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Let say you have a ring $R={\left(\begin{array}{cccc} \mathbb{C}[x,y] & y^2\mathbb{C}[x,y]\\ xy\mathbb{C}[x,y] & \mathbb{C}[x,y] \end{array} \right)}$

It it enough to prove that it's a subring of $M_2(\mathbb{C}[x,y])$ as then from that being a finitely generated $\mathbb{C}$-algebra I.e. it's generated by 8 elements(Is it generated by 8 elements?). Isn't it automatic that the subring is noetherian because all finitely generated rings are Noetherian.

Also, it is true that a subring is finitely generated if the ring is finitely generated?

I fear it isn't enough. However, in my notes he makes weird arguments like this. In proof that

$R={\left(\begin{array}{cccc} \mathbb{Z} & 3\mathbb{Z}\\ \mathbb{Z} & \mathbb{Z} \end{array} \right)}$ is Noetherian.

He says that it contains the ring of scalars which is isomorphic to $\mathbb{Z}$ over which is 4 generated. From this he concludes it is Noetherian.

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He who? ${}{}$ –  Mariano Suárez-Alvarez Nov 17 '11 at 2:12
    
@MarianoSuárez-Alvarez My lectures notes. If the reasoning is unclear I could write the whole paragraph of him proving the bottom ring is Noetherian. –  simplicity Nov 17 '11 at 3:00
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  • It is not true that a subring of a finitely generated ring is finitely generated, as you ask in your 3rd paragraph. Consider (in the similar context of $\mathbb C$-algebras...), for example, the subring generated by $\{xy^i:i\geq0\}$ in $\mathbb C[x, y]$.

  • It is not true that all finitely generated rings are Noetherian, as you say at the end of your 2nd paragraph. The free algebra $k\langle x,y\rangle$ on two generators $x$ and $y$ is not left noetherian.

  • Let $R={\left(\begin{array}{cccc} \mathbb{Z} & 3\mathbb{Z}\\ \mathbb{Z} & \mathbb{Z} \end{array} \right)}$. As a $\mathbb Z$-module, this is Noetherian, because it is in fact free and finitely-generated. An increasing sequence of ideals of $R$ is in particular an increasing sequence of sub-$\mathbb Z$-modules, so it stabilizes after a finite number of steps. Therefore $R$ is Noetherian as a ring. You should try to come up with a general statement from this argument, and then try to use it to deal with your initial example.

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I can't see the last point I.e. that the increasing sequence of ideals of R is an increasing sequence of sub-$\mathbb{Z}$-modules. –  simplicity Nov 17 '11 at 2:55
    
An ideal of $R$ is in particular an abelian subgroup, that is, a sub-$\mathbb Z$-module. –  Mariano Suárez-Alvarez Nov 17 '11 at 2:57
    
I understand the module bit. However, I can't see how you can conclude it is stationary i.e. stabilizes after a finite number of steps argument. –  simplicity Nov 17 '11 at 3:02
    
Do you understand why $R$ is a noetherian $\mathbb Z$-module and, in particular, what this means? –  Mariano Suárez-Alvarez Nov 17 '11 at 3:20
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