Original, constructive proof of Hilbert’s Basis Theorem.

ayushkhaitan3437October 4, 2014November 30, 2016Uncategorized

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 $R$ is a Noetherian ring (every ideal has a finite number of generators), then so is the polynomial ring $R[x]$ .

Let $I\subset R[x]$ be an ideal. It contains polynomials and constants. Let us take the set of all leading coefficients of the polynomials in $I$ , and call it $S$ . This is clearly an ideal! Hence, $S=(s_1,s_2,\dots,s_n)$ . There are polynomials of the form $s_1x^{k_1}+\dots, s_2x^{k_2}+\dots,s_nx^{k_n}+\dots$ in $I$ .

Let $i$ be $\sup\{k_1,k_2,\dots,k_n\}$ .

Now take the set $A=\{0,1,2,\dots i\}$ . The **leading coefficients** of polynomials of degree $j$ for each $j\in A$ , $\cup \{0\}$ , form an ideal. Hence, there must be a finite number of generators for the ideal containing leading coefficients of polynomials of degree $j$ .

For example, take all polynomials of degree $1$ in $I$ . The leading coefficients of such polynomials $\cup\{0\}$ form an ideal (try adding these polynomials or multiplying them with elements from $R$ ). As $R$ is Noetherian, the ideal of leading coefficients has a finite number of generators- say $\{a_1,a_2,\dots,a_t\}$ . It follows that $(a_1x+c_1),(a_2x+c_2),\dots,(a_tx+c_t)$ belong to this ideal of polynomials of degree $1$ , and can generate the leading coefficient of **any** polynomial of degree $1$ in $I$ .

Proceeding by recursion, we will soon have the finite list of generators generating the leading coefficients of all polynomials of degree $\leq i$ . We can hence say that we have the list of polynomials that can be linearly added to generate all polynomials of degree $\leq i$ . This is not difficult to see. It is important to note that we can also linearly generate the polynomials with leading coefficients $\{s_1,s_2,\dots,s_t\}$ .

Important note on notation: let $a_{1j},a_{2j},\dots,a_{t_jj}$ be the finite number of coefficients which generate the leading coefficients of all polynomials of degree $j$ belonging to $I$ .

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 $I$ . For polynomials of degree $\leq i$ , we’ve already shown how. For polynomials of degree $\geq i+1$ , generate all leading coefficients using $\{s_1,s_2,\dots,s_n\}$ (which in turn are generated by $\text{Coeff}$ ), and then generate the rest using $\text{Coeff}$ right away.

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. View more posts

7 thoughts on “Original, constructive proof of Hilbert’s Basis Theorem.”

Abhishek Khetan says:

October 7, 2014 at 4:59 am

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.

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1. ayushkhaitan3437 says:
  
  October 7, 2014 at 6:31 am
  
  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.
  
  Reply
2. ayushkhaitan3437 says:
  
  October 8, 2014 at 4:40 am
  
  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 $\leq i$ . Now let us take an arbitrary polynomial of degree $1$ . Remember that the finite number of generators that we get are elements of $R$ . When we’re talking about the generators of $I\subset R[x]$ , we need to consider the corresponding **polynomials** of those finite number of generators. Those polynomials may be of degree higher than $1$ , and hence may not generate the arbitrary polynomial of degree $1$ .
  
  What does solve the problem is this: the leading coefficients of all polynomials of degree $r$ , $\cup \{0\}$ , 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 $r$ , all elements of each set are also generated by the same elements.
  
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Ishan Mata says:

October 7, 2014 at 9:25 am

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.

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1. ayushkhaitan3437 says:
  
  October 7, 2014 at 1:24 pm
  
  Ishan, I did not start the recursion from $1$ . I started it from $0$ itself. It is just that the explanation that I have given is for $1$ . The explanation could have been about $5$ too, for example, and not impacted the proof.
  
  As for your second point, I did not say that the coefficients themselves generate $I\subset R[x]$ . The **polynomials** of degree $j\leq i$ , with the specified leading coefficients, generate $I$ . That should hopefully clear any doubts you might have.
  
  Reply
  1. Ishan Mata says:
    
    October 7, 2014 at 2:03 pm
    
    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
  2. ayushkhaitan3437 says:
    
    October 7, 2014 at 2:21 pm
    
    Thanks for reading! If only I had been the first one to discover this proof :(. Found out today that someone beat me to it.

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7 thoughts on “Original, constructive proof of Hilbert’s Basis Theorem.”

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