Is Ring $\mathbb C[x]/(x^2+1) \cong \mathbb C$

Question

Is Ring $\mathbb C[x]/(x^2+1) \cong \mathbb C$

I had proved that Ring $\mathbb C[x]/(x^2+1)\cong \mathbb C$ is it right?

My attempt

$\mathbb C[x]/(x^2+1)=$ {$ax+b|a,b\in \mathbb C$} with $x^2+1=0$

also $x^2+1=0$

implies $x=i$

Putting this in above set

we are left with C

Is it correct?

I will be thankful if someone correct me

Any help will be appreciated

Answer 1

By the Chinese Remainder Theorem we have $$\mathbb C[X]/(X^2+1)\cong \mathbb C\times\mathbb C\not\cong \Bbb C.$$

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

Answer 2

$\mathbb{C}[X]/(X^2+1)$ is not isomorphic to $\mathbb{C}$ because it has divisor of zero, $X^2+1=(X-i)(X+i)$ we deduce that the image of $X+i$ by $\mathbb{C}[X]\rightarrow \mathbb{C}[X]/(X^2+1)$ is a divisor of zero. $\mathbb{C}$ does not have divisor of zero since it is a field.

Answer 3

Note that $\Bbb C$ is a field. That would imply $\langle x^2+1\rangle$ is a maximal ideal and hence a prime ideal in $\Bbb C$, which is not true (because $x^2+1=(x-i)(x+i)\in \langle x^2+1\rangle $, but $x-i, x+i\notin \langle x^2+1\rangle$).