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

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

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 $$\phi_r(I) = \phi_r((f(X)) = (f(r)) = 0$$

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 $$I$$.

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 $$r\in \mathbb C$$...

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

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. $$OE\Delta$$.