How two different elements belong to a polynomial implies the polynomial is a proper ideal? This is an example from Galois Theory.

If F is a field, then a nonzero polynomial p(x) ∈ F[x] is irreducible
  if and only if (p(x)) is a prime ideal.
proof: 
Suppose p(x) is irreducible. 
If ab ∈ p, then p | ab 
Euclid's lemma implies that p | a or p | b. 
Thus, a ∈ (p) or b ∈ (p). 
Finally, p is a proper ideal.

I'm ok with the prove till a ∈ (p) or b ∈ (p). 
But then why p is a proper ideal? The question only give us p(x) is a polynomial in a field.
 A: If $p$ is irreducible, then by definition it is not a unit, so $1\not\in (p)$ and $(p)$ is a proper ideal.
A: Your proof is making me nervous.  Irreducible usually means a non-unit $p$ with $p=ab \Rightarrow a\in U(R)$ or $b \in U(R)$ (for integeral domains).  Prime is a non-unit with $p\mid ab \Rightarrow p\mid a$ or $p \mid b$.  Euclid's lemma is for primes rather than irreducibles usually, but since the integers are a PID they are equivalent, so it is not that big of a deal, but outside of the integers, this causes some confusion I think.  Your use of Euclid's lemma seems to be essentially already assuming $p(x)$ is prime rather than irreducible.
Polynomial rings over a field are Euclidean Domains which are PIDs which are UFDs.  Even in a UFD, irreducible implies prime, but here I think the PID proof of irreducible implying prime is worth mentioning.
You can actually show if $p(x)$ is irreducible, then $(p(x))$ is maximal and hence prime.  Suppose $(p(x)) \subseteq I \subseteq F[x]$ with $p(x)$ irreducible.  Since $F[x]$ is a PID, $I=(f(x))$ for some $f(x)$.  But $p(x) \in I$, so $p(x)=f(x)g(x)$ for some $g(x)$.  Since $p(x)$ is irreducible, either $f(x)$ is a unit or $g(x)$ is a unit.  If $f(x)$ is a unit, then $I=(f(x))=F[x]$ since any ideal containing a unit, must be the whole ring.  If $g(x)$ is a unit, then $(p(x))=(f(x))$.  This shows $(p(x))$ is maximal.
