Edit: After reading proofs of this theorem, it seems to me that this is similar in spirit to an already established proof (Proof 2 on ProofWiki, for instance).

This is a constructive proof of Hilbert’s Basis Theorem.

Hilbert’s Basis Theorem says that if is a Noetherian ring (every ideal has a finite number of generators), then so is the polynomial ring .

Let be an ideal. It contains polynomials and constants. Let us take the set of all leading coefficients of the polynomials in , and call it . This is clearly an ideal! Hence, . There are polynomials of the form in .

Let be .

Now take the set . The **leading coefficients** of polynomials of degree for each , , form an ideal. Hence, there must be a finite number of generators for the ideal containing leading coefficients of polynomials of degree .

For example, take all polynomials of degree in . The leading coefficients of such polynomials form an ideal (try adding these polynomials or multiplying them with elements from ). As is Noetherian, the ideal of leading coefficients has a finite number of generators- say . It follows that belong to this ideal of polynomials of degree , and can generate the leading coefficient of **any** polynomial of degree in .

Proceeding by recursion, we will soon have the finite list of generators generating the leading coefficients of all polynomials of degree . We can hence say that we have the list of polynomials that can be linearly added to generate all polynomials of degree . This is not difficult to see. It is important to note that we can also linearly generate the polynomials with leading coefficients .

Important note on notation: let be the finite number of coefficients which generate the leading coefficients of all polynomials of degree belonging to .

It is easy to see that generates the whole of . For polynomials of degree , we’ve already shown how. For polynomials of degree , generate all leading coefficients using (which in turn are generated by ), and then generate the rest using right away.

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The fact that “the set leading coefficients of polynomials of degree 1 form an ideal of R” is not true. For any ideal should contain 0 and leading coefficient can’t be zero. Probably you mean “the set of coefficients of x of all the polynomials of degree at most 1”. Your proof may still work but I haven’t attempted to fix it myself. The attempt seems promising. I have some of ideas for a proof too which I am willing to discuss over email.

Abhishek, you’re right. And your suggested edit fixes the problem. The rest of the proof can proceed in the same manner. I found out this morning that wikipedia lists a similar proof as the “Second Proof” in its article on Hilbert’s Basis Theorem. So that was a letdown.

Abhishek, I was thinking about your suggested edit yesterday. It does not solve the problem. For example, let us take the ideal generated by the leading coefficients of polynomials of degree . Now let us take an arbitrary polynomial of degree . Remember that the finite number of generators that we get are elements of . When we’re talking about the generators of , we need to consider the corresponding **polynomials** of those finite number of generators. Those polynomials may be of degree higher than , and hence may not generate the arbitrary polynomial of degree .

What does solve the problem is this: the leading coefficients of all polynomials of degree , , form an ideal. As all elements of the union of these two sets are generated by a finite number of elements, all of which are polynomials of degree , all elements of each set are also generated by the same elements.

Hello Ayush,

I have read your proof in a hurry and will read it again more carefully but for now I have two comments to make.

The first is that the recursion should start from degree zero and not from degree 1. However this is a minor and repairable error.

The second is, if I understand it correctly, you seem to be proving something stronger which is not necessarily true : every ideal of $R[X]$ can be generated by a finite set of elements of $R$.

The set Coeff need not be a subset of $I$. E.g. the ideal $(x)$ in $\mathbb{Z}[X]$ does not contain any non-zero integers. May be your proof still works but just needs to be re-written more correctly. Regards, Ishan.

Ishan, I did not start the recursion from . I started it from itself. It is just that the explanation that I have given is for . The explanation could have been about too, for example, and not impacted the proof.

As for your second point, I did not say that the coefficients themselves generate . The **polynomials** of degree , with the specified leading coefficients, generate . That should hopefully clear any doubts you might have.

Thanks for your clarifications. I read your proof more carefully and it works. It is an interesting proof, thanks for sharing ! 🙂

By the way what had confused me was your first statement in the last para “It is easy to see that \text{Coeff}=\{a_{10},\dots,a_{t_00};a_{11}\dots,a_{t_1,1};\dots,a_{1i},\dots,a_{t_ii}\} generates the whole of I.” The elements of this set are elements of $R$ (being coefficients of terms in polynomials and not terms themselves), and not polynomials. You may like to correct the typo. In the view of your clarification, I now understand what you meant. Thanks. With regards, Ishan

Thanks for reading! If only I had been the first one to discover this proof :(. Found out today that someone beat me to it.