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.

It seems that the two definitions 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 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?

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 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?

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 $G$.

Definition. $V$ is absolutely irreducible if for every $k$-algebra $A$ the representation $V_A = V\otimes_kA$ of the group $G_A = \mathrm{Spec}\, A\times_{\mathrm{Spec}\,k}G$ is irreducible (there are no $G_A$-stable $A$-submodules of $V_A$).

I think that I can prove the following result.

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

Now my question:

Suppose that $G$ is nonreduced. If $V$ is irreducible, then is $V$ absolutely irreducible?

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.

It seems that the two definitions 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 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?