## 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.

• $A=0$ is a counterexample. Please specify the restrictions that are necessary here. What is $n$ for example? Commented Dec 1, 2023 at 14:43
• Krull dimension of an ideal? Commented Dec 1, 2023 at 20:42
• I have edited this question. Commented Dec 2, 2023 at 2:33
• @MartinBrandenburg Sure, there's a Krull dimension of modules, or it could have been something like height or something. Glad to see it's been resolved now. Commented Dec 2, 2023 at 3:10
• @rschwieb Thank you for your suggestion . I'll avoid this issue next time Commented Dec 3, 2023 at 2:49

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 , $$dim_k I=1$$.

• Does $M=A\{b\}$ refer to generating $M$ with one element? I think also you can't say $b=x_1ay_1$: when $b\in \langle a\rangle$ you only know $b=\sum x_iay_i$ for some finite sum. Commented Dec 2, 2023 at 17:29
• The proof is not accurate but the idea is correct. The equations $b=x_1ay_1$ and $c=x_2by_2$ are not generally true. Instead, we have $b=\sum_i x_iay_i$ and $c=\sum_i w_ibz_i$ for some $x_i,y_i,w_i,z_i\in A$. The equation $b=x_1ay_1$ is never used and the equation for $c$ can be simplified as follows. Each $w_ib=\lambda_i b$ for some scalar $\lambda_i\in k$, and so $c=b(\sum_i\lambda_i w_i)$. We can re-write this as $c=bw$ for some $w\in A$. Since $cA=kc$ and $Ab=kA$, the proof ends as in the OP: $I=AcA=Ac=Abw=kbw=kc$. Commented Dec 2, 2023 at 21:31
• @rschwieb@Allen Bell Thank for your corrections.I have edited my answer. Commented Dec 3, 2023 at 2:48

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.