Twisted group algebra coincides with the group algebra?

Question

Let $\mathbb{F}$ be an arbitrary field and $G$ a finite abelian group. Then we can construct the group algebra $\mathbb{F}G$, where, under some conditions on the base field, it will be isomorphic to $\mathbb{F}^{|G|}$ (sum of $|G|$ copies of $\mathbb{F}$). Conversely, if we know that $\mathbb{F}G\cong\mathbb{F}^{|G|}$, then we derive some properties of the field $\mathbb{F}$.

A related construction is the twisted group algebra. Let $\alpha:G\times G\to\mathbb{F}^\times$ be a map satisfying $$ \alpha(u,v)\alpha(uv,w)=\alpha(u,vw)\alpha(v,w),\quad\forall u,v,w\in G. $$ Then $\alpha$ is called a factor set, or a 2-cocycle. We denote by $Z^2(G,\mathbb{F}^\times)$ the (abelian group) of all factor sets.

The twisted group algebra $\mathbb{F}^\alpha G$ is the $\mathbb{F}$-algebra having as a vector space basis $\{X_u\mid u\in G\}$, and the product is given by $X_uX_v=\alpha(u,v)X_{uv}$, for all $u,v\in G$. Note that $\mathbb{F}^\alpha G$ is an associative algebra with unity. Twisted group algebras are related to the so-called projective representations. Also, under some conditions, $\mathbb{F}^\alpha G$ is a direct sum of matrix algebras over $\mathbb{F}$.

My question concerns some kind of the converse of the last statement:

Question. Let $G$ be a finite abelian group, $\mathbb{F}$ an arbitrary field, and $\alpha\in Z^2(G,\mathbb{F}^\times)$. If $\mathbb{F}^\alpha G\cong\mathbb{F}^{|G|}$ (as ordinary $\mathbb{F}$-algebras), then is it true that $\mathbb{F}^\alpha G\cong\mathbb{F}G$ (as ordinary $\mathbb{F}$-algebras)?

Of course, if we assume that $\mathbb{F}$ is algebraically closed, then the answer is "yes".

Any answer, counter-example or reference would be appreciated.

Answer 1

The answer is 'yes', and the following proof is due to M. Kochetov.

Given a finite-dimensional unital commutative algebra $\mathcal{A}$ over $\mathbb{F}$, let $N(\mathcal{A})$ denote the number of algebra homomorphisms from $\mathcal{A}$ to $\mathbb{F}$.

Lemma. $\mathcal{A}$ is a direct sum of copies of $\mathbb{F}$ if and only if $N(\mathcal{A})=\dim\mathcal{A}$.
Proof. Indeed, $(\Rightarrow)$ is obvious, and $(\Leftarrow)$ follows from linear independence of distinct homomorphisms.

Note that both $N$ and $\dim$ are multiplicative over tensor products of algebras.

Now, write $G=\langle g_1\rangle\times\cdots\times\langle g_r\rangle$, direct product of cyclic groups, where each $g_i$ has order $m_i$. Then, we notice that $$ \mathbb{F}^\alpha G\cong\mathbb{F}[x_1]/(x_1^{m_1}-a_1)\otimes\cdots\otimes\mathbb{F}[x_r]/(x_r^{m_r}-a_r), $$ where $a_i=\prod_{j=1}^{m_i-1}\alpha(h_i,h_i^j)\in\mathbb{F}$. In the same way, $\mathbb{F}G\cong\bigotimes_{i=1}^r\mathbb{F}[x_i]/(x_i^{m_i}-1)$.

From the remarks above, $\mathbb{F}^\sigma G$ is a direct sum of copies of $\mathbb{F}$ if and only if each $\mathbb{F}[x_i]/(x_i^{m_i}-a_i)$ is a direct sum of copies of $\mathbb{F}$. The former condition is equivalent to $\mathbb{F}$ contains $m_i$ distinct $m_i$-roots of $a_i$; and this implies that $\mathbb{F}$ contains a primitive $m_i$-th root of $1$. Thus, each $\mathbb{F}[x_i]/(x_i^{m_i}-1)$ is a direct sum of $\mathbb{F}$'s. Hence, by the remarks above, $\mathbb{F}G$ is a direct sum of copies of $\mathbb{F}$.