The subring of $k[x, y]$ generated by $\{x y^{i}: i\geq 0\}$ is not finitely generated over $k$

Question

Let $R$ be the subring of $k[x, y]$ generated by $\{x y^{i}: i\geq 0\}$. Can someone explain why $R$ is not finitely generated as a ring over $k$ (i.e. finitely generated as a $k$-algebra)?

By definition, $R$ is the smallest subring over $k$ containing the set $\{x y^{i}: i\geq 0\}$. So it contains all expressions $x y^{i}$, but it also contains all polynomials in $x y^{i}$ with coefficients in $k$, such as $x y^4 + 7 x y^8$. Now, let's assume, to the contrary, that $R$ is generated by $n$ elements $f_1, f_2, …, f_n$. Then each $f_j$ is a polynomial in $xy^{i}$ (as $i$ ranges over finitely many choices). But where do we get a contradiction from here?

Source. This example is taken from Mariano's answer here.

Answer 1

The subring $R$ is a finitely generated $k[x,y]$-module since, as an ideal it is $(x)$.

However, as a $k$-algebra it cannot be finitely generated. Take $(f_1,...,f_r)$ to be a finite set in $R$ and set $S$ the algebra generated by $(f_1,...,f_r)$. Set $d_i$ the degree of $f_i$ in $x$. Now we are interested in the degree $1$ in $x$ part of elements of $S$. Since we have $\deg_x(f_if_j)>1$ any degree $1$ in $x$ part of elements of $S$ will come from :

$$\sum_{i=1}^r\lambda_if_i $$

Where $\lambda_i\in k$. Now project $S$ in $R/(x^2)$, we see (from the remark above) that the projection of $S$ on $R/(x^2)$ will be (as a $k$-vectorial space) of dimension $\leq r$. However it is clear that $R/(x^2)$ is (as a $k$-vectorial space) isomorphic to $k[y]$ hence infinite dimensional. So the projection of $S$ doesn't fill the whole space $R/(x^2)$, in particular $S\neq R$.

Edit : Based on Mariano's comment. When one talks about ring some (like me) might not ask for the ring to contain a unity and others (like Mariano) ask for a unity on the ring (most common convention). In the answer above, it is not supposed to contain one. If we want $R$ to contain the unity then, as $k$ vectorial space we have $R:=k\oplus (x)$. Since any element in $k\subseteq R$ is of degree $0$ the rest of the proof remains unchanged. The idea is still the following the degree $1$ part in $x$ $R/(x^2)$ is still isomorphic to $k[y]$ whereas the degree $1$ part of $S/(x^2)$ will be given by a $k$-linear combination of the degree $1$ of the generators of $f$. This gives a finite dimensional space which cannot coincide with the degree $1$ part of $R/(x^2)$.

Answer 2

Clément has already provided a nice answer. Here is an alternative solution, based on Mariano's comment here. Comments and criticism are welcome!

Let $I$ be the principal ideal in $R$ generated by the element $(xy)^2$. It suffices to show that $R/I$ is not a finitely generated $k$-algebra. It is not hard to see that a general element of $R/I$ is of the following form: $$ f(x)+xy g(x) + xy^2 h(y) \ \ \ \ \ \ \text{ (mod } I) $$ where $f, g\in k[x]$ and $h\in k[y]$. It is straightforward to compute a product of two arbitrary elements $p_1, p_2$ in $R/I$, namely: $$ p_1 \cdot p_2 = \left(f_1(x)+xy g_1(x) + xy^2 h_1(y)\right)\left(f_2(x)+xy g_2(x) + xy^2 h_2(y)\right) = $$ $$ =f_1(x)f_2(x)+xy\left(f_1(x)g_{2}(x)+f_2(x)g_1(x)\right)+xy^2\left(f_2(0)h_1(y)+f_1(0)h_2(y)\right) \ \ \ \ \text{ (mod } I) $$ Note that a term of the form $xy^{n}$ (for $n\geq 2$) can only come from $xy^2(f_2(0)h_1(y)+f_1(0)h_2(y))$, and so $\deg_{y} p_1 p_2 \leq \max(\deg_{y} p_1, \deg_{y} p_2)$.

Let $p_1, p_2, p_3, …, p_n$ be any $n$ elements of $R/I$. Write: $$ p_k = f_k(x)+xy g_k(x) + xy^2 h_k(y) \ \ \ \ \ \ \text{ (mod } I) $$ as above. Set $N = 3 + \max_{k} \deg_{y} h_k(y)$. Then by the remark above, we see that the algebra generated by $p_1, p_2, …, p_n$ cannot contain the element $x y^{N} \text{ (mod } I)$. Hence, $R/I$ is not finitely-generated algebra over $k$, as desired.