I was wondering the following two question :
Suppose $k$ is a vector space, $A$ is a $k$-algebra.
$1.$ If $A$ has an nondegenerate associative bilinear form; associative means $\langle ab,c\rangle=\langle a,bc\rangle$, then is $A$ necessarily finite dimensional?
$2.$ If $A$ has an associative bilinear form (not necessarily nondegenerate) such that $\langle 1,1\rangle\neq 0$, where $1\in A$ is the unit element with respect to its multiplication map, then is the given bilinear form necessarily nondegenerate? If not, what kind of examples do we have?
Thank you for any help... Hope you have a nice day!
Consider the algebra of Laurent polynomials in the variable $x$, and define a bilinear form putting $(x^n,x^m)$ to be $1$ or $0$ according to whether $n+m$ is equal to zero or not, and extending using bilinearity.
As for your second question, the answer is very much no. Consider a vector space V and the algebra $A=k\oplus V$ in which the summand $k$ is spanned by the unit and the product of any two elements of $V$ is zero. There is an associative bilinear form on this algebra which maps each pair of elements to the component in $k$ of their product. This is a counterexample.