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

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 ?

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    $\begingroup$ “Proof” is a noun; the verb is “to prove”. $\endgroup$ – Arturo Magidin Apr 28 at 21:19
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    $\begingroup$ 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 $(x+2)$... $\endgroup$ – Arturo Magidin Apr 28 at 21:23
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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

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

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  • $\begingroup$ thank you, this map (i mean $f(x) \rightarrow (f(0), f(-2))$) is what i was looking for $\endgroup$ – envy grunt Apr 28 at 21:35
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    $\begingroup$ @envygrunt no problem. This is a somewhat "typical" evaluation map that is worth just getting used to $\endgroup$ – Andres Mejia Apr 28 at 21:36
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Hint: Use the fact that $x^2+2x=x(x+2)$, and that $$\Bbb{C}[x]/(x)\cong\Bbb{C}[x]/(x+2)\cong\Bbb{C}.$$

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

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