### Original, constructive proof of Hilbert’s Basis Theorem.

#### by ayushkhaitan3437

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.

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.