Error in proof: $\mathbb{C} \cong \mathbb{C} \times \mathbb{C}$??

I've unintentionally "proved" the following: $$\mathbb{C} \cong \mathbb{C} \times \mathbb{C}$$ Can you help me tracing the error I made resulting to this non-proof? Here it is.

First of all, I recall an algebraic theorem about the complex plane: $\mathbb{C} \cong \mathbb{R}[X]/(X^2+1)$. The rest of the non-proof is about the apparent isomorphism $\mathbb{C} \times \mathbb{C} \cong \mathbb{R}[X]/(X^2+1)$.

The ring $R[X]$ is a commutative ring containing a unit, so we can write the ideal $(X^2+1)$ as a product of ideals $(X-i)(X+i)$. These ideals are indivisable: $(X+i)(X-i) \ni \frac{1}{2}(X+i)-\frac{1}{2}(X-i)=i$, so every polynomial $q = i^3q \cdot i$ is contained in the sum of ideals. Now we can use the generalisation of the Chinese remainder theorem: For a commutative ring R with unity, and indivisible ideals $I,J$ applies $R/IJ \cong R/I \times R/J$.

Now we have $\mathbb{R}[X]/(X^2+1) \cong \mathbb{R}[X]/(X+i) \times \mathbb{R}[X]/(X-i)$. I recall that for every homomorphism of rings $f: R \rightarrow S$ the isomorphism $R/\ker(f) \cong f(R)$ holds. Consider the mapping $s_r \ : \ R[X] \rightarrow S \ : \ \sum a_i X^i \mapsto \sum a_i r^i$. This substitution map is an homomorphism. A special case is the homomorphism $s_i \ : \ \mathbb{R}[X] \rightarrow \mathbb{C} \ : \ \sum a_i X^i \mapsto \sum a_i r^i$ that substitutes $i$ in $X$. Its kernel is exactly the ideal $(X-i)$. The image clairy contains $\{ a + bi: a,b \in \mathbb{R} \} = \mathbb{C}$, and the image is contained in $\mathbb{C}$ as well. So we obtain $\mathbb{C} \cong \mathbb{R}[X]/(X+i)$. The same trick with the substitution $s_{-i}$ shows that $\mathbb{C} \cong \mathbb{R}[X]/(X-i)$. This results into $\mathbb{C} \cong \mathbb{C} \times \mathbb{C}$. Here ends the "proof."

I feel a little bad about using $s_i$, because $i \notin \mathbb{R}$, but I remember that a similar mapping had to be used to proof that $\mathbb{C} \cong \mathbb{R}[X]/(X^2+1)$.

I'd be thankful if you help me to sift this through and find the error(s).

It's not too tough to spot! $i\notin\mathbb R$.

What your proof does in fact show is that by Chinese Remainder, $\mathbb C[X]/(X^2+1)\cong \mathbb C\times\mathbb C$.

The kernel of your homomorphism is $(X^2+1)$, not $(X-i)$, since $X-i\notin\mathbb R[X]$.

What does $\mathbb{R}[X]/(X+\operatorname{i})$ mean? The polynomial $X+\operatorname{i}$ does not belong to $\mathbb{R}[X]$, and so $\mathbb{R}[X]/(X+\operatorname{i})$ does not make sense. This is because $\operatorname{i}$ does not belong to $\mathbb{R}$.