How can a ring of polynomials with coefficients in a field $k$, and in infinitely many variables be a subring of $k[x,y]$? I am trying to answer this question:
Let $k$ be a field and $k[x,y].$ Define the subring $A \subset k[x,y]$ by $A = k[x, xy, xy^2, xy^3, ...].$ Show that $A$ is not Noetherian.
But I have the following question:
How can a ring of polynomials with coefficients in a field $k$, and in infinitely many variables, in our case $A$ be a subring of $k[x,y]$ which is a ring of polynomials with coefficients in a field $k$, and in finitely many variables ? should not $A$ be at least a finite set? could anyone give me an example to clarify this point please?
 A: An element of $A$ can be written as a finite linear combination $\sum p_\alpha X^\alpha$ of monomials $X^\alpha$ where each monomial is of the form
$$X^\alpha = (x y^{i_1})^{j_1} \cdots (x y^{i_n})^{j_n}$$
At the end, you get an element of $k[x,y]$.
The variables of $A$, namely $x, xy, \dots$ are not independent.
A: I think it's a notational issue:
Let $S\le R$ be (commutative) rings, and $A=\{a_1,a_2,\dots\}\subseteq R$.
Then $S[a_1,a_2,\dots]$ simply denotes the subring generated by $S\cup A$, i.e. the smallest subring of $R$ that contains both $S$ and $A$.
Note that, however, there's a homomorphism $\varphi$ from the polynomial ring $S[x_1,x_2,\dots]$ to $R$ that maps $x_i\mapsto a_i$, and $S[a_1,a_2,\dots]$ is just the image of $\varphi$.
A: Consider the polynomial ring $A = k[xy^i \,|\, i \in \mathbb Z_{\geq 0}].$ By definition, an element of $A$ is of the form $$c_1 (xy^{i_{1,1}})^{n_{1,1}} \cdots (xy^{i_{1, \ell_1}})^{n_{1, \ell_1}} + c_2 (xy^{i_{2,1}})^{n_{2,1}} \cdots (xy^{i_{2, \ell_2}})^{n_{2, \ell_2}} + \cdots + c_k (xy^{i_{k,1}})^{n_{k,1}} \cdots (xy^{i_{k, \ell_k}})^{n_{k, \ell_k}}$$ for some scalars $c_j$ in $k,$ some distinct integers $i_{j, m} \geq 0,$ and some integers $n_{j, m} \geq 0.$ By the fourth sentence of the following link, "it is possible to speak of a ring of polynomials in an infinite set of variables if it is assumed that each individual polynomial depends only on a finite number of variables." Considering that the variables of $A$ are $x, xy, xy^2, xy^3,$ etc., it follows that a polynomial in $A$ is a finite $k$-linear combination of products of powers of the variables $x, xy, xy^2, xy^3,$ etc., and so, we arrive at a concrete description of a typical element of $A.$
Bearing this all in mind, it follows that $-x$ and $3(xy^2)^3 - \frac 1 7 (xy^9)^2$ and $(xy^2)^3 (xy^3)^2 + (xy^2)^2$ are all elements of $A.$ But as these are all polynomials in the variables $x$ and $y$ with coefficients in $k,$ these are all elements of $k[x, y],$ as well.
Ultimately, the fact that $A \subseteq k[x, y]$ follows from the fact that all of the monomial generators of $A$ belong to $k[x, y]$: $xy^i$ is a monomial in $k[x, y]$ for all integers $i \geq 0.$ Consequently, any $k$-linear combination of powers of the generators of $A$ belongs to $k[x, y]$ so that $A \subseteq k[x, y].$
We may now use the subring test to prove that $A \subseteq k[x, y]$ as rings. By the previous paragraph, we have that $A \subseteq k[x, y].$ Considering that $1 = 1x^0$ is an element of $A,$ it follows that $A$ contains the multiplicative identity of $k[x, y].$ Clearly, $A$ is closed under subtraction: the difference of two polynomials in $A$ is again a polynomial in $A,$ as it is a $k$-linear combination of products of powers of the generators of $A.$ Last, by the distributive property of $A$ (indeed, this holds because $A$ is by definition a ring), any product of elements in $A$ is itself another element of $A$ because the product $(xy^i)^{n_i} (xy^j)^{n_j}$ of any powers of any two monomial generators of $A$ is itself a polynomial in $A.$ By the subring test, therefore, we conclude that $A$ is a subring of $k[x, y].$
