I tried to prove the Hilbert's Basis Theorem and don't know if it is correct or not.
Hilbert's Basis Theorem. If $R$ is a Noetherian ring, then $R[X]$ is a Noetherian ring.
Proof:
We know that $R$ is Noetherian iff every ideal is finitely generated i.e. for any ideal $I \subset R, \ \exists \ a_{1},\dots, a_{n} \in R$ such that $I=\langle a_{1},\dots, a_{k}\rangle:=\sum_{i=1}^{k}Ra_{i}$.
I'd like to prove that if $J$ is an ideal of $R[T]$, then it must have the form $I[T]\setminus J'$ with some ideal $I \subseteq R$ and $J'$ consisting of polynomials with coefficients in $I$, but it has a maximal degree (this is because if $J$ is an ideal of $R[T]$, that doesn't mean it must have polynominals of degree $1;...$ up to some $n$ but it must have $0$ since it is an additive subgroup).
We are going to assume that this ideal $J$ is not of the form that we've described. Now, since $J $ is an ideal of $R[T]$, $\forall f \in R[T]$, and, $g \in J, fg \in J$ but we could choose any $f$ and $g$ that we want, so $J$ can't have finite order (PS: If $R$ is not an integral domain, we should still get the same thing since if we choose $f \in R[T]$ such that the last coefficient is $1_{R}$ and we have any other $a \in R \backslash \{0\}$, then $a *1_{R}=a \neq 0$).
But now, a polynominal in $R[T]$ could have any coefficients in $R$ and so if we define$f(T):= \sum_{\nu=0}^{n}r_{\nu}T^{\nu} \in R[T]$ and $g(T):=\sum_{\nu=0}^{m}a_{\nu}T^{\nu} \in J \subseteq R[T]$, then: $(fg)(T)=\sum_{\nu=0}^{n} \sum_{\mu=0}^{m} r_{\nu} a_{\mu} T^{\nu+\mu} \in J$, but since $J$ is also an additive subgroup of $R[T]$, all the coefficients of the polynominals in it form an additive subgroup of $R$ and hence they form an ideal.
Therefore, we get that $J$ must have the form $J:=I[T]\backslash J'$ as described at the beggining of this part and if we want to write it as a set (to make no confusion on why we chose that $J'$), $J:=\{0;f;...\}$, where $deg(f)=n+1$. Let's call $n+1$ the minimal degree of $J$.
$The\, second\, part$: In this part we want to prove that $J$ is finitely generated by using the fact that $I$ is so.
We first see that every polynominal in $f \in J$ has the form: $f:=\sum_{\nu=n+1}^{n+1+m}a_{\nu}T^{\nu}=T^{n+1}(a_{n+1}+...+a_{n+m+1}T^{m})$. This tells us that every polynominal in $J$ is just a polynominal in $I[T]$ multiplied by that extra $T^{n+1}$. Now we will do induction on the degree of any $f \in I[T]$ to prove that $I[T]$ is finitely generated which implies that so is $J$(note that the induction part on the smallest degree of $f \in I[T]$ is the same thing as the induction on the minimal degree of $J$):
$P(0)$: For $deg(f)=0 $ we get that $f$ could be written as $\sum_{i=1}^{n}r_{i}a_{i}$ since $I=<a_{1};...;a_{k}>$. So $I[T]$ has the form $<a_{1};...;a_{k};f_{1};...;f_{l}>$ where $f_{1};...;f_{l}$ are polynominals but we don't know if $l \longmapsto \infty$ or not.
$P(1)$: For $deg(f)=1$, we define $f$ as $f:=aT+b \in I[T]$ and the idea is similar as for in $P(0)$( write a $a$ and $b$ as in $P(0)$).
$P(m) \Longrightarrow P(m+1)$: Let's assume that every polynominal of degree $m$ is finitely generated, namely $f:=\sum_{i=1}^{l'}p_{i}g_{i} \in I[T]$, where whe chose the polynominals $g_{i}$ and $l'$ for abreviation( $I[T]=<g_{1};...;g_{l'}>$) and $p_{i} \in R[T]$. Now let's take the polynominal $q \in I[T]$ of degree $m+1$ such that $q:=f+aT^{m+1}$ but $aT^{m+1}=aT^{m}T$ and $deg(aT^{m})=m$ and hence $aT^{m}$ has the form $\sum_{i=1}^{l'}p'_{i}g_{i}$ with $p'_{i} \in R[T]$ and hence $aT^{m+1}$ has the form $ \sum_{i=1}^{l'}(Tp'_{i})g_{i}$. Hence, $q$ has the form $\sum_{i=1}^{l'}p_{i}g_{i}+\sum_{i=1}^{l'}(Tp'_{i})g_{i}= \sum_{i=1}^{l'}(p_{i}+Tp'_{i})g_{i} $.
Hence, since the induction holds, this concludes the proof of the Hilbert's basis theorem
Please tell me if I made any mistakes.