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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 $\phi_r(f(x))=f(r)$

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    $\begingroup$ What is $\phi_r$ $\endgroup$ – ÍgjøgnumMeg Aug 1 at 23:00
  • $\begingroup$ Just edited my question to include that, it is the evaluation homomorphism $\endgroup$ – Carlos Seda Aug 1 at 23:02
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    $\begingroup$ Remember: $\Bbb C$ is a field, there aren't so many choices for ideals in $\Bbb C$ $\endgroup$ – ÍgjøgnumMeg Aug 1 at 23:08
  • $\begingroup$ Yes but $I$ is a proper ideal of $\mathbb{C}[X]$ which is not a field since $\mathbb{C}$ is a field. $\endgroup$ – Carlos Seda Aug 1 at 23:18
  • $\begingroup$ You misunderstand my hint $\endgroup$ – ÍgjøgnumMeg Aug 1 at 23:19
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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 $\phi_r(I) = \phi_r((f(X)) = (f(r)) = 0$

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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 $I$.

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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 $r\in \mathbb C$...

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    $\begingroup$ so since $\phi_r(J)= (0)$ and $J \supset I$ we get what we want. Thank you very much! $\endgroup$ – Carlos Seda Aug 1 at 23:20
  • $\begingroup$ @CarlosSeda It's a pleasure :) $\endgroup$ – Scientifica Aug 2 at 7:37
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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. $OE\Delta$.

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