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Multiple textbooks that I've looked at make some comment about how the fact that the evaluation map $\phi_c: F[x] \rightarrow F$ is a homomorphism "allows us" to evaluate the polynomial at a specific $c \in F$. However I'm having trouble understanding why it is important that this map be a homomorphism, or even why we need the map at all. Without it, what stops us from just considering, for any polynomial $a_nx^n + \dots + a_1x + a_0$ in $F[x]$, the element $a_n c^n + \dots + a_1 c + a_0$ in $F$? Why do we need the theorem that the evaluation map is a ring homomorphism?

nilcit
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    Being a homomorphism is a good property. For example, you can use isomorphism theorems, and this turns out to be useful in a lot of aspects. – Aloizio Macedo Oct 27 '15 at 23:20
  • But specifically, in this particular case, what does the fact that the evaluation map is a homomorphism allow us to do regarding polynomial rings that we could not otherwise justifiably do? – nilcit Oct 28 '15 at 00:37
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    If the base ring is not commutative, the evaluation map is not a homomorphism. – Matt Samuel Oct 28 '15 at 01:08

3 Answers3

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TL;DR: An essential property of polynomials is that they can be associated to* functions on their coefficient rings. The fact that the evaluation map is a homomorphism is what makes this possible.

In school, one is often taught that polynomials are expressions in which we can "plug in" numbers and "everything works out the same". The evaluation map being a ring homomorphism is how one takes this informal, intuitive understanding of polynomials and makes it rigorous and formal.

*(not necessarily in a bijective manner, see this question)

In some sense, the fact that the evaluation map is a ring homomorphism is the defining property of a polynomial ring. Such a property is also called a universal property.

UNIVERSAL PROPERTY OF POLYNOMIAL RINGS (See also)

In other words, the fact that the evaluation map is a ring homomorphism is the very reason to even bother thinking about polynomial rings in the first place.

The evaluation map being a homomorphism means that it is possible to associate abstract polynomials with actual functions on the ring. If the evaluation map were not a homomorphism, multiplication of polynomials would not correspond to multiplication of their "values" at every point of the ring, and addition of polynomials would not correspond to the addition of their "values" at every point of the ring. Since we expect multiplication and addition of functions to be defined pointwise, the evaluation map being a homomorphism is a necessary condition for the elements of the polynomial ring to correspond to functions on the coefficient ring. (It is also sufficient.)

For example, see this quote from the above-linked page:

The requirement that the rings be commutative is vital. A fundamental difference for polynomials over non-commutative rings is additional difficulty identifying polynomial forms and functions using this method. [emphasis mine]

So perhaps one could say that the book you are reading has the order backwards: instead of defining polynomial rings in terms of the universal property that evaluation maps are ring homomorphisms, they have defined a polynomial ring in terms of a specific construction, and then showed that this construction satisfies the universal property.

This is an understandable approach, because defining something in terms of universal properties may not be a good idea if no structure actually exists which satisfies the universal property.

Still, thinking in terms of universal properties, and subsequently proving that a structure exists which satisfies the universal property, may be more straightforward to some people.

Thinking in terms of universal properties first and constructions second is somewhat akin to declarative programming: we "declare" what properties we want a certain object to have, without worrying as much about the "implementation" (how to construct an object satisfying the property).

The approach used by your textbook is somewhat more like imperative programming: we specify how to construct an object, therefore being assured that it exists, and then retroactively examine what properties this constructed object has.

In either case, we have to worry about both whether an "implementation" exists as well as what properties we want, but in some sense we have the liberty to choose which concern to address first, or at least worry about more, depending upon our preferences and/or personal strengths.

Chill2Macht
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    You can identify polynomials and polynomial functions only in infinite case. For example, $X^p-X\in\mathbb Z/p\mathbb Z[X]$ is not zero polynomial, yet $x\mapsto x^p-x\colon \mathbb Z/p\mathbb Z\to \mathbb Z/p\mathbb Z$ is zero function. This is one good reason why we need to study abstract polynomial rings instead of just polynomial functions. – Ennar Jul 22 '17 at 10:19
  • @Ennar When I said "identify", I didn't mean in a one-to-one fashion. Your point is that the identification is not one-to-one in the finite case. We seem to be talking past each other. Since "identify" is most often used to mean a bijection, I have edited the post to say "associate with" instead of "identify", which hopefully does not connote the incorrect statement you mention. (Also note that it only said "identify" in the TL;DR part, not the part underneath, where it did originally say "associate with".) – Chill2Macht Jul 22 '17 at 10:27
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    I'm glad that you edited your answer because it can be dangerous for someone who is just learning algebra to miss this subtle point. In general, I agree with your point. – Ennar Jul 22 '17 at 10:29
  • @Ennar I agree -- I have edited the post to reflect this, although I didn't want to write about the issue too much in order to avoid going too far afield and making my answer more needlessly confusing than it already it is. – Chill2Macht Jul 22 '17 at 10:31
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    This answer is fantastic and, after over a year, fully answers my original question =) – nilcit Jul 23 '17 at 20:59
  • @nilcit I'm glad I was able to help you – Chill2Macht Jul 24 '17 at 06:43
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One reason why it is a very important homomorphism is that when $F$ is algebraically closed and $\mathfrak{m} \subseteq F[x]$ is a maximal ideal, the projection map $F[x] \to F[x]/\mathfrak m \simeq F$ is precisely the evaluation map, so this map is not innocent at all ; it appears naturally. The correspondence $$ \{ \mathfrak m \subseteq F[x_1,\cdots,x_n] \text{ maximal } \} \longleftrightarrow F^n $$ follows from the Nullstellensatz, a very famous and powerful theorem in algebraic geometry, so seeing the naive case of one variable at work is not so useless.

Note that in the algebraically closed case, we conclude $K \overset{def}= F[x]/\mathfrak m \simeq F$ since $K/F$ is an algebraic field extension and $F$ is algebraically closed. If $F$ is not closed, there are some maximal ideals of $F[x]$ whose projection $F[x] \to F[x]/\mathfrak m$ does not correspond to the evaluation map since $F[x]/\mathfrak m \not\simeq F$. However, since $F[x]$ is a principal ideal domain, $\mathfrak m = (p)$ is principal and $F[x]/(p)$ is a finite field extension of $F$. One could start digging into this correspondence further, but this is a good starting point to many areas of algebra : Galois theory, algebraic geometry, algebraic number theory... so I'll stop here.

Hope that helps,

Patrick Da Silva
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It's not that it is chosen to be a homomorphism, it is that by definition it has to be one. It would be a good exercise for you to check that it is in fact a homomorphism.

AnotherPerson
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    I know it's a homomorphism, and I understand the proof that it is one, but I'm wondering why textbooks draw our attention to this fact. What does the fact that it is a homomorphism give us? – nilcit Oct 28 '15 at 00:35