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

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

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 ?

• “Proof” is a noun; the verb is “to prove”. – Arturo Magidin Apr 28 at 21:19
• Well, evaluation at $0$ maps $\mathbb{C}[x]$ to $\mathbb{C}$ with kernel $(x)$; and evaluation at $-2$ maps $\mathbb{C}[x]$ to $\mathbb{C}$ with kenrnel $(x+2)$... – Arturo Magidin Apr 28 at 21:23

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

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

• thank you, this map (i mean $f(x) \rightarrow (f(0), f(-2))$) is what i was looking for – envy grunt Apr 28 at 21:35
• @envygrunt no problem. This is a somewhat "typical" evaluation map that is worth just getting used to – Andres Mejia Apr 28 at 21:36

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

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 $$(0, 4) + (0, -4) = (0, 0)$$.