Question
$k$ is a field, $A$ is a $k$-algebra with finite $k$-dimension, $J(A)$ is its Jacobson radical.
If $A/J(A) \simeq k^n$ for some $n\geq 1$ , is it true that A has an ideal of $k$-dimension $1$?
Of course $A/J(A)$ must have an ideal of $k$-dimension $1$, I want to find an ideal in $A$ of $k$-dimension $1$.
This question comes from a problem about coalgebra , see 《Hopf algebras and Their Actions on Rings》 6.3.4.
Let $D$ is a non-zero pointed coalgebra with finite $k$-dimension , then $D$ has a subcoalgebra $E$ with codimension 1.
Proof: Let $E$ be a maximal proper subcoalgebra of $D$ , then $E^{\perp}$ is a minimal non-zero ideal of $D^{*}$.
Using the theorey of coalgebra, one can see that
$$D^{*}/J(D_0) = D^{*}/D_0^{\perp} \simeq (D_0)^{*} \simeq k^n$$
Then the author claims that $E^{\perp}$ must has $k$-dimension 1 , but I cannot prove it.