yⁿ to y^n-1. Do this as often as necessary, until the lowest term of p becomes a constant c. Then scale by c inverse, so the constant is 1.

Now that p has a constant 1, postmultiply by x-1 one more time, then by y, and subtract p, leaving (x-1)y-1. From this polynomial alone, derive the following expressions.

1. xy-y-1

2. y*(1) = y²-y

3. (2)*(x-1) = y-1

4. x*(3) - (1) = y-x+1

5. (3) - (4) = x-2

Let y-1 and x-2 span an ideal H. Characterize H as polynomials times y-1 plus x-2 times polynomials.

Multiply by y on the right, and move y past y-1 (since they commute), and the characterization remains valid. Multiply by y on the left and y(x-2) becomes -y+1. If there is another factor of x, (y-1)x becomes y-x+1, or (y-1)-(x-2). Spin off x-2 times the remaining term, and we have once again y-1 times a term. When there is no x, scoot y-1 past the power of y.

Multiply by x on the left and slide it past x-2, and there is no trouble. Multiply by x on the right and (y-1)x becomes (y-1)-(x-2), as described earlier. Spin off the term times y-1 and the term times 2, leaving a term times x, where y has lower degree. Repeat this process until the factors of y are consumed.

Now H has been completely characterized, and does not span 1. It is a proper ideal, and it is contained in every other nonzero ideal. Our arbitrary polynomial may span R, as a two sided ideal, (we saw this earlier with y), but it definitely spans H.

Let's try to fold something else into H. Given a nonconstant polynomial, adjust it by polynomials in H. for instance, x³y⁶ - (x-2)x²y⁶ becomes 2x²y⁶. We basically replace each x with 2. Similarly, y is replaced with -1. The result is a unit in F. This holds for any polynomial, hence H is maximal. There is but one ideal between 0 and R, namely H.

Remember that H does not contain y, or any of the powers of y. Mod out by H and find a simple ring that is not artinian, and not semisimple. (I'll define simple and semisimple rings later; just remember this as an interesting example.)