Does $R[x]$ Noetherian imply $R[x,y]$ is Noetherian

Question

I am studying Noetherian rings at the moment and I know by the Hilbert Bases Theorem that if a ring $R$ is Noetherian then $R[x]$ is also Noetherian.

My question is: Does this imply $R[x,y]$ is Noetherian?

For example is $\mathbb{Z}[x,y]$ Noetherian?

My thoughts are that you can just view $R[x]$ as the original Noetherian ring and then use the Hilbert Bases theorem again on $(R[x])[y]$ and say it is therefore Noetherian.

But I feel like I'm missing something, thanks for any help.

Answer 1

Seems that what you're missing is that $(R[x])[y]$ is actually $R[x,y]$. But this is actually quite clear: any polynomial in 2 variables $$ P(X,Y)=\sum_{i,j}a_{i,j}X^iY^j $$ can be reordered according the degree of $Y$ and the resulting polynomials in $X$ thought of as "coefficients", namely $$ P(X,Y)=\left(\sum_{i}a_{i,0}X^i\right)+\left(\sum_{i}a_{i,1}X^i\right)Y+ \left(\sum_{i}a_{i,2}X^i\right)Y^2+\cdots. $$

Answer 2

Yes, you are correct - it follows from the Hilbert Basis Theorem that $R[X_1,\dots,X_n]$ is Noetherian for any $n$.