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Say we have an ideal $I\subset R[X]$. We select a set of polynomials $f_{1},f_{2},f_{3},\dots$ such that $f_{i+1}$ has minimal degree in $I\setminus (f_{1},f_{2},f_{3},\dots f_{i})$.

Can't $\deg (f_{i})$ be a strictly decreasing sequence?

For example, let the ideal $I\setminus (f_{1},f_{2},f_{3},\dots f_{i})$ contain polynomials of degree $2$ or greater. Say it contains the polynomials $x^2+2$ and $(x^2+2)^2+(x+1)$. Then $f_{i+1}=x^2+2$ and $f_{i+2}=x+1$!

Motivation: In this proof of Hilbet's Basis Theorem, we have constructed a polynomial $g=u_{1}f_{1}x^{n_{1}}+\dots u_{n}f_{n}x^{n_{N}}$, where $n_{i}=\deg (f_{N+1})-\deg(f_{i})$. I feel that then $n_{i}$ shoud be negative; and as $x$ raised to negative powers is not defined in $R[X]$, I'm having trouble understanding how $g=u_{1}f_{1}x^{n_{1}}+\dots u_{n}f_{n}x^{n_{N}}$ has been defined.

2 Answers2

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No. If $\deg f_{i+1}<\deg f_i$, then you would have chosen $f_{i+1}$ in place of $f_i$ in the previous step.

Going through the example in the OP.

Assume that the polynomials $p_1=(x^2+2)$ and $p_2=(x^2+2)^2+(x+1)$ belong to the ideal $I$, but do not belong to the smaller ideal $J_{i-1}=(f_1,f_2,\ldots,f_{i-1})$ generated by the polynomials $f_i$ picked earlier. The polynomial $p_3=x+1=p_2-p_1^2$ belongs to the ideal $I$. There are two possibilities.

I) If $p_3$ does not belong to the ideal $J_{i-1}$, then we will not be allowed to pick $f_i=p_1$, because $p_3$ has a lower degree and does not belong to the ideal $J_{i-1}$ either.

II) If $p_3$ belongs to the ideal $J_{i-1}$, then it will never be picked. Not now, not in the future.

So the scenario that we might pick $p_1$ in this round, and $p_3$ in the next is impossible.

Jyrki Lahtonen
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  • Neither $f_i$ nor $f_{i+1}$ was in the ideal $(f_1,\ldots,f_{i-1})$, so... Or yet in other words, the pool of polynomials you can select the next one from SHRINKS all the time. – Jyrki Lahtonen Jun 30 '13 at 18:19
  • Could you kindly explain how could we have chosen $(x+1)$ before choosing $x^2+1$ in the example that I have given? I suppose I'm still a little confused. –  Jun 30 '13 at 18:28
  • Oh god. I suppose $(x+1)$ is anyway part of the ideal as $(x^2 +1)^2 + (x+1)-(x^2+1)(x^2+1)=(x+1)$. I really don't know what's happening to me I'm asking the stupidest of questions. Thanks a lot for your time! –  Jun 30 '13 at 18:50
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    @Ayush: Don't worry about it. Happens to all of us. – Jyrki Lahtonen Jun 30 '13 at 18:52
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Moreover, $\deg(f_i)$ is a sequence of natural numbers (choose your own convention on what the degree of $0$ is here). Can you envision a strictly decreasing sequence of positive integers?

citedcorpse
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  • No. But that is the point! The essence of the proof is that there is no such infinite decreasing sequence of positive integers, and hence the ring $R[X]$ is neotherian. –  Jun 30 '13 at 18:29
  • @AyushKhaitan exactly, our supposition leads to a contradiction. it is a very nonconstructive proof (big names might even call it theology). if explicitness (err, the SFW kind) is what you seek, i recommend reading about Gröbner bases – citedcorpse Jun 30 '13 at 18:34