(All my rings and $R$-algbras are commutative and unital.)

Question. I think it makes sense to speak of the "roots" of an element of an arbitary $R$-algebra; a definition is given below. Does it actually make sense, or am I just confused? If it does make sense, where can I learn more? If not, why not; where does my reasoning go wrong?

First, some notation.

  1. Given a function $f : R \leftarrow \{x_1,\ldots,x_{n-1}\}$, let us write $\Phi(f)$ for the corresponding $R$-algebra homomorphism $R \leftarrow R[x_1,\ldots,x_{n-1}].$ Explicitly, we require that $\Phi(f)$ agrees with $f$ when restricted to the domain of $f$.

  2. Given a polynomial $p \in R[x_1,\ldots,x_{n-1}],$ let us write $\mathbf{Ev}_p$ for the corresponding function $$R \leftarrow R^{\{x_1,\ldots,x_{n-1}\}}$$ given by "evaluation." Explicitly, we define that $\mathbf{Ev}_p(f)= \Phi(f)(p)$ whenever $f$ is a function $R \leftarrow \{x_1,\ldots,x_{n-1}\}.$

Definition. Suppose $P$ is an arbitrary $R$-algebra and consider $p \in P$. I was thinking that we can define that a root of $p \in P$ is just a homomorphism $\varphi : R \leftarrow P$ satisfying $\varphi(p) = 0_R.$

If I'm not confused, this (essentially) generalizes the familiar case where $P$ is a polynomial ring. In particular, if $P =R[x_0,\ldots,x_{n-1}],$ then homomorphisms $\varphi : R \leftarrow R[x_0,\ldots,x_{n-1}]$ satisfying $\varphi(p)=0$ are "essentially the same" as functions $f : R \leftarrow \{x_0,\cdots,x_{n-1}\}$ satisfying $\Phi(f)(p)=0,$ which are precisely the functions $f : R \leftarrow \{x_0,\cdots,x_{n-1}\}$ satisfying $\mathbf{Ev}_p(f)=0$, which are precisely the roots of the polynomial $p$.

Is this correct? If so, where can I learn more? If not, where does my reasoning go wrong?

  • $\begingroup$ Seems good to me (since a homomorphism from a polynomial ring in $n$ variables is "the same as" picking $n$ elements in the codomain). I think I have seen $p^{eval}$ denoted as $\operatorname{Ev}_p$ (to emphasize that one can vary the $p$ as well). $\endgroup$ – Tobias Kildetoft Feb 10 '15 at 11:54
  • $\begingroup$ @TobiasKildetoft, thanks. I've edited with your suggested notation. $\endgroup$ – goblin Feb 10 '15 at 12:31
  • $\begingroup$ You can push this even further by considering more general algebras (in the universal algebra sense), see e.g. arxiv.org/abs/1205.6235 $\endgroup$ – Klaus Draeger Feb 10 '15 at 13:16

You have essentially found an explicit description of the hom-set $$\mathbf{Alg}_R (P / (p), R)$$ namely as the set of $R$-algebra homomorphism $\phi : P \to R$ such that $\phi (p) = 0$. More generally still, you might ask about "roots" of $p$ in another $R$-algebra $Q$, which would amount to asking about the hom-set $$\mathbf{Alg}_R (P / (p), Q)$$ so the conclusion is that the functor $F : \mathbf{Alg}_R \to \mathbf{Set}$ defined by $$Q \mapsto \{ \phi \in \mathbf{Alg}_R (P, Q) : \phi (p) = 0 \}$$ is a representable functor. Proceeding in this direction (for a very long time) eventually leads to the functor of points approach to scheme theory.


I just wanted to add the following: if $R$ is a commutative ring and $P$ is an $R$-algebra, then an $R$-algebra homomorphism $R \leftarrow P$ is sometimes referred to as a character of $P$. The following is well-known: if $R$ is an integral domain, then the characters of $P$ form an $R$-linearly independent subset of the $R$-module of functions $R \leftarrow P$, see here.


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