Bijective correspondence between the set of all $K$-algebra homomorphisms and the set of all zeros of $f$ in $A.$ I am studying Embeddings and zeros from the lecture notes given by our instructor. Here a theorem is mentioned without proof. Here's it $:$ 
Theorem $:$ Let $K$ be a field. Let $A$ be a $K$-algebra and $f \in K[X].$ Then  there is a bijective correpondence between the set $\text {Hom}_{K-\text {alg}} \left (\frac {K[X]} {\langle f \rangle} , A \right )$ of all $K$-algebra homomorphisms from $\frac {K[X]} {\langle f \rangle}$ into $A$ and the set $V_A(f)$ of all zeros of $f$ lying inside $A.$
That means for any given $\sigma \in \text {Hom}_{K-\text {alg}} \left (\frac {K[X]} {\langle f \rangle} , A \right )$ we need to produce a zero of $f$ lying inside $A.$  How do I find that?
Any help in this regard will be highly appreciated. Thank you very much for reading.
 A: Let $K$ a field, $A$ a $K$-algebra, and $f \in K[X]$.  Let $\pi: K[X] \rightarrow K[X]/(f)$ be the canonical projection.
That $A$ is a $K$-algebra means that $A$ is a ring and there is a ring homomorphism $\eta: K \rightarrow A$.  The structure map $\eta$ extends to a morphism $\eta: K[x] \rightarrow A[x]$ (by slight abuse of notation) defined for $g = \sum g_i X^i \in K[X]$ by $\eta(g) := \sum \eta(g_i)X^i$.  
In talking about $V_A(f)$, the set of zeros of $f$ in $A$, we already were implicitly talking about the set of zeros of $\eta(f)$ — there's no way around that.   
Another way to view evaluation of polynomials is as a morphism in and of itself.  In general, if $R$ is a ring and $a \in R$ then we have an 'evaluation-at-$a$' ring homomorphism $\phi_a: R[x] \rightarrow R$ defined by $\phi_a(f) := f(a)$.  When $R$ has the structure of a $K$-algebra, $\phi_a$ also has the structure of a $K$-algebra homomorphism.  
Putting the evaluation and $K$-algebra structure morphism together, we have a canonical $K$-algebra homomorphism that evaluates polynomials of $K[x]$ at a fixed element $a \in A$:  if $\phi_a: A[x] \rightarrow A$ is the relevant 'evaluation-at-$a$' homomorphism, then $\phi_a \eta: K[x] \rightarrow A$ is the homomorphism we want.  
In this notation, $V_A(f) = \{ a \in A \mid \phi_a \eta(f) = 0 \}$. 
 So given an element of $a \in V_A(f)$, we see that dually $f$ is a zero of the $K$-algebra homomorphism $\phi_a \eta$.  
This is important because now the first isomorphism theorem implies that $\phi_a \eta$ factors uniquely through $K[x]/(f)$. 
In other words, we produced a canonical way to map an element $a \in V_a(f)$ to a $K$-algebra homomorphism $K[x]/(f) \rightarrow A$.  
Now we address the inverse of this assignment.  Given any $K$-algebra homomorphism $\sigma: K[X]/(f) \rightarrow A$, consider the composition $\sigma \pi: K[X] \rightarrow A$.  From the definition of a $K$-algebra homomorphism, we deduce that for any $g \in K[X]$, $\sigma \pi(g) = \sum \eta(g_i) \sigma \pi(X)^i$.  In other words, $\sigma \pi$ is just the 'evaluation-at-$\sigma \pi(X)$' homomorphism.  Moreover since $\pi(f) = 0$, we have $\sigma \pi(X) \in V_A(f)$.  

Putting it all together, the bijection $V_A(f) \longleftrightarrow \operatorname{Hom}_{K-\text{Alg}}(K[X]/(f), A)$ is as follows: an element $a \in V_A(f)$ is sent to the unique map $\sigma$ such that $\sigma \pi = \phi_a \eta$ (guaranteed by the first isomoprhism theorem); an element $\sigma \in \operatorname{Hom}_{K-\text{Alg}}(K[X]/(f), A)$ is sent to $\sigma \pi(X)$.

