Finitely generated k-algebra is Noetherian

Question

If any finitely generated algebra is Noetherian, does that mean that $k[x]$ itself is Noetherian? But in this case we can take ideal $k[x^2] \subset k[x]$ which is not finitely generated.

Answer 1

Firstly, I assume that $k[x] = k[y_1,...,y_n]$ a polynomial rings in finitely many variables? At least that's what I would use to define a finitely generated $k$-algebra.

Now this statement above is an application of Hilbert's basis theorem. Proofs can be found in the wiki-article. However, maybe it's a good exercise to try this on your own. Once you have this theorem, the rest follows quickly by induction over $n$ and the fact that the quotient of a noetherian ring is noetherian again.