If $\alpha$ is transcendental, why is the evaluation homomorphism $k[t] \to k(\alpha)$ not onto? What does a general element in $k(\alpha)$ look like?

Question

Let $k$ be a field of characteristic $0$.

Isn't every element in $k(\alpha)$ a polynomial in $\alpha$ over $k$?

If so, then I don't see why the evaluation homomorphism is not onto.

If every element of $k(\alpha)$ is a not a polynomial in $\alpha$ over $k$, then I must ask: What does a general element look like?