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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.

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  • $\begingroup$ 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 $f$. $\endgroup$ – reuns Dec 7 '19 at 7:41
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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 $\sigma \pi(X)$.

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