# Find image and kernel of $\varphi: \mathbb{Z}[x] \to \mathbb{C}$ given by $x \mapsto i$

Consider the homomorphism $\varphi: \mathbb{Z}[x] \to \mathbb{C}$ given by $x \mapsto i$. Find:
(a) the image;
(b) the kernel;
(c) exhibit the bijection of the Correspondence Theorem explicitly for $\mathbb{Z}[x] \twoheadrightarrow \text{im}(\varphi)$. (Hint for (c): you should probably use the fact that $\mathbb{Z}[i]$, the Gaussian integers, is a PID!)

Intuitively, the image seems to be all Gaussian integers $\mathbb{Z}[i]$. How do I prove this?

EDIT: $(x-i) \subseteq \text{ker}(\varphi)$. The kernel seems to be the ideal generated by $(x^2+1)$. $\mathbb{Z}[x]/(x^2+1) \cong \mathbb{Z}[i]$. How do I prove this?

Thanks!

• Is $x-i \in \mathbb{Z}[x]$? In other words, would you say that $i \in \mathbb{Z}$? – Shalop May 14 '15 at 3:51
• Note that a polynomial $p(X)$ goes to $p(i)$. You should convince yourself the kernel is $(X^2+1)$, and the image is $\Bbb Z[i]$, which gives the isomorphism $\Bbb Z[X]/(X^2+1)\simeq \Bbb Z[i]$. – Pedro Tamaroff May 14 '15 at 3:55
• They are already telling you the image is the Gaussian integers implicitly. Also, saying "consider the homomorphism that maps $x$ to $i$" is meaningless unless you have already seen that such a homomorphism always exists and is unique, I think this is sometimes called the universal property of the ring of polynomials. – Jorge Fernández Hidalgo May 14 '15 at 3:55
• any more help? particularly c? – clay May 14 '15 at 5:28

## 1 Answer

how it is possible that x-i in kernel,because x-i is not in domain,but x^2+1 is in kernel .now if a polynomial p(x) is in kernel then p(i)=0, similarly p(-i)=0, now you can see that p(x) must have a factor x^2+1 in domain ring. this proves the the statement

• Wlecome to math.SE! This answer needs you to edit it. Your first observation is correct (but poorly written when it comes to puntuaction). The second one is always false: $p\in\ker\varphi\Leftrightarrow p(i)-1=-1$, obviously. – user228113 May 14 '15 at 10:02