Associative bilinear form that is unital I encountered a vector space with an associative bilinear map that is also unital... And I feel like this implies that this vector space is actually an algebra over the underlying field. Am I right?
Associative is the condition $u(u(x,y),z)=u(x,u(y,z))$ and unital is the condition that there exists $e\in V$ the underlying vector space $V$ such that $u(x,e)=u(e,x)=x$ for all $x\in V$.
Thank you. Have a nice day!
 A: This is actually a definition of an algebra (which makes it apparent how to define a coalgebra).
Def 1: Let $K$ be a field. An (associative unital) $K$-algebra is a triple of data $(A, \mu, \eta)$ such that $A$ is a $K$-vector space, $\mu: A\times A\to A$ is a bilinear map called multiplication map, and $\eta: K\to A$ a linear map called unit. Such that


*

*Associativity: For all $x,y,z\in A$, one has $\mu(\mu(x,y), z)=\mu(x, \mu(y,z))$.

*Unital: For all $x\in A$ and $\alpha\in K$, $\mu(\eta(\alpha),x)=\alpha x=\mu(x, \eta(\alpha))$.


with this definition the element $e\in A$ that you mentioned is $e=\eta(1)$. Sometimes, using universal property of tensor product people instead define $\mu: A\otimes A\to A$ (linear) instead. The latter comes in handy actually.
Def 2: Let $K$ be a field, $A$ a ring equipped with a homomorphism $f: K\to A$. Then $A$ is called a $K$-algebra.
To see the equivalence of these definitions, say we start from def 2: Define $\eta$ as the underlying $K$-linear map of $f$. Also since $A$ is a ring, it already comes equipped with a associative bilinear map $\mu: A\times A\to A$. Moreover $A$ is obviously a $K$-vector space with scalar product $\alpha x=\mu(f(\alpha),x)=\mu(x, f(\alpha))=x\alpha$. So done.
Conversely, say we start with def 1: Then together $(A,\mu)$ is a ring. Now consider the map $\eta: K\to A$. Note that if $\alpha, \beta\in K$, then
$$
\eta(\alpha\beta)=\alpha \eta(\beta)=\mu(\eta(\alpha), \eta(\beta))
$$
Hence $\eta$ is actually a ring homomorphism, not just a $K$-linear map.
