# How to prove that the evaluation map is a ring homomorphism?

This is a really easy question, but I'm stuck in the logic of it...

Let $F$ be an integral domain and $F[x]$ its polynomial ring. Let $a\in F$ fixed, define $\phi: F[x]\to F$ as $\phi(p(x))=p(a)$. Prove that $\phi$ is a homomorphism.

I daresay this is trivial and follows from its own definition... but I'm not sure. I'm seeing it as a "change of variable" or something like that. So I ask... what would be the argument? Thanks a lot, I'm sorry for this easy one.

• Just check that it satisfies the properties of a homomorphism. It's not clear to me what you're struggling with. – Michael Albanese Sep 18 '14 at 2:15
• Hi Michael. See the answer below, maybe you get the point I'm stuck with. Thanks. – Daniel Sep 18 '14 at 2:19
• The operations on $F[x]$ are designed so that the map $p[x]\mapsto \sum_{k} \alpha_k a^k$ is a homomorphism of rings. It would not work if we had the definition say:$(\sum \alpha_k x^k) \cdot (\sum_k \beta_k x^k) = \sum_k \alpha_k \beta_k x^k$ – orangeskid Sep 18 '14 at 2:24

You can find a proof that evaluation maps are homomorphisms here. Here I have retyped it using the notation given in your exercise:

Let $f(x) = a_nx_n + \ldots + a_0x_0$, and $g(x) = b_nx_n+ \ldots + b_0x_0$, where the $a_i,b_i ∈ \mathbb R$. (We'll also allow leading coefficients to be zero in order to make it easier to add $f$ and $g$ formally.) We then check the ring homomorphism conditions:

• We have: \begin{aligned} \phi(f+g) &=\phi((a_n+b_n)x_n + \ldots + (a_0 + b_0)x^0) \\ &= (a_n + b_n)k^n + \ldots + (a_0 + b_0)k^0 \\ &= (a_nk^n + \ldots + a_0k^0) + (b_nk^n + \ldots + b_0k^0) \\ &=\phi(f) + \phi(g) \end{aligned} So $\phi$ is an additive group homomorphism.

• Since we know that $\phi$ is an additive homomorphism, we only need to check that it is multiplicative on monomials. But that's easy: $$\phi((ax^n)(bx^m))=\phi(abx^{n+m}) = abk^{n+m}=\phi(ax^n)\phi(bx^m)$$

• Great! you got the spirit of the question. Thank you very much for the answer, and the reference. – Daniel Apr 15 '16 at 14:35

(This is too long for a comment, but may be useful for someone like me who has learned the universal property of polynomial rings from a commutative algebra course but wants examples of how to understand how to apply it in concrete situations like the one OP has above.)

If you assume the universal property of polynomial rings as given (which you shouldn't, since it's basically just a more general version of the result you want to show, i.e. this accept should not be accepted), then the result you want follows by choosing the right data to input into the result.

Let $$R,S$$ be commutative and unitary rings. Let $$\langle s_j \rangle_{j \in J}$$ be an indexed family of elements of $$S$$. Let $$\psi: R \to S$$ be a ring homomorphism. Then there exists a unique ring homomorphism $$\phi: R[\{X_j: j \in J \}] \to S$$ such that:

1. $$\forall r \in R$$, one has that $$\phi(r) = \psi(r)$$, i.e. $$\phi$$ extends $$\psi$$,
2. $$\forall j \in J$$, one has that $$\phi(X_j) = s_j$$.

A fairly rigorous proof of this result is given here, although the notation on the page explaining the properties of $$\phi$$ is inconsistent with the notation on the original page, hence why I re-wrote the result in its entirely so that it is more easily understandable.

• Note the integral domain $$F$$ is by assumption a commutative and unitary ring. Therefore, we can take both $$R = F$$ and $$S = F$$.

• Take $$J$$ to be any one-element set, and let $$s:= s_1 = a \in S = F$$.

• It follows also from this that $$R[\{X_j: j \in J \}] = F[X_1] =: F[X]$$.

• Finally, let the ring homomorphism $$\psi: R \to S$$ just be $$Id_F: F \to F$$, the identity on $$F$$.

Then in this special case, the above result says that:

There exists a unique ring homomorphism $$\phi: F[X] \to F$$ such that:

1. $$\forall x \in F$$, one has that $$\phi(x) = x$$.
2. Moreover, $$\phi(X) = a$$.

It follows from this, as @orangeskid noted in a comment on the original post, that $$\phi(p) = "p(a)"$$ for all polynomials $$p \in F[X]$$.

While this approach is certainly more abstract, it does have at least the benefit that we do not need to assume a priori that polynomials do in fact form an algebra of functions on $$F$$, instead we can prove that such an identification of polynomials with an algebra of functions on $$F$$ is valid.