In the standard proof of the Hilbert basis theorem, we make the inductive construction that $I_0 = 0$ and $I_{i+1} = \langle f_0, \ldots, f_i, f_{i+1} \rangle$ where $f_{i+1}$ is the polynomial in $R[X] - I_i$ of least degree, and make the claim that $f \in I_i$ iff $\text{deg}(f) \leq i$. Why is that true? This plays a central role later on when a polynomial is constructed that contradicts this assertion.
If we consider $R[X]$ as a module it is true, since addition of polynomials and multiplication of a polynomial by an element of the coefficient ring does not change the degree. However, then we would just prove it for modules and not for rings. If $R[X]$ is a ring though, the degree can be changed by multiplication. e.g. $x^3 + 2x^2 + 2x +1 = (x+1)(x^2+x+1) \in \langle x+1, x^2+x+1 \rangle$.
Is it maybe that the basis theorem actually states that if $R$ is Noetherian, then the module $R[X]$ is Noetherian?
Thank you in advance!
The claim is actually that if $\deg f < \deg f_{i+1}$ then $f \in I_i$. (The opposite direction is not claimed).
This is clearly true, since when we are choosing $f_{i+1}$, if such an $f$ existed which was not in $I_i$, we would have chosen that $f$ rather than $f_{i+1}$.
Meanwhile, the point mentioned in the comments - that $f_{i+1} \in I_i$ is a contradiction - is because $f_{i+1}$ was chosen such that $f_{i+1} \notin I_i$.