# 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.

• If $\sigma$ is a $K$-homomorphism $K[X] \to A$ with $\sigma(f(X))=0$ then $f(\sigma(X))=0$ ie. $\sigma(X)$ is a root of $f$ in $A$. Do you see why it is quite the same as a $K$-homomorphism $K[X]/(f(X))\to A$ ? How do you find such a root of $f$ in $A$ ? Well it depends on what is your $A$ and $f$. – reuns Dec 7 '19 at 7:41

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 $$\sigma \pi(X)$$.