# On the definition of an algebra

Mathematics Asked by HaKuNa MaTaTa on December 3, 2020

An algebra is given by a triple $$(A,mu,nu)$$ where $$A$$ is a vector field over $$k$$, and
$$begin{gather*} mu colon A otimes A rightarrow A, \ nu colon k rightarrow A end{gather*}$$
are such that:

$$mu$$ is a linear map satisfying $$mu circ (mu otimes mathrm{id}) = mu circ (mathrm{id} otimes mu)$$; which is to say that $$mu$$ is associative.

The map $$nu$$ satisfies $$mu circ (nu otimes mathrm{id}) = gamma$$ where $$gamma$$ is the canonical isomorphism $$gamma colon k otimes A rightarrow A$$; this shows that $$nu(1)$$ is a unit in $$A$$

Can somebody explain to me how the constraint on $$nu$$ is equivalent to saying that $$nu(1)$$ is a unit in $$A$$? Thanks!

Both $$mu circ ( nu otimes mathrm{id} )$$ and $$gamma$$ are linear maps from $$k otimes A$$ to $$A$$. These maps coincide if and only if they are equal on simple tensors, i.e. we have $$mu circ (eta otimes mathrm{id}) = gamma$$ if and only if $$( mu circ (nu otimes mathrm{id}) )(lambda otimes a) = gamma(lambda otimes a)$$ for all $$lambda in k$$ and $$a in A$$. In other words, we need that $$nu(lambda) cdot a = lambda a$$ for all $$lambda in k$$ and $$a in A$$. Here we denote on the left hand side by $$cdot$$ the multiplication on $$A$$ coming from $$mu$$, and on the right hand side we have the scalar multiplication coming from the vector space structure of $$A$$. By the linearity and $$nu$$ and the bilinearity of the multiplication $$mu$$ we can rewritte the left hand side of this required equation as $$nu(lambda) cdot a = nu(lambda 1) cdot a = ( lambda nu(1) ) cdot a = lambda ( nu(1) cdot a ) ,.$$ We can hence rewrite the above equality as $$lambda ( nu(1) cdot a ) = lambda a$$ for all $$lambda in k$$ and $$a in A$$. We see that this holds for all $$lambda$$, $$a$$ if and only if it holds for $$lambda = 1$$ and all $$a$$. In other words, the identity $$mu circ (eta otimes mathrm{id}) = gamma$$ holds if and only if $$nu(1) cdot a = a$$ for all $$a in A$$. But this condition means precisely that $$nu(1)$$ is a multiplicative neutral element of $$A$$ with respect to $$mu$$.

Correct answer by Jendrik Stelzner on December 3, 2020

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