Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I don't understand a step in this proof of the Hilbert Basis Theorem. Here is the proof Planeth Math.

I don't understand why $ \mathrm{deg} (f_{N+1}-g)< \mathrm{deg}(f_{N+1}) $. This can only happen if $ g $ has the same degree as $ f_{N+1} $ and also the same leading term, but I don't know why.

share|cite|improve this question
Did you follow the construction of $g$? More specifically, the elements $u_k$ and the exponents $v_k$ were not chosen randomly. – Dylan Moreland Mar 20 '12 at 0:23
They constructed $g$ in such a way that it cancels the lead coefficient of $f_{N+1}$. – Daniel Montealegre Mar 20 '12 at 0:46
yes you are right, i saw it, thanks – Evolution Mar 20 '12 at 2:16

I'll walk you through the proof (and will answer your question in a comment to this answer):

We claim that if $R$ is a Noetherian ring then so is the ring of polynomials $R[x]$.


Our idea to prove this is to show that an arbitrary ideal of $R[x]$ is finitely generated.

To this end, let $I$ be an ideal of $R[x]$. Now we construct a sequence of polynomials $f_k$ as follows: Pick $f_1$ to be of minimal degree in $I$. Pick $f_{k+1}$ to be of minimal degree in $I \setminus \langle f_1 , \dots , f_k \rangle$. Note that by construction, $\mathrm{deg}(f_{k+1}) \geq \mathrm{deg}(f_i)$ for all $i \in \{1, \dots, k\}$.

Now if $f_k (x) = r_n x^n + \dots + r_1x + r_0$ then we denote by $a_k$ its leading coefficient $r_n$. Consider the set $J = \{a_1, a_2, \dots \}$. It's easy to see that this an ideal in $R$: If you add two of its elements $a_i + a_j$ then $a_i + a_j$ is the leading coefficient of $f_i + f_j$ and hence $J$ is closed with respect to addition. Also, if you multiply $a_i$ with an element in $R$, $ra_i$ is the leading coefficient of $r f_i$, so $J$ is also closed with respect to multiplication by elements of $R$. By assumption, $R$ is Noetherian, hence $J$ is finitely generated: $J = \langle a_1, \dots , a_N \rangle$.

We claim that $I = \langle f_1, \dots, f_N \rangle$.

By construction, $I \supset \langle f_1, \dots, f_N \rangle$.

For the other direction of inclusion assume that $I \supsetneq \langle f_1, \dots, f_N \rangle$. Then by the way we constructed $f_k$ we have $f_{N+1} \in I \setminus \langle f_1, \dots, f_N \rangle$. Now we construct a polynomial $g(x)$ with the same leading coefficient as $f_{N+1}$ as follows:

Note that $a_{N+1} = \sum_{i=1}^N a_i s_i$ for $s_i \in R, a_i \in J$ since $J$ is finitely generated. Define $g(x) := \sum_{i=1}^N a_i s_i f_i (x) x^{\mathrm{deg}(f_{N+1}) - \mathrm{deg} (f_i)} $. Then $g(x)$ has leading coefficient $a_{N+1}$ and the same degree as $f_{N+1}$. Hence $f_{N+1} - g(x)$ has degree strictly less than $f_{N+1}$ hence must be in $\langle f_1, \dots, f_N \rangle$, for if not, $f_{N+1}$ would not be of minimal degree in $I \setminus \langle f_1, \dots, f_N \rangle$. So we have $ f_{N+1}(x) - g(x) \in \langle f_1, \dots, f_N \rangle$.

By construction, we have $g(x) \in \langle f_1, \dots, f_N \rangle$. Therefore we also have $f_{N+1} \in \langle f_1, \dots, f_N \rangle$. But this is a contradiction since $f_{N+1} \in I \setminus \langle f_1, \dots, f_N \rangle$.

So we can conclude that we must have $I \subset \langle f_1, \dots, f_N \rangle$ and hence equality.

share|cite|improve this answer
So by construction, the leading term of $g$ is the same as that of $f_{N+1}$. – Rudy the Reindeer Jul 22 '12 at 12:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.