I just don't understand the point of terming the evaluation of a polynomial by a map like this? And what's more, the map is going into a larger field than the field the polynomial is in anyway. What is the point?
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7Isn't it just pleasing to know that a polynomial, as a formal linear combination of powers of a variable, "plays nicely" with polynomial functions? This is one way to view the evaluation map, and the fact that we have a (family of) homomorphism(s) is formally stating that the two ideas coexist peacefully. Of course, the homomorphism is more useful than that, but I find it to be aesthetically pleasing on a basic level. – pjs36 May 13 '15 at 17:42
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Why does a homomorphism mean they play nicely? Maybe my intuition of the point of homomorphisms period is a little sketchy. – user121615 May 13 '15 at 18:07
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Here is a (small) example that turns out to be rather important, take:
$\phi_i: \Bbb R[x] \to \Bbb C$, given by:
$\phi_i(f(x)) = f(i)$ (here, by $i$, we mean a square root of $-1$).
Of particular note is the fact that $\phi_i(x^2 + 1) = 0$. This allows us to "evaluate" real polynomials that don't have real solutions.
Note that viewing $\Bbb R[x]$ as a collection of functions $\Bbb R \to \Bbb R$ doesn't allow us this wider view.
The above example hinges on the fact that $\Bbb C = \Bbb R[i]$, that is, any complex number can be viewed as a "polynomial in $i$". Since $i^2 = -1$, we can reduce higher powers of $i$ than $1$ to lower powers, so all of these polynomials in $i$ reduce to the form $a + bi$.
David Wheeler
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It doesn't go to a larger field, the evaluation map goes from a polynomial ring over a ring $R$ to $R$ itself. $R[X]\rightarrow R$. It's a very useful function that comes in handy quite often. Why do you think it's useless?
Gregory Grant
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Because I don't understand what's the point of versing it with a map. Why not just evaluate the polynomial and check it's value. And according to my book, it definitely does go into an extention field of the field the polynomial was originally in – user121615 May 13 '15 at 17:28
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Well then they're evaluating it at a value that's not in the base ring to begin with. Anyway, one needs the fact that evaluation is a ring homomorphism, that comes up useful in many proofs. If you keep studying algebra it won't be long before you will encounter it in that context. – Gregory Grant May 13 '15 at 17:32
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6@user121615 "evaluate the polynomial and check its value" IS the evaluation map, and it is not just a map but also a ring homomorphism. This is also a way mathematicians think: polynomials are just formal expressions and are not dependent on any evaluation, but of course they are isomorphic to the space of polynomial functions(as rings). Also what you said doesn't make much sense, since a polynomial lives in a ring e.g. you can't take inverses. – Qidi May 13 '15 at 17:41
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5@Qidi "...but of course they are isomorphic to the space of polynomial functions (as rings)." This is not true in general. Over $\mathbb{F}_p$ the polynomial $f(x) = x^p - x$ induces the zero function since $a^p = a$ for all $a \in \mathbb{F}_p$, but $f$ is not $0$ in $\mathbb{F}_p[x]$. – Viktor Vaughn May 13 '15 at 22:13
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But it can be out side the ring $R$ right? Because $x$ doesn't have to be in $R$ – Nabs Aug 02 '24 at 18:57
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It is not true that polynomial evaluation maps must take values in the coefficient ring $R$. Rather, we can evaluate them into any ring that contains a central image of $R$, i.e. any $R$ algebra, e.g. rings of polynomials, power series, matrices, etc over $R$. This is the universal property of polynomial rings, which in fact characterizes them. – Bill Dubuque May 05 '25 at 02:35
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The main usefulness I know is for the polynomial ring $R[x],$ if you take the minimal polynomial $f$ of an algebraic element $\alpha,$ then it induces an isomorphism $$R[x]/(f)\cong R[\alpha].$$
This is nice for (study of) extensions.
