Hilbert's basis theorem

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In mathematics, Hilbert's basis theorem states that every ideal in the ring of multivariate polynomials over a field is finitely generated. This can be translated into algebraic geometry as follows: every algebraic set over a field can be described as the set of common roots of finitely many polynomial equations. The theorem is named for the German mathematician David Hilbert who first proved it in 1888.

Hilbert produced an innovative proof by contradiction using mathematical induction; his method does not give an algorithm to produce the finitely many basis polynomials for a given ideal: it only shows that they must exist. One can determine basis polynomials using the method of Gröbner bases.

[edit] Proof

A slightly more general statement of Hilbert's basis theorem is: if R is a left (respectively right) Noetherian ring, then the polynomial ring R[X] is also left (respectively right) Noetherian.

For , if with a_n not equal to 0, then degf: = n and a_n is the leading coefficient of f.
Let I be an ideal in R[x] and assume I is not finitely generated.
Then inductively construct a sequence f₁,f₂,... of elements of I such that f_{i + 1} has minimal degree among elements of , where J_i is the ideal generated by f₁,...,f_i.
Let a_i be the leading coefficient of f_i and let J be the ideal of R generated by the sequence a₁,a₂,....
Since R is Noetherian there exists N such that J is generated by a₁,...,a_N.
Therefore for some .
We obtain a contradiction by considering where n_i = degf_{N + 1} − degf_i, because degg = degf_{N + 1} and their leading coefficients agree, so that f_{N + 1} − g has degree strictly less than degf_{N + 1}, contradicting the choice of f_{N + 1}.
Thus I is finitely generated.
Since I was an arbitrary ideal in R[x], every ideal in R[x] is finitely generated and R[x] is therefore Noetherian.

Or a constructive proof:
Given an ideal J of R[X] let L be the set of leading coefficients of the elements of J.
Then L is clearly an ideal in R so is finitely generated by a(1),...,a(n) in L, and there are f(1),...,f(n) in J with a(i) being the leading coefficient of f(i).
Let d(i) be the degree of f(i) and let N be the maximum of the d(i)'s.
Now for each k=0,...,N-1 let L(k) be the set of leading coeficients of elements of J with degree atmost k.
Then again, L(k) is clearly an ideal in R so is finitely generated by a(k,1),...,a(k,m(k)) say.
As before, let f(k,i) in J have leading coefficient a(k,i).
Let H be the ideal in R[X] generated by the f(i)'s and f(k,i)'s.
Then surely H is contained in J and assume there is an element f in J not belonging to H, of least degree d, and leading coefficient a.
If d is larger or equal to N then a is in L so, a=r(1)a(1)+...+r(1)a(n) and g= r(1)X^{d − d(1)}f(1) + ... + r(n)X^{d − d(n)}f(n) is of the same degree as f and has the same leading coefficient.
Since g is in H, f-g is not, which contradicts the minimality of f.
If on the other hand d is strictly smaller than N, then a is in L(d), so a=r(1)a(d,1)+...+r(m(d))a(d,m(d)).
A similar construction as above again gives the same contradiction.
Thus, J=H, which is finitely generated.
QED.

[edit] Other

The Mizar project has completely formalized and automatically checked a proof of Hilbert's basis theorem in the HILBASIS file.

[edit] References

• Cox, Little, and O'Shea, Ideals, Varieties, and Algorithms, Springer-Verlag, 1997.