# No Prime Ideal Of An Integral Domain Contains 1

• October 2nd 2010, 12:33 PM
mathematicalbagpiper
No Prime Ideal Of An Integral Domain Contains 1
Let $A$ be a commutative integral domain with identity 1. Let $X$ be the set of all prime ideals of $A$, and let $V(E)$ be the set of all prime ideals of $A$ containing the subset $E$ of $A$.

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

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

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

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

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

Thus there is no prime ideal that contains 1, since 1 is not contained in any proper ideal."
• October 2nd 2010, 01:38 PM
HappyJoe
I guess you are asking for feedback on your proof.

For the implication in this direction ==>, you write that 1 is an additive generator for A. But there may be elements in A that cannot be written as a finite sum of 1's. Instead, remember that an ideal is closed under multiplication with elements from the ring. So since 1 is in the ideal, then for all r in A, we have also that r = r1 is in the ideal (so all elements of A are in I, hence I=A).

The other direction is good.

As you yourself note, this really hasn't got much to do with prime ideals. Rather, it is true that the only ideal containing 1 is the entire ring - no proper ideals contain the identity.