Galois descent for $K$-algebras I'm $\newcommand{\Qbar}{\overline{\mathbb{Q}}}$ trying to understand Galois descent, so here's a simple question.
Let $K$ be a number field. It's clear that $K[x]\times K[y]$ has the property that tensoring over $K$ with $\Qbar$ yields $\Qbar[x]\times\Qbar[y]$.
There surely must exist other $K$-subalgebras $A\subset\Qbar[x]\times\Qbar[y]$ with the same property right? 
Can someone give some examples? In particular I'd like an example where Spec $A$ is connected (if it exists)
 A: One can actually see geometrically where Qiaochu's answer is coming from, and actually classify twists of two copies of the line.
Namely, we think of your object as geometrically a disjoint union of lines. The automorphisms of $\mathbb{A}^1_{\bar{k}}\sqcup\mathbb{A}^1_{\bar{k}}$ are 
$$\mathrm{Aut}(\mathbb{A}^1_{\bar{k}}\sqcup\mathbb{A}^1_{\bar{k}})=\mathrm{Aff}_2(\bar{k})\rtimes (\mathbb{Z}/2\mathbb{Z})$$
which I leave to you as an exercise—the $\mathbb{Z}/2\mathbb{Z}$ coming from the 'switching map'. Moreover, this is actually an isomorphism of $G_k$-modules (where the switching map has trivial $G_k$-action since it's defined rationally), and thus we can classify twists of $\mathbb{A}^1_{\bar{k}}\sqcup\mathbb{A}^1_{\bar{k}}$ as 
$$H^1(G_k,\mathrm{Aut}(\mathbb{A}^1_\bar{k}\sqcup\mathbb{A}^1_{\bar{k}}))=H^1(G_k,\mathrm{Aff}_2(
\bar{k})\rtimes (\mathbb{Z}/2\mathbb{Z}))$$
To compute the this right-hand object we write the short exact sequence
$$1\to \mathrm{Aff}(\bar{k})\to M\to \mathbb{Z}/2\mathbb{Z}\to 1$$
where $M=\mathrm{Aff}_2(\bar{k})\rtimes (\mathbb{Z}/2\mathbb{Z})$. Taking the long-exact sequence in cohomology gives
$$H^1(G_k,\mathrm{Aff}_2(\bar{k}))\to H^1(G_k,M)\to H^1(G_k,\mathbb{Z}/2\mathbb{Z})\to H^2(G_k,\mathrm{Aff}_2(\bar{k}))$$
But, note that 
$$\mathrm{Aff}_2(\bar{k})=\bar{k}^2\rtimes \mathrm{GL}_2(\bar{k})$$
and so taking the long exact sequence in cohomology (and applying Hilbert's theorem 90 and the normal basis theorem) we see that 
$$H^1(G_k,\mathrm{Aff}_2(\bar{k}))=H^2(G_k,\mathrm{Aff}_2(\bar{k}))=0$$
which gives us the isomorphism
$$\{\text{twists of }\mathbb{A}^1_k\sqcup\mathbb{A}^1_k\}=H^1(G_k,M)=H^1(G_k,\mathbb{Z}/2\mathbb{Z})=\mathrm{Hom}_\text{cont.}(G_k,\mathbb{Z}/2\mathbb{Z})$$
so that twists of $\mathbb{A}^1_k\sqcup\mathbb{A}^1_k$ correspond to quadratic extensions. In particular, they correspond to irreducible quadratic polynomials $p(x)$ with the twist being $k[x]/(p(x))$ where the map switching the roots of $p(x)$ becomes the switching map representing $\mathbb{Z}/2\mathbb{Z}$.
A: One way to write $K[x] \times K[y]$ is as $K[x][e]/(e^2 - e)$ (the isomorphism sends $(f(x), g(y))$ to $f(x) e + g(x) (1 - e)$). It's easy to identify other forms of this: we can take $K[x][e]/f(e)$ where $f(e) \in K[e]$ is a quadratic polynomial with distinct roots in $\overline{\mathbb{Q}}$. This will be connected if $f(e)$ is irreducible over $K$, in which case we'll get $L[x]$ where $L$ is a quadratic extension of $K$. 
