# Existence of Principal Ideal

My first question is DOES a Principal Ideal always exist in a ring? (My thoughts: By it's very definition any element can generate a PI and hence it exists)

Follow-up question: (If answer to 1st question is YES) Please have a look at the alternate proof I have in mind for the following question:

Question: Prove that a commutative ring R with unity whose only ideals are <0> and R itself is a field.

PROOF: Let "a" be any element belonging to R. (Assume a to be non-zero) Consider < a >.

Now since only ideals are <0> and R itself, and since a is non-trivial, < a > must be equal to R.

< a > contains elements of the form {r1a, r2a, ...} for all r belonging to R. But since < a > = R, one of these elements MUST be UNITY!

Hence there exists ri such that r1a = 1. Hence showing the existence of inverse of a.

Furthermore since a is ANY element belonging to R we have shown that for every element an inverse exists. Hence it's a field (other properties already satisfied) (Left inverse proof is trivial from above)

What is the problem in this PROOF?

• What makes you think there is a problem? – Eric Wofsey Nov 16 '18 at 5:35
• Because the solution given in the book and the professor is "supposedly" the shortest solution, but is longer than this. Hence the doubt. – PLAP_ Nov 16 '18 at 5:36
• – Angina Seng Nov 16 '18 at 5:44
• Your proof looks fine to me. To answer the question in your first paragraph, yes, every element generates a principal ideal. – Bungo Nov 16 '18 at 5:46
• Better to think of $\langle a\rangle$ as $aR$. This is the right principal ideal generated by $a$, but with a commutative ring it's just the principal ideal. Then your proof becomes $1\in R=aR$, so $a$ has a right inverse. – vadim123 Nov 16 '18 at 5:54

[ The group-theoretic analogy is stronger for related objects called "modules" than for rings per se. In fact, this is no accident: abelian groups are precisely the same as modules over $$\Bbb Z$$. ]