Is $\mathbb Z[\sqrt[3]2] \cong \mathbb C$? 
Is $\mathbb Z[\sqrt[3]2] \cong \mathbb C$?

In Artin's Algebra, the the kernel of substitution homomorphism from $\mathbb Z[x] \to \mathbb C$ defined by $x \mapsto \sqrt[3]2$ is $(x^3-2)$. So, by first isomorphism theorem, this means $\mathbb Z[x]/(x^3-2) \cong \mathbb C$?
But $\mathbb Z[x]/(x^3-2)\cong \mathbb Z[\sqrt[3]2]=\{a+b\sqrt[3]2+c\sqrt[3]2^2\mid a,b,c \in \mathbb Z\}$. Therefore, 
$$\mathbb Z[\sqrt[3]2] \cong \mathbb C.$$
I know this sounds elementary but for some reason I'm having doubts because besides the Guassian integers, I've never seen complex numbers related to integers.
 A: These rings cannot be isomorphic because the underlying sets do not have the same cardinality. Alternatively, if you have not seen cardinality before, in the group $\mathbb{C}$, for any $z \in \mathbb{C}$ there is a solution $w$ to the equation $w + w = z$, but there is no solution to this equation for the element $z=1$ in $\mathbb{Z}[\sqrt[3]{2}]$. Your mistake is to assume that the substitution homomorphism is surjective in this case. It's still true that it (like any map) gives an isomorphism of the source mod the kernel to the image. 
A: Suppose that there exists a ring homomorphism $f:\Bbb{C} \to \Bbb{Z}[\sqrt[3]{2}]$.  Then $$0 = f(0) = f(i^2+1) = f(i^2)+f(1) = f(i)^2+1 \in \Bbb{Z}[\sqrt[3]{2}].$$  So the polynomial $X^2+1$ has a root $f(i)$ in $\Bbb{Z}[\sqrt[3]{2}]$.  Therefore, $$X^2+1 = (X-f(i))(X-a) = X^2-(a+f(i))X+af(i).$$  Consider the coefficient of $X$, we have $a = -f(i)$.  This gives $1 = af(i) = -f(i)^2 \le 0$, which is absurd.  Therefore, no homomorphism can exist between $\Bbb C$ and $\Bbb{Z}[\sqrt[3]{2}]$.
A: We cannot have $\mathbb Z[\sqrt[3]2] \cong \mathbb C$ because $\mathbb Z[\sqrt[3]2]$ is not a field but $\mathbb C$ is a field.
