An argument in Hilbert's basis theorem

Question

Hilbert's basis theorem : If $R$ is a Noetherian ring, then $R[X]$ is a Noetherian ring.

The following proof is from Fulton's algebraic curves. I am skipping some parts of the proof and coming directly to my doubt.

Proof: We first choose a set $J$ which contains all the leading coefficients of any ideal $I$ which we want to prove, is finitely generated. We prove that $J$ is an ideal in $R$ and hence finitely generated, say, $(a_1,\dots,a_r)$ with the corresponding polynomials $f_1,\dots, f_r$. To generate the elements of lower degree in $I$, we take other 'similar' sets $J_m \; \forall m \leq N $ where $N $ is greater than the $\max\{ deg(f_1), \dots, deg(f_r) \}$. We denote the corresponding polynomials for $J_m$ by $\{ f\}_{mj}$.

Next, we claim that $I$ is generated by $f_1,\dots, f_r$ and all $\{ f\}_{mj}$. Proof given in book: Let's call the latter as $I'$. We can easily see that $I'\subset I$. Suppose $I'$ were smaller than $I$; let $G$ be an element of $I$ of lowest degree that is not in $I'$. If $deg(G) > N$, we can find polynomials $Q_i$ such that $\sum Q_iF_i$ and $G$ have the same leading term.

argument: But then $deg(G - \sum Q_iF_i ) < deg G$, so $G - \sum Q_iF_i \in I'$, so $G \in I'$.

Doubt: I don't understand why the following is true? If $deg(G - \sum Q_iF_i ) < deg G$, then $G - \sum Q_iF_i \in I'$. It might be possible that both $I$ and $I'$ might have polynomials of all degree.

Answer 1

It seems that you don't want to proceed with lowest degree element. Even if you don't start with that, assume $G \in I \backslash I'$ whose degree is, say, $m >0$. Based on the similar argument, you can say that the degree of $G - \sum F_iQ_i$ is less than degree of $G$. You can keep doing this until you get to 0 in which case, we have a $constant \in I \backslash I'$. Hence, $I = k[X_1,\dots, X_n]$. Contradiction.

Answer 2

We know $I' \subseteq I$ and we suppose they are not equal. Then there is an $x \in I$ that is not in $I'$. Each such $x$ has a degree, and thus I may choose such an $x$ of minimal degree and call it $G$. However, the element $G-\sum Q_iF_i$ is obviously in $I$, and, by construction, has smaller degree than $G$. Thus, by the choice of $G$, $G-\sum Q_iF_i$ must be in $I'$.

Just to illustrate the point about choosing an element of smallest degree: if $R=k[x,y]$, $I=(x,y^3)$ and $J=(x,y)$, then $I \subsetneq J$. Even though both of these ideals contain polynomials of every (nonzero) degree, I can still choose an element of $J$ not contained in $I$, e.g., $y^2$. To choose an element of smallest degree of $J$ that is not in $I$ could be to take $y$ or $x+y$, there are many, but the degree of such an element must be $1$.