Let $A = k[z_0, z_1, z_2, \ldots]$ be the polynomial ring in countably many variables over a field $k$. Let $I$ be the ideal generated by the polynomials $z_{i+1}z_j - z_{i}z_{j+1}$ and let $R = A/I$.
I'm trying to get a better sense of $R$. For starters I thought it would be good to check that $R$ is a domain but already I'm struggling to formalize the proof.
One way I was thinking to pursue it was to first check that the homogeneous components of degree $d > 1$ of $R$ are freely generated as $k$-vector spaces by the elements $\bar{z}_0^{d-1} \bar{z}_i$ but I didn't like the way the proof was looking for $d > 2$. Anyway, this approach is already unsatisfactory for me because I'm also interested in quotients of $k[z_0, z_1, z_2, \ldots]$ by homogeneous quadratic ideals that look like $I$ but are generated by $n$-sums of instead of $2$-sums.
Intuitively, if we let $z_1/z_0 = y$, then we find $z_i = z_0 y^i$ in $A/I$.
This suggests that we define a morphism $\varphi \colon A \to k[x,y]$ by $z_i \mapsto xy^i$.
It is easy to see that $I \subseteq \ker \varphi$. Conversely, let $f \in \ker \varphi$. Applying the relations in $I$, the polynomial $f$ is equivalent mod $I$ to some polynomial $g$ of the form $$ h(z_0) + \sum_{i = 1}^n z_i g_i(z_0).$$ The fact that $g \in \ker \varphi$ then clearly implies $g = 0$, hence that $f \in I$. Thus $\ker \varphi = I$, and we have $$A/I \approx k[x, xy, xy^2, \dots] = k \oplus xk[x,y].$$