$k[x]$ and its subfield as a finitely generated $k[t]$ Module

Question

The following is from "Aluffi Chapter 0" V.1.3

1.3. Let $k$ be a field, and let $f \in k[x], f \notin k$. For every subring $R$ of $k[x]$ containing $k$ and $f$, define a homomorphism $\varphi: k[t] \rightarrow R$ by extending the identity on $k$ and mapping $t$ to $f$. This makes every such $R$ a $k[t]$-algebra (Example III.5.6).

• i) Prove that $k[x]$ is finitely generated as a $k[t]$-module.
• ii) Prove that every subring $R$ as above is finitely generated as a $k[t]$-module.
• iii) Prove that every subring of $k[x]$ containing $k$ is a Noetherian ring.

My question is about my approach:

i) For first we can define trivial isomorphism of rings as:

$$g:k[t]\to k[x]\\f(t)\mapsto f(x)$$

Which makes $k[x]$ a $k[t]$-Module

Since they are isomorphic module $k[t]$ as $k[t]$-Module is the same as $k[x]$ as $k[t]$ module which is then finitely generated by $1\in k[x]$. However this is a different $k[t]$ module structre on $k[x]$ than what question asked. Then I guess we should use $\varphi: k[t] \to R$ by extending the target as $$\varphi: k[t] \to R\subset k[x]$$ so via this $k[x]$ is a $k[t]$ module

I think I am so confused, any hint, answer would be appreciated.

Answer 1

Your function $g$ is not an isomorphism of rings. What if $f(x)=x^2$? Then you're only getting even degree polynomials in your image. But in that case $k[x]$ has a generating set with $2$ elements, $\{1,x\}$, because you can get every polynomial from these two elements by linear combinations with coefficients that are even degree polynomials. You need to generalize this. Try the division algorithm.

Answer 2

1.3(i) we need to prove $\forall x \in k[x]$ x is a $k[f]$-linear combination.the idea is to use division with remainders.see here.then as a finitely generated module over a Noetherian ring $k[f]$,$k[x]$ is a Noetherian $k[f]$-module.

1.3(ii) $R$ is a $k[f]$-submodule of $k[x]$ since $k[f]\subseteq R$,then $R$ is finitely generated as $k[f]$-module since $k[x]$ is a Noetherian $k[f]$-module and $R$ is a Noetherian $k[f]$-module.

1.3(iii)$\forall$ subring $R$,if $k=R$,k is Noetherian. elsewise $\forall$ ideal $I$ of $R$,we have $k\subsetneq I$,$\exists$ monic polynomial $f\in I$.since $k[f]\subseteq I\subseteq R$,use(ii):$I$ is finitely generated as $k[f]$-module since $R$ is a Noetherian $k[f]$-module which means $I$ is a $k[f]$-linear combination hence $I$ is a finitely generated ideal