Semi-simple elements of reductive linear algebraic group is dense. I have seen several claims of the fact that the set of semi-simple elements of a connected, reductive linear algebraic group over an algebraically closed field is dense,
moreover that the set of regular semi-simple elements is open dense.
How can I show these facts?
For simplicity it is enough to assume characteristic zero if it simplifies the argument.
 A: Let me indicate a somewhat simple proof, but which contains some exercises (which I can help with later if you’d like).
NB: For simplicity, all fields in the following are assumed perfect.

Definition: Let $G$ be a reductive group over the field $F$. Then, we call an element $\gamma$ in $G(F)$ regular semisimple if
$G_\gamma^\circ$ (the connected component of the centralizer of
$\gamma$ in $G$) is a torus.

Here is an example/exercise to try and familiarize yourself with the notion.

Example/exercise: For an element $\gamma$ of $\mathrm{GL}_n(F)$ show that the following conditions are equivalent:

*

*$\gamma$ is regular semisimple,

*$\gamma$ has distinct eigenvalues (over $\overline{F}$).


To generalize this example/exercise to arbitrary groups let us define the Harish-Chandra discriminant
$$D_G\colon G\to \mathbb{A}^1_F$$
of a reductive group $G$ to be such that for an $F$-algebra $R$ and an $R$-point $\gamma\in G(R)$ one has $D_G(\gamma)$ is the coefficient of $t^r$ in $\det(1+t-\mathrm{Ad}(\gamma))$ (where, as usual, $\mathrm{Ad}(\gamma)\colon \mathrm{Lie}(G)_R\to\mathrm{Lie}(G)_R$ is the adjoint action).
One can check the following more down-to-earth definition of the Harish-Chandra discriminant.

Example/exercise: Let $G$ be a reductive group over the field $F$ and let $T$ be a maximal torus of $G$. Then, for $\gamma\in
 T(F)\subseteq G(F)$ one has that the equations
$$D_G(\gamma)=\det\left(1-\mathrm{Ad}(\gamma)\mid
 \mathrm{Lie}(G)/\mathrm{Lie}(T)\right)=\prod_{\alpha\in
 R(G,T)}(1-\alpha(\gamma))$$ where $R(G,T)$ is the set of (geometric)
roots of $G$ relative to $T$ (i.e. roots of $G_{\overline{F}}$
relative to $T_{\overline{F}}$).

One can then show the following:

Exercise: Let $G$ be a reductive group over a field $F$ and let $\gamma$ be an element of $G(F)$. Then, the following conditions are
equivalent:

*

*$\gamma$ is regular semisimple,

*$D_G(\gamma)\ne 0$,

*there exists a unique maximal torus $T$ of $G$ such that $\gamma\in T(F)$.


I claim that these exercises handle your question, and more! Indeed, Let us denote by $G^\mathrm{rss}$ the open subset of $G$ (as a variety) obtained as the non-vanishing locus of $D_G$ (which is nothing but a regular function in $\mathcal{O}_G(G)$). Note then that $G^\mathrm{rss}(F)$ consists of the regular semisimple of $G(F)$ which, in particular, is contained in the set of semisimple elements of $G(F)$. Note though that since $G$ is connected, it’s actually (geometrically) irreducible and so $G^\mathrm{rss}$ is dense in $G$. But, since $G$ (and thus $G^\mathrm{rss}$) is unirational we deduce that $G^\mathrm{rss}(F)$ is dense in $G$.
In conclusion, we deduce the following:

Theorem: Let $G$ be a reductive group over $F$. Then, the set $G^\mathrm{rss}(F)$ of regular semisimple elements of $G(F)$ is Zariski dense in $G$. A fortiori, the set of semisimple elements of $G(F)$ is Zariski dense in $G$.

