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Let $R$ be a (commutative) ring with $1$. For every $R$-Algebra $A$ we define $F_A: A^n \rightarrow A$ with the property that

\begin{equation}\varphi(F_A(a_1,\ldots,a_n)) = F_B(\varphi(a_1),\ldots,\varphi(a_n)) \qquad (1)\end{equation}

, where $\varphi: A \rightarrow B$ is a homomorphism between $R$-Algebras.

Show that there is a unique polynomial $P \in R[X_1,\ldots,X_n]$ such that $P_A = F_A$.

I understand that $P_A$ suffices the condition $(1)$, but $P \rightarrow P_A$ is in general not bijective, so our $P$ is not unique. I do not know what to do now. Could you please help me?

I have learned the following definitions:

$R$-Algebra: Let $A$ and $R$ be rings. Then $A$ is called an $R$-Algebra if there exists a ring-homomorphism $s: R \rightarrow A$.

$R$-Homomorphism: Let $A$ and $B$ be $R$-Algebras. A ring-homomorphism $f: A \rightarrow B$ is called an $R$-homomorphism if $f \circ s_A = s_b$.

($P_A$ is the map $P_A:A^n \rightarrow A, a \mapsto P(a)$, where $P \in R[X_1,\ldots,X_n]$)

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    $\begingroup$ Isn't $F_A$ defined for all $R$-algebras $A$? $\endgroup$
    – Berci
    Oct 13 at 23:51
  • $\begingroup$ Yes, sorry. I will make an edit. $\endgroup$
    – 3nondatur
    Oct 13 at 23:52
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    $\begingroup$ If yes, you can choose $A:=R[X_1,\dots,X_n]$ and set $P:=F_A(X_1,X_2,\dots,X_n)$. $\endgroup$
    – Berci
    Oct 13 at 23:56
  • $\begingroup$ Sorry to ask again, but why can we be sure that $P$ is then indeed unique? $\endgroup$
    – 3nondatur
    Oct 14 at 0:06
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Indeed we might not be able to recover the polynomial of a single function $R^n\to R$, however the assumption includes all $R$-algebras.

Consider $A:=R[X_1,\dots,X_n]$.
The trick is that, for any polynomial $P\in A$, if we substitute the elements $X_i\in A$ by themselves, we obtain $$P_A(X_1,\dots,X_n)=P$$ which shows exactly how we can uniquely recover a polynomial.

So if $F_A=P_A$ (still for this particular $A$) for some polynomial $P$, then by the above, we must have $$P=F_A(X_1,\dots,X_n)\,.$$ To see then $F_B=P_B$ holds for any $R$-algebra $B$, just use the hypothesis and that $R$-algebra homomorphisms from the polynomial ring $A$ correspond exactly to the evaluations of $X_i$.

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