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In this answer to an old post, it is claimed: For an algebra $A$ with an augmentation $\epsilon:A\to k$, let $I$ be the kernel of $\epsilon$, then elements that can not be written as products, can be thought of as elements of $I/I^2$.

Does this mean, every element $a\in A$ such that $a\neq bc$, for all nonunits $b,c \in A$, lies in the kernel $I$?? But it seems I can always define a $k$-homomorphism $\epsilon$ such that $\epsilon(a)\neq 0$. Which is true? Or maybe I misunderstood something?? Thanks.

user760
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    Yeah that answer is at best misleading/confusing. But it's not true that you can always define a homomorphism $\varepsilon$ such that $\varepsilon(a) \neq 0$. What if $a^2 = 0$ for example? – diracdeltafunk Nov 18 '23 at 16:03
  • @diracdeltafunk Thanks. It seems to me, we can only say the quotient $I/I^2$ contains all the "irreducibles" in the kernel, which could be none at all. – user760 Nov 18 '23 at 16:12
  • @diracdeltafunk maybe not all the irreducibles in the kernel even, if some of these equal the sum of reducibles. – user760 Nov 18 '23 at 16:17

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