Proof of Hilbert's Basis Theorem: won't $\deg (f_{i})$ be a strictly decreasing sequence? 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. 
 A: 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.
A: 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?
