# Why is this polynomial unique?

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 $$P \in R[X_1,\ldots,X_n]$$)

• Isn't $F_A$ defined for all $R$-algebras $A$? Oct 13 at 23:51
• Yes, sorry. I will make an edit. Oct 13 at 23:52
• If yes, you can choose $A:=R[X_1,\dots,X_n]$ and set $P:=F_A(X_1,X_2,\dots,X_n)$. Oct 13 at 23:56
• Sorry to ask again, but why can we be sure that $P$ is then indeed unique? Oct 14 at 0:06

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 $$X_i$$.