[This should be a comment, I guess]
This is the kind of question you can answer yourself by simply checking the conditions in the definition are satisfied. You really should do it, independently of what anyone answers here or elsewhere!
You should know, though, that a field can be an algebra over itself in many different ways, as soon as you have a field $K$ such that there are at least two ring homomorphisms $K\to K$ (there is always one such ring homomorphism, the identity map, so to get something interesting, we need two).
It is easy to give examples of fields which do have such non-identity homomorphisms. For example, every Galois extension of $\mathbb Q$, or the field of rational functions $\mathbb C(x)$ (there is a non-identity homomorphism $\mathbb C(x)\to \mathbb C(x)$ which maps $x$ to $x^2$, and in fact many similar other ones), &c.