"$L/K$ forms of each other"

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In section $4$ of these notes, the author says two algebraic groups $G$ and $H$ defined over a field $K$ are "$L/K$ forms of each other" if they are "isomorphic over $L$", where $L$ is a finite field extension of $K$. I'm trying to understand what this means.

Given an algebraic group $G$ defined over $K$, is the product of varieties $G\times L$ an algebraic group over $L$? Here, we consider $L$ to be $\mathbb{A}_K^n$ where $[L:K]=n$. If this is true, does this phrase "isomorphic over $L$" mean that $G\times L\cong H\times L$ as algebraic groups over $L$? If so, can someone give me an example of two non-isomorphic algebraic groups $G$ and $H$ defined over $\mathbb{R}$ such that $G\times\mathbb{C}\cong H\times\mathbb{C}$? Examples in positive characteristic would be great too if they are readily available.

If I'm way off track, please help me understand section $4$ of the linked notes. Thanks!

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This is not quite right. Instead of the product $G\times \mathbf A^1_L$, you should take the base-change $G_L:=G\times_{\operatorname{Spec}K}{\operatorname{Spec}L}$. The algebraic group structure carries over to $G_L$, which makes it into an algebraic group over $L$.

For example, consider the elliptic curves

$$E: \quad y^2=x(x+1)(x+2),$$ $$E': \quad -y^2=x(x+1)(x+2).$$

They are both defined over $\mathbf R$, but they are not isomorphic over $\mathbf R$. However, over $\mathbf C$, we have an isomorphism $E_\mathbf C \to E_\mathbf C'$ given by $(x,y) \mapsto (x,iy)$:

$$-y^2=x(x+1)(x+2)\quad \Rightarrow \quad(iy)^2=x(x+1)(x+2).$$

Thus, $E$ and $E'$ are $\mathbf C/\mathbf R$-forms of each other.

Of course, there are many other examples.

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For an example in positive characteristic, let $k$ be a finite field and let $k^\prime$ be an extension of degree $d>1$. The restriction of scalars $T=\mathrm{Res}_{k^\prime/k}(\mathbf{G}_m)$ of the multiplicative group over $k^\prime$ is an algebraic group over $k$ characterized up to $k$-isomorphism by the equality $T(R)=(R\otimes_kk^\prime)^\times$ for all $k$-algebras $R$. It is an example of a non-split torus. More precisely, $T$ becomes isomorphic to $\mathbf{G}_m^d$ after base change to $k^\prime$, but it is not isomorphic to $\mathbf{G}_m^d$ over $k$, as can be seen by the Galois action on the character groups (the character group of $\mathbf{G}_m^d$ has trivial $\mathrm{Gal}(\bar{k}/k)$-action, while the character group of $T$ is isomorphic to $\mathbf{Z}[\mathrm{Gal}(k^\prime/k)]$ as a $\mathrm{Gal}(\bar{k}/k)$-module).

This example works more generally with $k^\prime$ a non-trivial finite separable extension of any field $k$ ($T$ will then become isomorphic to a product of copies of $\mathbf{G}_m$ over any extension of $k$ containing a normal closure of $k^\prime$).

A different sort of example arises from a finite-dimensional central simple algebra $B$ over a field $k$ of dimension $n^2$. Such an algebra satisfies $B\otimes_k\bar{k}\cong\mathrm{Mat}_n(\bar{k})$, and in fact becomes isomorphic to the $n\times n$ matrix algebra after base change to an appropriate finite extension of $k$. There is an algebraic $k$-group $G_B$ characterized up to isomorphism by the equality $G_B(R)=(R\otimes_kB)^\times$ for all $k$-algebras $R$, and if $k^\prime/k$ splits $B$, then $G_B\times_kk^\prime\cong\mathrm{GL}_n$.