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 … a_n.

For each generator a_i of H, pick a polynomial p_i(x) in W with the lead coefficient a_i. 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 a_n, e.g. c₁a₁ + c₂a₂ + c₃a₃ + … c_na_n. Apply c₁ through c_n, times various powers of x, to p₁ through p_n, and subtract from q. The result is a polynomial of lesser degree, that is spanned by p₁ through p_n iff q is spanned. In other words, the generators p₁ through p_n 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 c_ix^kp_i 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 p_n, 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 … x_n, 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 H_i is H.

By acc the chain is finite, so that H_m = 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 H_m. 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 x^m.

How might we span a series q that starts with x^m+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 x^m+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 x^m+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.