# Is there a quotient of $k[X,Y]$ which is an integral domain but not a subring of $k[x_1, \dots, x_n]$

Question:

Let $k$ be the field $\Bbb Q$ or $\Bbb R$. Can we find a prime ideal $P$ of $k[X,Y]$ such that $k[X,Y]/P$ is not isomorphic to a subring of $k[x_1, \dots, x_n]$ (whatever $n \geq 0$ is)?

My question above comes from the following observation: the ring $\Bbb C[x,y]/(y^2-x^3)$ is isomorphic to the subring $\Bbb C[t^2,t^3]$ of $\Bbb C[t] \subset \Bbb C[t,y]$ (but not isomorphic to $\Bbb C[t]$ itself, however). The quotients of $\Bbb C[x_1, \dots, x_m]$ occur frequently in algebraic geometry, and I was a bit surprised to see that such a quotient could be… a subring of $\Bbb C[x_1, \dots, x_m]$.

Another example the quotient $\Bbb C[x,y]/(xy-1)$, i.e. the localization $\Bbb C[x]_{x}$, is clearly (isomorphic to) a subring of $\Bbb C(x)$. Thinking about this, I ended up with the following surprising result:

The quotient ring $\Bbb C[x_1, \dots, x_m]/I$ is an integral domain if and only if it is isomorphic to a subring of $\Bbb C[x_1, \dots, x_n]$ for some $n \geq 0$.

Proof: The direction $\Longleftarrow$ is easy, since a subring of an integral domain is an integral domain.

But the direction $\implies$ is less clear. Assume $K[x_1, \dots, x_m]/I$ is an integral domain, where $K$ is any field of characteristic $0$ of cardinality $2^{\aleph_0}$ (e.g. $K=\Bbb C$). Denote by $L$ the algebraic closure of the fraction field of $K[x_1, \dots, x_m]/I$. Since the algebraically closed field $L$ has characteristic $0$ (because $I \cap \Bbb Z \subset I \cap K = \{0\}$) and its cardinality is $2^{\aleph_0}$, it is isomorphic to $\Bbb C$ (see for instance here). Therefore, $K[x_1, \dots, x_m]/I \subset L \cong \Bbb C$, and we are done (with $n=0$). $\hspace{5cm}\blacksquare$ (In particular, $\Bbb R[x]$ is isomorphic to a subring of $\Bbb C$ ! This is less difficult to see for $\Bbb Q[x] \cong \Bbb Q[\pi] \subset \Bbb R \subset \Bbb C$).

However, I should mention that I didn't find an explicit embedding $\Bbb C[x,y](y^2-x(x+1)) \hookrightarrow \Bbb C[x_1, \dots, x_n]$ for instance. (Notice that $Q(x,y):=y^2-xP(x) \in \Bbb C[x,y]$ is irreducible whenever $P \in \Bbb C[x]$ satisfy $P(0) \neq 0$, i.e. $x \not\mid P$, so that $(Q)$ is a prime ideal of the UFD $\Bbb C[x,y]$. It is indeed sufficient to apply Eisenstein critertion, $x$ being irreducible in the UFD $\Bbb C[x]$).

The argument above strongly relies on uniqueness properties of algebraically closed fields, so it can't be applied to my original question about $k=\Bbb Q,k=\Bbb R$. The quotient $k[x,y]/P$ is a finitely generated $k$-algebra, but I don't see how to relate this to some polynomial algebra.

I tried to work with $k[x,y]/(y^2-x^3-x)$ and $n=1$. In particular, I looked for the existence of polynomials $P,Q \in k[t]$ such that $P(t)^2 = Q(t)^3+Q(t)$. It was a bit long, and anyway I'm not if this is useful. Notice that the group of units of $k[x,y]/(xy-1)$ is $k^* \times \Bbb Z$, which is isomorphic to a subgroup of $k^*$.

• I found another interesting example: let $k$ be $\Bbb Q$ or $\Bbb R$. Then $k[x,y]/(xy-1) \cong k[t,t^{-1}]$ doesn't embed as ring in $k[x_1,\dots,x_n]$, for if $$g : k[t,t^{-1}] \hookrightarrow k[x_1,\dots,x_n]$$ is an injective ring morphism, then the units $\{a t ^n \mid a \in k^*, n \in \Bbb Z\}$ of $k[t,t^{-1}]$ must be mapped to the units $k^*$ of $k[x_1,\dots,x_n]$, so we actually get a morphism $g : k[t,t^{-1}] \hookrightarrow k$. Its restriction to $k$ is the identity (see here), and then $g(t) = g(g(t)) \in k$, so that $g$ is not injective. Nov 4, 2016 at 8:01
The field $\mathbf{R}[X]/(X^2+1)$ is in fact a field but not contained in $\mathbf{R}[X_1,X_2,\ldots,X_n]$. Here is the idea for producing more such examples. Hilbert's Nullstellenstatz gives bijective correspondence between maximal ideals and points of affine space only for algebraically closed fields. For other fields $K$ we can find maximal ideals which correspond only to a point in an extension field. SO quotients by such maximal ideals will not be a subring of a polynomial ring over $K$.
• Yes, of course… Do you have an example with a prime ideal $P$ that is not maximal (then it must be in $k[X,Y]$, not in the PID $k[X]$)? Yes I have: $P=(X^2+1) \triangleleft \Bbb R[X,Y]$… Nov 3, 2016 at 13:57