= Solution
For $\alpha\in L$ with $f(\alpha)=0$, evaluation $K[t]\to L$, $g\mapsto g(\alpha)$, has $(f)$ in its kernel. It therefore induces a $K$-algebra homomorphism $\sigma_\alpha:K[t]/(f)\to L$. Conversely, such a homomorphism is determined by $\alpha=\sigma(\bar t)$, since every class is a <polynomial> in $\bar t$. Applying it to $f(\bar t)=0$ shows $f(\alpha)=0$. These constructions are inverse, giving the <root-embedding correspondence>
$$
\boxed{\operatorname{Root}_f(L)\longleftrightarrow\operatorname{Hom}_K(K[t]/(f),L)}.
$$
Irreducibility ensures the quotient is a field, so each unital homomorphism is injective.
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