Injective Ring Homomorphism I seem to be having the wrong impression of what $p$ stands for; is $p(x)=x(x+1)(x+2)$ or is it something else? Clarification would be appreciated so that I can complete the lemma below.

Consider the ring $R:=\mathbb{C}[x]/I$ where $I$ is the ideal in $\mathbb C[x]$ generated by $x(x-1)(x-2)$. Show there exists an injective ring homomorphism $R \to \mathbb{C} \times \mathbb{C} \times \mathbb{C}$. 

I realize that the first order of business is to show that this is indeed a well-defined ring homomorphism. 
Lemma
For each $\alpha \in \{0,1,2\}$ the evaluation map $\mathbb C[x] \to \mathbb{C}$, $p(x) \mapsto p(\alpha)$, induces a well-defined ring homomorphism $R \to \mathbb{C}$.
The induced map takes $p+I$ to $p(\alpha)$. Thus, for the uniqueness part we only need to check that $I$ is contained in the kernel of the evaluation map of $\mathbb{C}[x]$. That way, if $p=f+r$ for some $r \in I$, so that $p+I=f+I$ in $\mathbb{C}[x]/I$, then $p(\alpha)=f(\alpha)$. It is clear $p(\alpha)=0$ for $\alpha \in \{0,1,2\}$ and so 
 A: HINT: Consider the map
$$
\Bbb C[X]\longrightarrow\Bbb C\times\Bbb C\times\Bbb C
$$
given by $q(X)\mapsto(q(0),q(1),q(2))$. Convince yourself that is a homomorphism, that is surjective, and compute the kernel.
A: This is a special case of the Chinese Remainder Theorem:

If $R$ is a commutative ring with $I, J$ coprime ideals (i.e.  $I+J = R$), then $IJ = I \cap J$ and $$\frac{R}{IJ}\cong \frac RI\times \frac RJ$$

via the homomorphism induced by $$R \to R/I \times R/J\\x\mapsto (x+I, x+J)$$
Note that $\mathbb C[X]/(x + \alpha)\cong \mathbb C$ for any $\alpha \in \mathbb C$.
A: The evaluation map $e_a:p\mapsto p(a)$, where $p\in\mathbb{C}[x]$ and $a\in\mathbb{C}$ is a homomorphism $e_a\colon \mathbb{C}[x]\to\mathbb{C}$. By the properties of products, also
$$
f=e_0\times e_1\times e_2\colon\mathbb{C}[x]\to
\mathbb{C}\times\mathbb{C}\times\mathbb{C}
$$
defined by
$$
f(p)=(p(0),p(1),p(2))
$$
is a (ring) homomorphism. You can also check it directly.
What is $\ker f$? It consists of all polynomials $p$ such that $p(0)=0$, $p(1)=0$ and $p(2)=0$. Thus if $p\in\ker f$, we have
$$
p(x)=x(x-1)(x-2)q(x)
$$
by using the fact that $p(a)=0$ if and only if $p(x)$ is divisible by $x-a$.
Conversely, it is obvious that $x(x-1)(x-2)q(x)\in\ker f$, for all $q\in\mathbb{C}[x]$. Hence
$$
\ker f=I
$$
where $I$ is the ideal generated by $x(x-1)(x-2)$.
Thus $f$ induces an injective homomorphism $\bar{f}\colon\mathbb{C}[x]/I\to\mathbb{C}\times\mathbb{C}\times\mathbb{C}$.

In addition, we can also note that $f$ (and so also $\bar{f}$) is surjective. Indeed, given any $(a,b,c)\in\mathbb{C}\times\mathbb{C}\times\mathbb{C}$, there exists a polynomial $p(x)$ such that
$$
p(0)=a,\quad p(1)=b\quad p(2)=c
$$
by Lagrange-Newton interpolation.
