Open subfunctors of Affine scheme is not affine

Question

So I was given this problem by my teacher, let $\text{Aff}_k$ be the opposite category of commutative $k$-algebras or $\text{CAlg}_k^{op}$. One can define a functor $\text{Aff}_k\to Fun(\text{CAlg}_k, Set)$ by the definition $$A\mapsto h_A=\hom_{\text{CAlg}_k}(A, -)$$ Call functors of this form affine schemes. Now we can make a definition -

Definition : Given an affine scheme $X=h_A$ define an open subfunctor corresponding to a subset $I\subset A$ as the functor $$D(I)(R)=\{a:A\to R | Ra(I)=R\}\subset h_A(R)=\hom_{\text{CAlg}_k}(A, R)$$

Now the problem states - If $A=k[X, Y]$ and $I=\{X, Y\}$ then show that $D(I)$ is not an affine scheme. That is, there is no $k$-algebra $B$ such that $D(I)\cong h_B$.

To me it seems like a different version of the problem - "The scheme $S=\mathbb{A}_k^2\setminus \{(0,0)\}$ is not affine." As I understand that in that case one shows that the coordinate ring of $S$ is isomorphic to $k[X, Y]$ again which leads to a contradiction. Now what I tried was consider $X=h_A$ to be an affine scheme. Then $$\hom_{Fun(\text{CAlg}_k, Set)}(X, \mathbb{A}^1)=\hom_{Fun(\text{CAlg}_k, Set)}(h_A, h_{k[T]})=\hom_{\text{CAlg}_k}(k[T], A)\cong A$$ So if I can show $\hom_{Fun(\text{CAlg}_k, Set)}(D(I), \mathbb{A}^1)\cong k[X, Y]$ then we are done. But I am not sure how to proceed with this proof. Any help would be appreciated.

Answer 1

For all algebras $C$ let $$U(C):=\{f:k[x,y]\to C:(f(x),f(y))=C\}.$$ Let $B$ be an algebra, for all algebras $C$ let $$h(C):=\hom(B,C),$$ and suppose there is an isomorphism of functors of algebras $\phi:h\to U$. Yoneda's lemma tells us that if $f_0:=\phi_B(1_B):k[x,y]\to B$, an element of $U(B)$, then for all $C$ and all $f:B\to C$ we have $\phi_C(f)=f\circ f_0\in U(C)$.

If $u$, $v:B\to C$ are two maps of algebras and $u\circ f_0=v\circ f_0$, then $\phi_C(u)=\phi_C(v)$ and since $\phi$ is an isomorphism of functors the map $\phi_C$ is injective and therefore $u=v$. We see that $f_0$ is an epi of algebras.

Now the inclusion $p:k[x,y]\to k[x^{\pm1},y]$ is in $U(k[x^{\pm1},y])$, so there is a map $i_p:B\to k[x^{\pm1},y]$ such that $p=i_p\circ f_0$. Similarly, the inclusion $q:k[x,y]\to k[x,y^{\pm1}]$ is in $U(k[x,y^{\pm1}])$, so there is a map $i_q:B\to k[x,y^{\pm1}]$ such that $q=i_q\circ f_0$.

Now let $r:k[x^{\pm1},y]\to k[x^{\pm1},y^{\pm1}]$ and $s:k[x,y^{\pm1}]\to k[x^{\pm1},y^{\pm1}]$ be the inclusions. As $$r\circ i_p\circ f_0=r\circ p=s\circ q=s\circ i_q\circ f_0$$ and $f_0$ is an epi, we have that $$r\circ i_p=s\circ i_q.$$ This implies that $i_p$ and $i_q$ have images contained in $k[x,y]$. By correstricting $i_p$ we obtain two maps $$k[x,y]\xrightarrow{f_0} B\xrightarrow{i_q|^{k[x,y]}}k[x,y]$$ whose composition is the identity, so that the identity of $k[x,y]$ is in $U(k[x,y])$.

It follows from this that the ideal $(x,y)$ of $k[x,y]$ is equal to $k[x,y]$: that is absurd.