How to prove ring isomorphism ${\mathbb{C}[x]}/{(x^2 + 2x)} \cong \mathbb{C} \oplus \mathbb{C}$

Question

$${\mathbb{C}[x]}/{(x^2 + 2x)} \cong \mathbb{C} \oplus \mathbb{C}$$

I want to use isomorphism theorem here, so I need to give a map:

$$\phi: \mathbb{C}[x] \rightarrow \mathbb{C} \oplus \mathbb{C} \bigm| \ker\phi = (x^2 + 2x)$$

But $\ker\phi = (x^2 + 2x)$ means that all polynomials with roots $0, -2$ go to $(0,0)$.

Can you give me an example of this map ?

Answer 1

There is a nice map you can construct $\phi: \mathbb C[x]/(x^2+x) \to \mathbb C \oplus \mathbb C$ by noting that $x^2+2x=x(x+2)$. It is given by

$$f(x) \mapsto (f(0),f(-2)).$$

Whata is the kernel of $\phi$?

There is an alternative approach, which is by the chinese remainder theorem. The trick is that $(x)$ is coprime to $(x+2)$ since $x-(x+2)=2 \in \mathbb C$. The chinese remainder theorem ensures then that

$$\mathbb C[x]/[(x) \cdot (x+2)] \cong \mathbb C[x]/(x) \otimes \mathbb C[x]/(x+2) \cong \mathbb C \oplus \mathbb C$$

Answer 2

Hint: Use the fact that $x^2+2x=x(x+2)$, and that $$\Bbb{C}[x]/(x)\cong\Bbb{C}[x]/(x+2)\cong\Bbb{C}.$$

Answer 3

One can use the Chinese remainder theorem, here - since $(x^2 + 2x) = (x)(x+2)$ and these two ideals sum to all of $\mathbb{C}[x]$. This means your map should send $x$ to $(0, -2)$ and of course $1$ to $(1, 1)$ and then the homomorphism requirement determines the rest.

As a sanity check, note that $x^2 + 2x$ maps to $(0, 4) + (0, -4) = (0, 0)$.