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I was trying to solve this question from Artin:

Exercise 11.6.1. Let $φ : R[x] \rightarrow C\times C$ be the homomorphism defined by $φ(x) = (1, i)$ and $φ(r) = (r, r)$ for $r ∈ R$. Determine the kernel and the image of $φ$.

This is what the first part of the solution says: $φ$ is the evaluation map $f\mapsto(f (1), f (i))$. So $f \in \ker(\phi) \Leftrightarrow f(1) = f (i) = 0 \Leftrightarrow f \in ((x − 1)(x^2 + 1))$, i.e., $\ker(\phi) = ((x − 1)(x^2 + 1))$.

I have not come across this terminology yet and do not understand it. What is $f$ and where did it come from? Whether or not in this context, please can someone explain what an 'evaluation map' is?

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  • $\begingroup$ I fixed a few typos. The claim they make is that the map $\phi$ is defined by $\phi(f) = (f(1), f(i))$. If takes an element $f$ of $R[x]$, evaluates it at $1$ and at $i$ and makes that pair $(f(1), f(i))$. This is a claim that needs verification. It follows from the properties of a homomorphism and the definition of the operations in the two rings: If $f(x)=a_0+a_1x+...+a_nx^n$ then $\phi(f)=\phi(a_0)+\phi(a_1)\phi(x)+...+\phi(a_n)\phi(x)^n=(a_0,a_0) + (a_1,a_1)(1,i)+...+(a_n,a_n)(1,i)^n=(f(1),f(i))$. $\endgroup$
    – Hellen
    Commented Sep 23, 2017 at 1:54
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    $\begingroup$ Is $R$ the set of all real numbers and $C$ the set of all complex numbers? $\endgroup$
    – choco_addicted
    Commented Sep 23, 2017 at 2:07

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It’s just that $f$ is a typical (“generic”) element of the domain, $R[x]$. An evaluation map generally is something that takes as its input a function and evaluates it at a previously given point.

Typically, you have a space, $\mathfrak X$, and a family $\mathfrak F$ of functions on $\mathfrak X$, say all with values in a ring $R$. Then for $x\in\mathfrak X$, you have the evaluation map $\text{ev}_x:\mathfrak F\to R$, defined by: for every $f\in\mathfrak F$, $\text{ev}_x(f)=f(x)$.

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