Modular block central characters correspond to blocks

Let $\lambda :Z(kG)\to k$ be a unital $k$-algebra homomorphism. Each block idempotent maps to an idempotent of a field, hence zero or one. Orthogonality prevents two values from being one, and their sum is one by F1, so exactly one value is one, at an idempotent $b$. For any $z$, $\lambda (z)=\lambda (bz)$; thus $\lambda$ factors through the single center $Z(B)$.

Fix a block $B\mathrm{\ne }0$. Among dimensions of proper left ideals of $B$ there is a largest, since zero is proper and dimensions are integers less than ${\dim }_{k}B$. A left ideal of that dimension is maximal, and its quotient $S$ is a nonzero simple $B$-module. Extending the action through $kG\to B$ makes it a simple $kG$-module. Each $z\in Z(B)$ acts as a module endomorphism of $S$, hence as a unique scalar by F3. These scalars define a unital $k$-algebra map $\chi :Z(B)\to k$. It is surjective because scalar multiples of $b$ act by those scalars.

The kernel of $\chi$ is a maximal ideal and therefore is F2's unique maximal ideal $J$. Any other unital $k$-algebra map from $Z(B)$ to $k$ has the same kernel. For $z\in Z(B)$, step 1.2 gives $z-\chi (z)b\in J$, so that map must send $z$ to $\chi (z)$. Thus $\chi$ is independent of $S$ and is the unique such map. Extending by ${\lambda }_b(z)=\chi (bz)$ and using step 1.1 proves the bijection and the stated characterization. The group algebra is nonzero, so its block set is not empty; a sole block gives a sole map. Selecting one finite-dimensional ideal for an existence proof uses no AC.