Let $X$ be a scheme over a field $k$ (if necessary, locally of finite type). Let $E$ be a connected component of $X$. Then $E$ is closed. For example, $X$ could be a group scheme over $k$, and $E = X^0$ the connected component of the identity point in $X$.
There are many closed subscheme structures one can place on $E$.
I've seen notes that mention the possibility of $E$ being geometrically connected. A scheme $Y$ over $k$ is said to be geometrically connected if $Y \times_k \overline{k}$ is connected. However, $E$ is just a closed subset of $X$, not yet a scheme.
When people talk about $E$ being geometrically connected, are they saying with respect to a fixed closed subscheme structure on $E$? Or is it the case that the underlying space of $E \times_k \overline{k}$ does not depend on the choice of closed subscheme structure on $E$?
In the case where $X = \textrm{Spec } A$ is affine, it seems what I am doing is comparing the spectra of $A/I \otimes_k \overline{k}$ and $A/\sqrt{I} \otimes_k \overline{k}$.
We can identify $A/I \otimes_k \overline{k} = A \otimes_k \overline{k}/(I \otimes \overline{k})$, and the same for $\sqrt{I}$. I believe that the span of $I$ and $\sqrt{I}$ in $A \otimes_k \overline{k}$ are ideals with the same radical, which tells me that the prime spectra of $A/I \otimes_k \overline{k}$ and $A/\sqrt{I} \otimes_k \overline{k}$ are the same topological space. So if $X$ is affine, there is no problem.
