A Frobenius algebra is a vector space that is both an algebra and a coalgebra in a compatible way. (See here for a precise definition.) I guess that a subalgebra of a Frobenius algebra is not again a Frobenius algebra? What is an instructive example that demonstrates this?
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asked Oct 19, 2022 at 12:25
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19$\begingroup$ Matrix algebras over a field are Frobenius and every finite dimensional algebra embeds in a matrix algebra but many of them are not Frobenius $\endgroup$– Benjamin SteinbergCommented Oct 19, 2022 at 12:34
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2$\begingroup$ The ring $K[x]/x^{n+1}$ is a Frobenius algebra, with the coproduct sending $x^i$ to $\sum_{j+k=n+i}x^j\otimes x^k$, but the subalgebra generated by $\{x^i:i>1\}$ is not. $\endgroup$– Neil StricklandCommented Oct 19, 2022 at 15:04
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2 Answers
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Any finite dimensional $k$-algebra $A$ is also a subalgebra of its trivial extension $$T(A)=A \oplus \rm{Hom}_k(A,k),$$ which is Frobenius. In this way you get examples where the Frobenius algebra is not semi-simple.
The (symmetric) $k$-bilinear form on $T(A)$ is given by $$\langle (a,f),(b,g) \rangle = g(a)+f(b),$$ for $(a,f),(b,g) \in T(A)$, see section 3 in:
Bessenrodt, Christine; Holm, Thorsten; Zimmermann, Alexander, Generalized Reynolds ideals for non-symmetric algebras., J. Algebra 312, No. 2, 985-994 (2007). ZBL1119.16001.
answered Oct 19, 2022 at 14:45
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$\begingroup$ Wow, that's a handy way to produce a Frobenius algebra I haven't run across. What's the associated form? I was guessing something like $(a,f)(b,g)=f(b)g(a)$... maybe the $A$ module action has to be like this: $(a\cdot f)(x)=f(ax)$? $\endgroup$– rschwiebCommented Oct 19, 2022 at 17:21
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$\begingroup$ @rschwieb, math.stackexchange.com/questions/229412/… proves it symmetric although it doesn’t spell out the form. It shows it is isomorphic to its dual as a bimodule. $\endgroup$ Commented Oct 19, 2022 at 17:31
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$\begingroup$ @BenjaminSteinberg Looks like my conjecture about the form was incorrect, although it looks like there's echoes of that in the post linked... I guess sometime I'll have to unravel it. $\endgroup$– rschwiebCommented Oct 19, 2022 at 17:47
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2$\begingroup$ @rschweib The symmetric bilinear form is $f(b)+g(a)$. I will add a reference. $\endgroup$ Commented Oct 19, 2022 at 18:08
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$\begingroup$ @DagOskarMadsen ooh, I was close. Please use autocomplete to get my name's vowels in the right order :) I hate to miss updates... $\endgroup$– rschwiebCommented Oct 20, 2022 at 15:27
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To make Benjamin Steinberg's (very apt) observation concrete, one may take $T_2(F)$, the upper triangular matrix ring over a field.
It is not Frobenius because it is not self-injective, but it's certainly a subring of a Frobenius algebra, namely $M_2(F)$.
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$\begingroup$ One can find many finite dimensional algebras that aren't Frobenius in this DaRT query; however, there's no way to filter down to which ones are actually f.d. algebras over fields explicitly. $\endgroup$– rschwiebCommented Oct 19, 2022 at 13:54