What does this Ring-Homomorphism do to constants: $\mathbb{Z}[X]\rightarrow \mathbb{Z}[i], X\mapsto i$? I have been given this definition for a Ring homomorphism:
$$\varphi:\mathbb{Z}[X]\rightarrow \mathbb{Z}[i], X\mapsto i$$
Where $\mathbb{Z}[i]$ is the ring of the Gaussian numbers (complex numbers with integer components).
With the properties of ring homomorphisms, I've found 
$$\varphi(X)=i, \varphi(X^2)=-1, \varphi(X+X^2)=i-1$$
and so on. But what does this homomorphism map 
$$\varphi(X^2+2), \varphi(X+5)$$
to? These are just examples, I don't need to calculate those specifically.
Ultimately I need to show that this homomorphism induces an isomorphism $\mathbb{Z}[X]/(X^2+1)\cong\mathbb{Z}[i]$, and I don't see how I can do that without the definition of $\varphi$ for constants.
 A: Using the polynomial of the form $aX+b\in \Bbb Z[X]$ we have $\varphi$ is surjective.
Next, to find kernel of $\varphi$ divide any $f\in \Bbb Z[X]$ by $X^2+1$ and look at the remainder. This will give $\ker \varphi=\langle X^2+1\rangle$. For instance, $g\in \ker\varphi$ with $g(X)=(X^2+1)q(X)+r(X)$ and $\text{deg}(r)<\text{deg}(X^2+1)=2$, means $r(X)=cX+d$ for some $c,d\in \Bbb Z$. Now, $0=g(i)=0\cdot q(i)+ci+d\implies c=0=d$ i.e. $g$ is divisible by $X^2+1$. Hence, $\ker\varphi\subseteq \langle X^2+1\rangle$.
So by first Isomorphism Theorem you are done.
A: When you have a substitution homomorphism on a polynomial ring, the substitution is as it would usually be if you "plugged in" the element into the polynomial function. Thus if
$$f(X) = X^3+b$$
then the element is sent to
$$f(i) = i^3+b = -i+b$$
In other words, constants are sent to themselves. 
This kind of substitution doesn't always work in general rings, but many homomorphisms can be defined this way. For a polynomial ring over a field, they are characterized by the fact that any homomorphism can be defined in this way. Homomorphisms in $\mathbb Z[X]$ need not be of this form, however, because for example we can send $2\mapsto 0$.
A: Recall the following theorem
Proposition 11.3.4[Artin p.g 329] let $\phi:R \to R'$ be a ring homomorphism and let $\alpha \in R'$. Then there is a unique ring-homomorphism $\Phi:R[x] \to R'$ such that $\Phi(x) = \alpha$ and for every $a \in R$, $\Phi(a) = \phi(a)$.
Can you see how you can apply this theorem?
A: Any positive integer can be written as a sum of 1s:
$$
n = \underbrace{1 + \cdots + 1}_{n\text{ times}}.
$$
For example, $5 = 1 + 1 + 1 + 1 + 1$. Since $\phi$ is a homomorphism, you know that
$$
\phi(n) = \phi(1 + \cdots + 1) = \phi(1) + \cdots + \phi(1) = 1 + \cdots + 1 = 5.
$$
We also know that $\phi(0) = 0$, and that $\phi(-x) = -\phi(x)$ for any $x$, and combining these facts we get that for any integer $n$ we have $\phi(n) = n$. Thus in this case, a constant is just taken to itself (or, the version of itself as an element of $\mathbb Z[i]$). So now you can see what happens with an arbitrary polynomial: for instance,
$$
\phi(X^2 + 5) = \phi(X^2) + \phi(5) = \phi(X)^2 + 5 = i^2 + 5 = 4.
$$
Note that this doesn't work for general rings $R[X]$ if only $\phi(X)$ is specified, but it works for $\mathbb Z$ because any element of $Z$ can be written in terms of 1 and the ring operations.
