## Chains of Modules, Hilbert's Basis Theorem

### Hilbert's Basis Theorem

Let R be a noetherian ring and let S be the polynomial extension R[x]. We will prove S is also noetherian. Obviously S isn't a noetherian R module, since it has an infinite basis, namely the powers of x. But S is a noetherian S module, and hence a noetherian ring.

Note that S is not artinian, since the powers of x generate an infinite descending chain of principal ideals. This is one of those noetherian only theorems.

Consider any ideal W in R[x] and let H be the set of extant lead coefficients. Verify that H is an ideal in R, hence finitely generated by a1 a2 a3 … an.

For each generator ai of H, pick a polynomial pi(x) in W with the lead coefficient ai. These will act as the first n generators of W.

Let m be the maximum degree of these polynomials, and let q be a polynomial in W with degree at least m. Write its lead coefficient as a linear combination of a1 through an, e.g. c1a1 + c2a2 + c3a3 + … cnan. Apply c1 through cn, times various powers of x, to p1 through pn, and subtract from q. The result is a polynomial of lesser degree, that is spanned by p1 through pn iff q is spanned. In other words, the generators p1 through pn span, and remove, the leading term of q, whatever it may be. Hence this is called hilbert's basis theorem.

Notice that none of the intermediate expressions cixkpi exceeds the degree of q. The polynomials line up at the left edge of q, combine, and strip off the leading term.

Repeat this process until you are left with a polynomial in W of degree less than m. If this is spanned, then our original polynomial q is spanned.

These polynomials live in a free R module of rank m, generated by the powers of x. This module is noetherian courtesy of R, and the submodule produced by intersecting with W is a finitely generated R module, and a finitely generated R[x] module. Select generators that span this portion of W, as an R module. This may be more generators than we need, but it is still a finite set. Combine these generators with p1 through pn, and W is finitely generated. Therefore R[x] is noetherian.

If we choose our generators carefully, we can make sure the intermediate expressions never exceed the degree of q. This was the case when q had degree ≥ m. The generators line up at the left and kill the lead term; we are making progress. For degrees below m, select a finite set of generators that span the constant polynomials of W, then find generators that span the linear polynomials, then the quadratic polynomials, and so on up to degree m-1. The result is still a finite collection of generators. Now, given q in W, with degree k less than m, use the generators that span the polynonmials of degree k to kill off the lead term, and so on down to the constant. We never kill off the lead term while introducing a higher term. Each iteration reduces the degree of q.

Note that R need not be commutative. If R is left noetherian then R[x] is left noetherian.

### Finitely Generated Ring

Let R be noetherian and let S be a finitely generated R algebra. This isn't a finitely generated module; it's a finitely generated ring. Also, the generators adjoined to R commute with each other and with R.

Repeatedly adjoin indeterminants to R and apply hilbert's basis theorem at each step. The resulting polynomial ring, in several variables, is noetherian. Since S is a quotient thereof, it too is noetherian.

If R is itself finitely generated, and commutative, it is noetherian. This because R is the quotient of Z adjoin x1 x2 x3 … xn, which is noetherian.

### Formal Power Series

A similar proof shows R[[x]] is noetherian. Build H using the lead coefficients in the power series of an ideal W. (In this case the lead coefficient has the lowest degree, not the highest.) Let H0 be the ideal of constant coefficients. Let H1 be the ideal of coefficients on x when the constant term is 0. Build H2, H3, H4, and so on, so that the union of Hi is H.

By acc the chain is finite, so that Hm = H.

Select generators a1 a2 a3 etc, and corresponding power series p1 p2 p3 etc, spanning H0, then H1, then H2, and so on up to Hm. These are all the generators we will need.

Given a series q in W, subtract a linear combination of generators to kill off the constant term. Then kill off the linear term, then the squared term, and so on. We only need consider series in W that start beyond xm.

How might we span a series q that starts with xm+7? A linear combination of generators at level m reproduces the lead coefficient of q. Multiply this linear combination by x7, and subtract from q. This leaves a series that starts with xm+8. Again, a linear combination of generators at level m yields the desired coefficient, and when this is multiplied by x8, q loses another term. Each generator at level m is multiplied by some c7x7+c8x8, using appropriate coefficients c7 and c8 for that particular generator. When the term xm+9 is subtracted away, the generators are multiplied by linear combinations of x7, x8, and x9. This continues forever, building a power series for each generator. Thus q is spanned by the generators at level m, and W is finitely generated.

By induction, the formal power series in several variables is noetherian, and so is any quotient ring thereof.

Once again R need not be commutative. If R is left noetherian, the resulting power series is left noetherian.

### Noncommuting Indeterminants

If x and y do not commute, R[x,y] is not noetherian.

Let Z be the base ring, and consider the polynomials, or power series, in x and y. Start with the ideal, or left ideal, generated by x2y2. Strings with single powers of y are conspicuously absent. In particular, there is no xy, or x2y. For notational convenience, let a = xy and let b = x2y. Adjoin a2b2 to produce the second ideal in the chain. The strings ab and a2b are missing, so label them c and d respectively. Adjoin c2d2 to get the third ideal, and so on. This continues forever, building an infinite ascending chain of ideals.