What polynomial maps to $i$ under $\mathbb{Q}[x] \to \mathbb{Q}[x]/(x^2+1) \simeq \mathbb{Q}[i]$? The rings $\mathbb{Q}[i]$ and $\mathbb{Q}[x]/(x^2+1)$ are isomorphic, and there is a surjective ring homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[x]/(x^2+1)$. Can someone give me an example of something in $\mathbb{Q}[x]$ that would map to $i$?
 A: The polynomial $x$ is mapped to $i$ by the ring homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[x]/(x^2+1)$ followed by the usual isomorphism.
A: We can work in the setting of groups. Let $f\colon G\to H$ be a homomorphism that sends a specific element $g_0$ of $G$ to a specific element $h_0$ of $H$.
Then, for all elements $x$ in the kernel of $f$, the elements $xg_0$ are mapped to the same $h_0$. And every element mapped to $h_0$ is of this form.
In your case,  for arbitrary $f(x)$,  the elements $x+f(x)(x^2+1)$ are all mapped to $i$. (That is all those polynomials giving $x$ as remainder  upon division by $x^2+1$.)
A: The quotient ring $\mathbb{Q}[x]/(x^2+1)$ consists of the set of all cosets of the form $ax+b+(x^2+1)$ for $a,b \in \mathbb{Q}$.  The natural ring homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[x]/(x^2+1)$ is defined by $f(x) \mapsto \hat{f}(x) + (x^2+1)$, where $\hat{f}$ is the remainder when dividing $f$ by $x^2+1$.  The kernel of this homomorphism is the set $(x^2+1)$ of all multiples of the polynomial $x^2+1$.  
Consider the map from $\mathbb{Q}[x]/(x^2+1)$ to $\mathbb{Q}[i]$ defined by $ax+b+(x^2+1) \mapsto ai+b$. It can be verified that this map is an isomorphism.  It is then clear that the preimage of $i$ is the coset $x+(x^2+1)$. The preimage of this coset in the surjective ring homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[x]/(x^2+1)$ (which was defined in the previous paragraph) is the set of all polynomials in $\mathbb{Q}[x]$ of the form $x + g(x) (x^2+1)$ (i.e. $x$ plus any multiple of the polynomial $x^2+1$). So, an answer to your question is: the polynomial $x+(12 x^3+x^2+4) (x^2+1)$, or simply $x$. 
