Chains of Modules, Twisted Polynomial Ring

Twisted Polynomials

A twisted polynomial over a ring r is r adjoin two or more indeterminants, where some or all of the indeterminants do not commute with each other. (They do however commute with r.) For any two indeterminants x and y, yx = fxy for some f in r. If f = 1 then those two indeterminants commute.

This is really a quotient ring, r[x₁,x₂,x₃,…] mod the twisting relations. Thus the result is going to be a ring.

By assumption, the indeterminants can be well ordered, as suggested by the x₁ x₂ x₃ notation above. Each twisting relation puts the indeterminants in order. For instance, x₇x₅ is out of order, but we can put things back in order by replacing this pair with fx₅x₇, where f is the twist associated with these two variables. This creates a canonical representation for the polynomials of the twisted ring. The variables always appear in order. And if you have to multiply two polynomials together, you always normalize the result, so that the variables are once again in order.

x²y³ * x⁴y⁵ = x²y³x⁴y⁵ = f¹²x⁶y⁸

Twisted Power Series

Given a set of twisted indeterminants, as described above, create the ring of formal power series in these variables, with coefficients in r. Again, the variables appear in order in each term of the series, and products are normalized when two series are multiplied together.

Consider the product of two geometric series.

(1 + x + x² + x³ + x⁴ + …) * (1 + y + y² + y³ + y⁴ + …) =
1 + x + y + x² + xy + y² + x³ + x²y + xy² + y³ +
x⁴ + x³y +x²y² + xy³ + y⁴ + …

No twisting is necessary, because the variables are already in order - but what if we swap the operands.

(1 + y + y² + y³ + y⁴ + …) * (1 + x + x² + x³ + x⁴ + …) =
1 + x + y + x² + fxy + y² + x³ + f²x²y + f²xy² + y³ +
x⁴ + f³x³y +f⁴x²y² + f³xy³ + y⁴ + …

Twisted Polynomials and Series are Noetherian

If d is a division ring, the ring of twisted polynomials over d, in finitely many variables, is noetherian. You need to be familiar with the details of hilbert's basis theorem, because this proof is similar.

Clearly d is noetherian, and so is d[x]. We want to bring in y, such that yx = fxy, where f is a nonzero constant in d.

Given a polynomial in the twisted ring, regroup its terms, so that it looks like a polynomial in y, with coefficients in d[x]. Here is an example.

p₇(y) = (x²+8)y³ - (6x²-x-1)y² + (9x⁴+2x²-x-3)y + x⁵+47

Following Hilbert's proof, let a₁ through a_n generate the ideal of lead coefficients in an arbitrary ideal w, hence a₁ through a_n are elements of d[x]. Select a polynomial p_i(y) for each a_i, e.g. p₇ above, exhibiting the lead coefficient a₇ = x²+8.

Remember, we are trying to use the polynomials p₁ through p_n to kill off the lead term of a high degree polynomial q(y). As you consider the span of these generators, only multiply on the left, to create left ideals, as we are trying to show this noncommutative ring is left noetherian. So, to break down the leading term of q(y), multiply the generators by appropriate elements from d[x], to make the lead coefficient work out, and by powers of y on the left, so that everything lines up with q. Unfortunately the powers of y introduce a twist.

For instance, let q have degree 50, whence p₇ must be multiplied by y⁴⁷. Normalize y⁴⁷p₇, and its lead term becomes (f⁹⁴x²+8)y¹⁰⁰. We were going to multiply p₇ by some c₇ on the left, where c₇ is a member of d[x], because that is what we need to do to span the lead coefficient of q. But we need to multiply by c₇y⁴⁷, and that twists the result. To compensate, divide c₇ by f⁹⁴ on the right. This can be done, since d is a division ring. With this adjustment, the resulting linear combination of generators removes the highest power of x from the highest power of y in q. Note that y⁵⁰ might still be present in q. We haven't cleared the leading term completely, but we have broken it down. Its coefficient, drawn from d[x], is smaller.

There is an important detail here, that you don't want to miss. If q begines with x⁴y⁵⁰, and our linear combination creates +x⁵y⁵⁰ and -x⁵y⁵⁰, these may not cancel after they are twisted by f. Our adjustments guarantee x⁴y⁵⁰ will go away - nothing else. However, x⁵y⁵⁰ never enters the picture. Why? Because the generators a₁ through a_n, in d[x], were chosen so that they can span any polynomial in d[x] without going beyond the degree of that polynomial. (This was described in the previous section.) Therefore p₁ through p_n do indeed span, and remove, x⁴y⁵⁰, and they do so without going beyond x⁴y⁵⁰.

Continue this process until the term of y⁵⁰ is gone, then attack the term y⁴⁹, and so on. Eventually the degree of q drops below m, where it is spanned by the extra generators from the d[x] module of rank m. Thus d[x,y] twisted is noetherian.

If you want a third variable, bring in z, where xy and xz may have their own twists. The above proof holds, and d[x,y,z] is noetherian. This remains valid through finitely many variables.

A similar proof shows the twisted power series over d is noetherian. Like the above, this is an extension of hilbert's basis theorem. I'll leave the details to you.

A Twist of 0

Let yx = 0xy; in other words, yx = 0. Once again the variables are in order within each term; something in r times a power of x times a power of y. After multiplication, normalization culls any terms containing yx.

Let xy generate the first left ideal, producing polynomials whose terms all end in xy. Bring in xy² for the next ideal, and xy³ for the next, and so on. This is an ascending chain, hence the polynomial ring is not left noetherian. Note that the union of this chain is the two sided ideal generated by xy.