How to show that $\mathbb{G}_a$ and $\mathbb{G}_m$ are connected?

Question

Let $K$ be a field and $\mathbb{A}^1$ the affine line. Let $\mathbb{G}_a$ be the affine line $\mathbb{A}^1$ with group law $\mu(x, y)=x+y$. Let $\mathbb{G}_m$ be the affine open subset $K^* \subset \mathbb{A}^1$ with group law $\mu(x,y)=xy$. How to show that $\mathbb{G}_a$ and $\mathbb{G}_m$ are connected? I think that we have to show that $\mathbb{G}_a$ (resp. $\mathbb{G}_m$) is not the union of two disjoint proper closed subsets of $\mathbb{G}_a$ (resp. $\mathbb{G}_m$). Here $\mathbb{G}_a$ and $\mathbb{G}_m$ are considered as algebraic groups. Thank you very much.

Answer 1

Recall that a ring $R$ is said to be connected if $R$ doesn't contain any idempotent besides $0$ and $1$. It's a classical result that an affine variety is connected if and only if its coordinate ring is connected as an algebra (you can easily find this on google).

The coordinate rings of $G_a$ and $G_m$ are $K[X]$ and $K[X,Y]/(XY-1)$, respectively. Those are integral domains, and of course an integral domain is connected (check). Ask if you need more details.