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Suppose that we have a short exact sequence of $K$-algebras, for a field $K$. $$0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0,$$ where $A$ is isomorphic to some ideal of $B$ and $B/A \simeq C$ as $K$-algebras.

If this SES splits does that automatically mean that $B \simeq A \oplus C$ as $K$-algebras? I know that, as vector spaces ($K$-modules) they are certainly split.

I vaguely recall that this 'split' iff 'direct sum' property is not necessarily true, but I'm unable to find a resource that actually shows this, and I'm starting to wonder if I just made it up because rings and algebras are poorly behaved mathematical structures.

dbossaller
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    What is a short exact sequence in the category of $K$-algebras? – dan_fulea Oct 14 '21 at 17:40
  • Are you working with unital $K$-algebras? If so, then short exact squences aren't very interesting, because for $B/A$ to make sense, $A$ needs to be an ideal in $B$. But if it's also supposed to be a unital $K$-algebra, then $A=B$ and $C=0$ – Lukas Heger Oct 14 '21 at 17:45
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    To address the 'split iff direct sum' property: When an exact sequence $0 \to A \to B \to C \to 0$ splits, then certainly you have $B \cong A \oplus C$. However the converse is not necessarily true, there are sequences $0 \to A \to A \oplus C \to C \to 0$ that are not split, see for example this question to have a counterexample when working with modules. – Rushy Oct 14 '21 at 17:49
  • @dan_fulea. In the category of $K$-algebras, a short exact sequence consists of two unital algebras $B$ and $C$ and a non-unital algebra $A$ such that: 1) $A \simeq I$ for some ideal of $B$, 2) $B/I \simeq C$, 3) There exist unital homomorphisms $\phi: A \rightarrow B$, and $\psi: B \rightarrow C$, such that $\ker(\psi) = \text{im}(\phi)$. – dbossaller Oct 15 '21 at 02:28

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This depends on your choice of algebra structure on $A \oplus C$. If one takes the standard product of the direct sum of algebras: $(a_1, c_1)(a_2, c_2) = (a_1a_2, c_1c_2)$, we don't have $B \cong A \oplus C$ in general.

To find the correct multiplication operation, we can first split the above sequence as a sequence of vector spaces and then transport the multiplication operation from $B$ to $A \oplus C$. Let us write $s : C \to B$ for the splitting of the short exact sequence. Note that $B$ is an $A$-bimodule, whose multiplication preserves the image of $s$. So, this makes $C$ into an $A$-bimodule. Then, we see that the product $$(a_1, c_1)(a_2, c_2) = (a_1a_2, a_1c_2 + c_1a_2 + c_1c_2)$$ makes $B$ isomorphic to $A \oplus C$.

Sven Holtrop
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