Prove that if $I$ is a proper ideal of $\mathbb{C}[X]$, then there exists $r \in \mathbb{C}$ with $\phi_r(I) \neq \mathbb{C}$ I am currently studying for my Algebra qualifying exam and stumbled upon this exercise in an old exam. 
I know since $I$ is proper then $1\notin I$ and I think that this is the key to finding the $r$ for this proof but I do not how to proceed on this one. 
Any help or hints will be greatly appreciated!
Edit: $\phi_r : \mathbb{C}[X] \rightarrow \mathbb{C}$ and it is defined as the evaluation homomorphism where $\phi_r(f(x))=f(r)$
 A: It is well known that $\mathbb{k}[X]$ is a PID when $\mathbb{k}$ is a field. Hence $I = (p)$ for some (non constant, as $I$ is proper) polynomial $p$. By the fundamental theorem of algebra, we know that $p$ has a root $r \in \mathbb{C}$. Thus, the evaluation in $r$ vanishes in $I$.
A: As $\mathbb C$ is algebraically closed, the maximal ideals of $\mathbb C[X]$  are the ones of the form $(X-a)$ where $a\in \mathbb C$.
Take $J$ a maximal ideal containing $I$. Then $J=(X-r)$ for some $r\in \mathbb C$...
A: Note that $\Bbb C[X]$ is a principal ideal domain, as $\Bbb C$ is a field. Thus, $I = (f(X))$ for some $f(X) \in \Bbb C[X]$. Since $\Bbb C$ is a field and $\phi_r(I) \leq \Bbb C$, either $\phi_r(I) = 0$ or $\phi_r(I) = \Bbb C$. Because we only want $\phi_r(I) \neq \Bbb C$, it must be that $I$ maps to $0$. Since $\Bbb C$ is algebraically closed, just pick $r$ such that $f(r) = 0$ so that $\phi_r(I) = \phi_r((f(X)) = (f(r)) = 0$
A: We begin with the observation that we may take
$I \ne \{0\} \tag 0$
without loss of generality, since trivially
$\phi_r (\{0\}) = \{0\} \ne \Bbb C. \tag{0.5}$
Now $\Bbb C$ being a field, $\Bbb C[X]$ is a principal ideal domain; thus every ideal
$I \subset \Bbb C[X] \tag 1$
is of the form
$I = (p(X)) \tag 2$
for some
$p(X) \in \Bbb C[X]; \tag 3$
we note that $I$ proper in $\Bbb C[X]$,
$I \subsetneq \Bbb C[X], \tag 4$
implies
$p(X) \notin \Bbb C, \tag 5$
that is, $p(X)$ is not a constant polynomial; otherwise, with
$0 \ne p_0 = p(X) \in I \subset \Bbb C, \tag 6$
for any
$c \in \Bbb C \tag 7$
we may write
$c = (cp_0^{-1})p_0 \in I, \tag 8$
and so
$\phi_r(I) = \Bbb C, \tag 9$
since
$c(r) = c \tag{10}$
for every $r$; by this argument we rule out the constancy of $p(X)$; thus
$\deg p(X) \ge 1, \tag{11}$
from which it follows via the fundamental theorem of algebra that
$\exists r \in \Bbb C, \; p(r) = 0; \tag{12}$
now every 
$g(X) \in I \tag{13}$
may be written in the form
$g(X) = h(X)p(X) \tag{14}$
for some
$h(X) \in \Bbb C[X], \tag{15}$
from which
$\phi_r(g(X)) = g(r) = h(r)p(r) = 0; \tag{16}$
so we affirm that
$\phi_r(I) = \{0\} \ne \Bbb C, \tag{17}$
the requisite result.  $OE\Delta$.
