Ring homomorphisms $\mathbb Z[x]\rightarrow \mathbb Q$

Question

I'm supposed to find all the ring homomorphisms $\mathbb Z[x]\rightarrow \mathbb Q$.

Here's my attmept: $\mathbb Z[x]$ is generated by $\left\{ x^n \right\} _{n\geq 0}$, so the images of the generators determines the homomorphism. Since ring homomorphisms also respect multiplication, the images of $1,x$ determine everything. Homomorphisms respect units, so the image of $x$ determines our homomorphism. Hence, the only homomorphisms are the evaluations: for each $q\in \mathbb Q$ we have $\mathrm{eval}_q:\sum_na_nx^n\mapsto \sum _n a_n q^n$.

Is this correct? If so, isn't the codomain completely irrelevant? It seems we can say the same for any ring $R$ in place of $\mathbb Q$.

Answer 1

I would like to expand on @Alexander's answer to give you a cleaner and more general version of what he's saying. However, to truly understand this you will need some basics in category theory.

What is happening here, is that you are working in a category $\mathcal{C}$, namely the category of commutative rings $\mathcal{C}=\bf{CRing}$. There is a forgetful functor $$U:\bf{CRing}\longrightarrow\bf{Set}$$ to the category of (small) sets, given simply by sending a commutative ring $C\in\mathcal{C}$ to its underlying set, forgetting about the algebraic structure. The nice thing is that this functor $U$ has a left adjoint, that is another functor $$F:\bf{Set}\longrightarrow\bf{CRing},$$ which we call the free commutative ring functor, such that $$\hom_{\bf{CRing}}(F(I),R)\cong\hom_{\bf{Set}}(I,U(R))$$ (in a natural way) for any $I\in\bf{Set}$ and any $R\in\bf{CRing}$. In fact, it is given by considering the polynomial ring $\mathbb{Z}[I]$ in the variables $i\in I$. Yours is the special case where $I=\{x\}$. The adjunction directly gives you the result. (But, of course, you have to prove your result in general to show that this is in fact an adjunction...)

Why do I tell you this? Well, the fact is that this story appears time and again a bit everywhere in mathematics (especially in algebra). Often you have a forgetful functor from your category of interest $\mathcal{C}$, and if you're luck you will be able to construct a free functor (or at least show that it exists). The image of this free functor is formed by the so-called free objects, that have all kinds of nice properties. Examples are free groups, free algebras, free modules...

Answer 2

What you say is correct. If you do it in the right generality, it becomes completely obvious that this holds regardless of the fact tht the codomain is $\mathbb{Q}$.

Let me give to the "like a boss" view point. Let $A$ be a commutative ring, $I$ a set. Then the polynomial ring $A[(X_i)_{i\in I}]$ is defined by the existence of a functorial bijection $$\mathrm{Hom}_{A-alg}(A[(X_i)_{i\in I}],B)\rightarrow\mathrm{Hom}_{Set}(I,B)$$ for any commutative $A$-algebra $B$. In the usual construction this gives exactly the evaluation maps you describe. Your answer is merely a special case of this, where we take $A=\mathbb{Z}$, $B=\mathbb{Q}$, $I$ to consists of single element.