Prove R conatins an ideal that is not finitely generated. $R = F[x,x^{2}y,\ldots,x^n y^{n-1},\ldots]$ and is a subring of $F[x,y]$ where $F$ is a field.
Seems like $R$ itself is not finitely generated and $R$ is an ideal of itself. Im just not positive this would work or really how to show it.
Say $I \le R$ is an ideal and if $I$ were finite then for some $n \in \mathbb{Z}^+$, $I = \{x^{k+1} y^k \mid0 \le t \le n \}$
this is where I get stuck
The ideal $R$ is generated by $(1)$. Remember that for this, you are allowed to multiply by any element in $R$. So for example $x^2y$ is in $(1)$ since $x^2y = (x^2y) \cdot 1$.
Your statement about $I$ is not correct. For example, $I = (x^2y + x^3y^2)$ is finitely generated but not of the form you claim.
To solve the exercise, you want to find some ideal that is not finitely generated. One choice here is $I = (x, x^2y, x^3y^2, \dots)$. The trick is to show that it is not finitely generated. Suppose $I$ is equal to $(x, x^2y, x^3y^2, \dots, x^{N+1}y^N)$. Then is $x^{N+2}y^{N+1}$ in $I$?