Proof
The following more general statement will be proved.
Theorem. If is a left- (respectively right-) Noetherian ring, then the polynomial ring is also a left- (respectively right-) Noetherian ring.
It suffices to consider just the "Left" case.
Proof (Theorem)
Suppose per contra that were a non-finitely generated left-ideal. Then it would be that by recursion (using the axiom of countable choice) that a sequence of polynomials could be found so that, letting of minimal degree. It is clear that is a non-decreasing sequence of naturals. Now consider the left-ideal over where the are the leading coefficients of the . Since is left-Noetherian, we have that must be finitely generated; and since the comprise an -basis, it follows that for a finite amount of them, say will suffice. So for example, some Now consider whose leading term is equal to that of moreover so of degree contradicting minimality.
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(Thm) |
Proof (Theorem):
Let be a left-ideal. Let be the set of leading coefficients of members of This is obviously a left-ideal over and so is finitely generated by the leading coefficients of finitely many members of say Let Let be the set of leading coefficients of members of whose degree is As before, the are left-ideals over and so are finitely generated by the leading coefficients of finitely many members of say with degrees Now let be the left-ideal generated by We have and claim also
Suppose per contra this were not so. Then let be of minimal degree, and denote its leading coefficient by
Case 1: Regardless of this condition, we have so is a left-linear combination of the coefficients of the Consider which has the same leading term as moreover so of degree contradicting minimality.
Case 2: Then so is a left-linear combination of the leading coefficients of the Considering we yield a similar contradiction as in Case 1.
Thus our claim holds, and which is finitely generated.
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(Thm) |
Note that the only reason we had to split into two cases was to ensure that the powers of multiplying the factors, were non-negative in the constructions.
Read more about this topic: Hilbert's Basis Theorem
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