Questions: Hilbert Basis Theorem

Question 1

The Hilbert basis theorem implies which of the following?

• A. Every ideal in k[x₁, ..., xₙ] is principal (generated by one element)
• B. Every ideal in k[x₁, ..., xₙ] is finitely generated
• C. Every ideal in k[x₁, x₂, ...] (infinitely many variables) is finitely generated
• D. Every polynomial ring over any ring is Noetherian

Answer: (B)

The Hilbert basis theorem says R Noetherian implies R[x] Noetherian. Since a field k is Noetherian, k[x₁] is Noetherian, then k[x₁, x₂] = k[x₁][x₂] is Noetherian, and by induction k[x₁, ..., xₙ] is Noetherian — meaning every ideal is finitely generated. Not every ideal is principal (the ideal (x, y) in k[x, y] requires two generators). The ring k[x₁, x₂, ...] of infinitely many variables is not Noetherian. And polynomial rings over non-Noetherian rings need not be Noetherian.

Question 2

Hilbert originally proved his basis theorem in the context of invariant theory. A professor states that the theorem was initially controversial because it was a pure existence result. Why?

• A. It used the axiom of choice, which was not yet accepted
• B. The proof shows that a finite generating set must exist (via contradiction using the ACC) without constructing specific generators
• C. It contradicted known results about polynomial rings
• D. It only applied to fields of characteristic zero

Answer: (B)

Hilbert's 1890 proof used an ascending chain / maximal ideal argument to show that every ideal in a polynomial ring is finitely generated, without exhibiting the generators. Gordan famously protested that 'this is not mathematics, it is theology.' The non-constructive nature was philosophically jarring at the time. Constructive versions (via Gröbner bases, developed much later) eventually provided explicit algorithms, but the original theorem's power lies precisely in its generality — it works regardless of whether you can compute the generators.

Question 3

If R is Noetherian, then R[x₁, ..., xₙ] is Noetherian for every finite n.

• T. True
• F. False

Answer: True

This is the iterated form of the Hilbert basis theorem. Since R is Noetherian, R[x₁] is Noetherian by the theorem. Then R[x₁][x₂] = R[x₁, x₂] is Noetherian by applying the theorem again. Inducting on n, R[x₁, ..., xₙ] is Noetherian for every finite n. The key word is 'finite' — the theorem does not extend to infinitely many variables.

Question 4

The converse of the Hilbert basis theorem holds: if R[x] is Noetherian, then R is Noetherian.

• T. True
• F. False

Answer: True

If R[x] is Noetherian, then every ideal of R[x] is finitely generated. Given any ideal I of R, the extended ideal I·R[x] (generated by I inside R[x]) is finitely generated; contracting back to R shows I is finitely generated. More directly: R is a quotient of R[x] (via x ↦ 0), and quotients of Noetherian rings are Noetherian. So the Hilbert basis theorem is actually an 'if and only if': R[x] is Noetherian ⟺ R is Noetherian.

Question 5

Outline the key steps of the proof that R Noetherian implies R[x] Noetherian.

Model answer: Let I be an ideal of R[x]. For each degree d, collect the leading coefficients of all degree-d polynomials in I into a set Lₐ ⊆ R. The sets L₀ ⊆ L₁ ⊆ ··· form an ascending chain of ideals in R. Since R is Noetherian, this chain stabilizes at some degree N, and each Lₐ is finitely generated. Choose polynomials in I whose leading coefficients generate each Lₐ for d ≤ N. Given any f ∈ I, use these polynomials to reduce f's leading coefficient to zero (degree ≥ N) or to reduce its degree (degree < N). The process terminates, showing f is in the ideal generated by the chosen finite set.

The proof works by reducing the problem from R[x] to R: the Noetherian property of R controls the leading coefficients, which is enough to control all of R[x]. This 'leading coefficient' strategy is the ancestor of Gröbner basis theory, where the same idea is systematized for computational purposes.