Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 122 / 3 / c

Solution

The algebra must be finite-dimensional, and there is no universal bound on its dimension. Let

e = ε m

n (1) = \sum_{i = 1}^{r} a_{i} \otimes b_{i} .

For any

v \in A

v = \sum_{i = 1}^{r} e (v, a_{i}) b_{i} .

Thus the finitely many

b_{i}

span

A

e

is nondegenerate in the other variable as well, so the maps to the dual space determined by this pairing are isomorphisms. In the standard algebraic terminology this is a Frobenius algebra.

Every positive finite dimension of a vector space occurs: take

A = k^{d}

with coordinatewise multiplication and

ε (a_{1}, \dots, a_{d}) = \sum_{i} a_{i}

. The coordinate idempotents

e_{i}

give

e (e_{i}, e_{j}) = δ_{ij}

and

n (1) = \sum_{i} e_{i} \otimes e_{i}

. If zero unital algebras are allowed, the zero object gives dimension zero too.

The pairing need not make

A

k [t] / (t^{2})

with

ε (a + b t) = b

(0110)

1 \mapsto 1 \otimes t + t \otimes 1

(t)

. Finite-dimensionality is the dimension conclusion; separability is not part of the given definition.

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