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

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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 $X_i$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.