A question about the Jacobson radical

Question

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 $1$.

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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.

Answer 1

Let $I$ be a non-zero minimal ideal of $ A$ (must exist since $dim_k \ A$ is finite) , then $I=\langle a \rangle $ for some $a \in A$ . We show that $dim_k \ I=1$ .

Notice that $A/J(A) \simeq k^n $ for some $n\geq 1$ , then simple left A-modules have $k$-dimension 1. Similarly, simple right A-modules have $k$-dimension 1.

$I$ is a non-zero left A-module, then it contains a simple left-submodule $M$ since $dim_k \ I$ is finite . Suppose that $M=A\{ b \}$ for some $b \in I$, then $b=\sum x_iay_i$ for some $x_i ,y_i\in A$. Notice that $xb \in kb$ for any $x\in A$ . Consider the ideal $\langle b \rangle$ , then
$$0 \subset \langle b \rangle \subseteq I$$ From the minimality of $I$ , we have $I=\langle b \rangle$ .

Repeat the operation in last paragraph, we can find an element

$$c=\sum u_ibw_i=\sum (u_ib)w_i=\sum (\lambda_ib)w_i=\sum b(\lambda_i w_i)=b(\sum \lambda_i w_i)=bw \in I$$

such that $w\in A$ and $cx \in kc $ for any $x\in A$ and $I= \langle c \rangle $.

Then we can see that $I= \langle c \rangle =kc$ , thai is , $dim_k I=1$.

Answer 2

Since $A$ is finite-dimensional over $\mathsf k$, its Jacobson radical $J=J(A)$ is nilpotent, i.e. there is some $m\geq 1$ with $J^m=0$. Let $d \in \mathbb Z_{>0}$ be the minimal such $m$, so that $J^{d-1}\supsetneq J^d=\{0\}$. Now $J^{d-1}$ is a 2-sided ideal in $A$, and the action of $A$ on $J^{d-1}$ factors through $A/J$ by our choice of $d$, and by assumption $A/J\cong \mathsf k^n$.

Let, for $i \in \{1,2,\ldots,n\}$ the element $e_i\in A$ be such that $\{e_i+J: 1\leq i \leq n\}$ are the primitive idempotents of $A/J$ so that $1+J = \sum_{i=1}^n e_i +J$ and $e_ie_j +J= \delta_{ij}+J$ (where $\delta_{ij} = 1$ if $i=j$ and is zero otherwise). Thus if $g \in J^{d-1}\backslash \{0\}$, then $$ g=1.g.1 = (\sum_{i_1=1}^n e_{i_1}).g.(\sum_{i_2=1}^n e_{i_2}) = \sum_{i_1,i_2} e_{i_1}ge_{i_2} $$ so there must exist $j_1,j_2 \in \{1,\ldots,n\}$ such that $e_{j_1} g e_{j_2} \neq 0$. Let $r = e_{j_1} g e_{j_2}$ and consider the two-sided ideal $ArA$. Now if $a_1,a_2 \in A$ then $a_1e_{j_1} =\lambda e_{j_1} +b_1$ and $e_{j_2}a_2 = \mu e_{j_2} + b_2$ where $\lambda, \mu \in \mathsf k$ and $b_1,b_2 \in J$. Thus we have $$ a_1 r a_2 = (a_1e_{j_1})g(e_{j_2} a_2) = (\lambda.e_{j_1}+b_1).g.(\mu.e_{j_2}+b_2)= \lambda\mu (e_{j_1} g e_{j_2}) = \lambda\mu.r $$ so that $\mathsf k.r$ is a two-sided ideal with $\dim_{\mathsf k}(\mathsf k.r) =1$ as required.