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

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.
