Let $G$ be an affine group scheme of finite type over an algebraically closed field $k$. Suppose that $V$ is a finite dimensional representation of $G$. 


For every $k$-algebra $A$ we have the base change representation $V\_A = V\otimes\_kA$ of the group $G\_A = \mathrm{Spec}\, A\times\_{\mathrm{Spec}\,k}G$.


**Definition 1.** *$V$ is absolutely irreducible if for every $k$-algebra $A$ there are no nontrivial subspaces $W\nsubseteq V$ such that $W\_A = A\otimes\_kW$ is $G\_A$-stable in $V\_A$.*


**Definition 2.** *$V$ is absolutely irreducible if for every field extension $L$ of $k$ the representation $V\_L$ of the group $G\_L$ is irreducible.*


Two definitions above capture the same notion. Indeed, if $W\_A$ is $G\_A$-stable for some $k$-algebra $A$, then pick a nonzero morphism $A\rightarrow L$ into a field $L$ over $k$. Then $W\_L$ is $G\_L$-stable. So if there are no stable subspaces in the second sense, there are also no stable subspaces in the first. The other implication is trivial.


**Fact.**
*Suppose that $G$ is reduced. If $V$ is irreducible, then $V$ is absolutely irreducible.*


**Sketch of proof.**
Since $G$ is reduced and $k$ is algebraically closed, we deduce that $V$ is an irreducible representation of the abstract group $G(k)$. Suppose that the dimension of $V$ is $n$ and consider a morphism $\rho:G(k)\rightarrow \mathbb{M}\_n(k)$ inducing action of $G(k)$ on $V$. Extend $\rho$ to a morphism of $k$-algebras $\tilde{\rho}:k[G(k)]\rightarrow \mathbb{M}\_n(k)$. By [Jacobson's density theorem](https://en.wikipedia.org/wiki/Jacobson_density_theorem) we derive that $\tilde{\rho}$ is surjective. This property is stable under base change. This implies that $V\_L$ is irreducible over $G\_L(L)$ and hence $V\_L$ is irreducible over $G\_L$. 


Now my question:



> 
> *Suppose that $G$ is nonreduced. If $V$ is irreducible, then is $V$ absolutely irreducible?*
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Absolutely irreducible representations of affine group schemes of finite type over a field
Suppose $G$ is an affine group scheme over an algebraically closed field $k$, and $V$ is a finite dimensional representation of $G$. Let $K$ be an extension of $k$, and assume that $V\_K$ is reducible as a representation of $G\_K$. Then I claim that $V$ is reducible as a representation of $G$.


Let $r$ be an integer with $0 < r < \dim V$, such that $V\_K$ has a $G\_K$-invariant $r$-dimensional sub-vector space. Let $\mathbb{G}(r,V)$ the Grassmannian of $r$-dimensional vector spaces of $V$, with its induced action of $G$. Since formation of scheme-theoretic fixed point loci commutes extension of base field, the fix point locus $\mathbb{G}(r,V)^G$ has a $K$-valued point. This means that is non-empty; since it is of finite type, and $k$ is algebraically closed, this means that $\mathbb{G}(r,V)^G$ has a $k$-valued point, so $V$ is reducible.


