From the 3rd edition of the book "The Linear Algebra a Beginning Graduate Student Ought to Know" by Jonathan S. Golan, we find the following exercise (number 287) under chapter 6:
"Let $p$ be a prime integer and let $F$ be a field of characteristic $p$. Let $(K,\bullet)$ be an associative and commutative unital $F$-algebra and let $\alpha : v \rightarrow v^p$. Show that $\alpha$ is an isomorphism of unital F-algebras."
Now an isomorphism is a bijective homomorphism. Clearly, our map is a homomorphism due to the freshman's dream (for any $a,b \in K$):
$$ (a+b)^p = a^p + b^P $$
And due to the fact that for any $c \in F$, we may observe that: $ c^p = c*c^{p-1} = c*e = c $, since our underlying multiplicative group for $F$ has order $p-1$ (since our zero is not a part of it). Hence, for all $v \in K$:
$$ (cv)^p = c^pv^p=cv^p$$
Therefore, we have a homomorphism. Now consider the case that the algebra $K$ is not finitely generated, but, let's say, is indeed generated by a single element plus the identity, i.e. the following is a basis for $K$ as a vector space for $F$:
$$\{e, v, v^2, v^3, v^4, ...\}$$
It seems, that with such an algebra, we will not have a surjective homomorphism and hence we cannot have an isomorphism. This can be noted as our mapping "$\alpha$" sends us to a subspace of $K$ strictly generated by:
$$\{e, v^p, v^{2p}, v^{3p}, ...\}$$
Furthermore, even if the algebra is finitely generated, it is possible that the statement doesn't apply. E.g. take $\{e, v, v^2, v^3, ..., v^{p-1}\}$ as a basis and define $v^p = e$. In this case, we are mapping to the underlying field $F$ as identified with the subspace $\{ ke \ | \ k \in F\}$.
So, the statement of the question does not appear to apply in general to either infinitely or finitely generated algebras (with the properties in question). So, can someone clarify was the likely intended meaning of the question must have been? Many thanks.