No Prime Ideal Of An Integral Domain Contains 1

Let $\displaystyle A$ be a commutative integral domain with identity 1. Let $\displaystyle X$ be the set of all prime ideals of $\displaystyle A$, and let $\displaystyle V(E)$ be the set of all prime ideals of $\displaystyle A$ containing the subset $\displaystyle E$ of $\displaystyle A$.

My task is to show that $\displaystyle V(\{1\})=\emptyset$.

Here's my thinking on this, and what I've done:

"To show that $\displaystyle V(\{1\})=\emptyset$, it suffices to show that an ideal of $\displaystyle A$ is proper if and only if it doesn't contain 1, since by definition all prime ideals are proper ideals.

$\displaystyle \Rightarrow)$ Proof by contrapositive. Suppose $\displaystyle 1\in I$ where $\displaystyle I$ is an ideal of $\displaystyle A$. Since an ideal is a subgroup under addition, and further since 1 is an additive generator for $\displaystyle A$, it follows that $\displaystyle I=A$ thus $\displaystyle I$ is not a proper ideal.

$\displaystyle \Leftarrow)$ Suppose that $\displaystyle 1\notin I$. Since $\displaystyle 1\in A$ by definition, it immediately follows that $\displaystyle I\subset A$ and thus $\displaystyle I$ is proper.

Thus there is no prime ideal that contains 1, since 1 is not contained in any proper ideal."