Let R be a domain with fraction field L, and assume a is algebraic over L, show that is a nonzero ideal in R.

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- October 22nd 2008, 10:38 PMterr13Integrality
Let R be a domain with fraction field L, and assume a is algebraic over L, show that is a nonzero ideal in R.

- October 22nd 2008, 11:01 PMNonCommAlg
- October 22nd 2008, 11:05 PMterr13
You're right, sorry about the confusion, it should be algebraic over L.

- October 23rd 2008, 12:19 AMNonCommAlg
you again repeated what you already posted! (Wondering) you should be more careful with posting your question! the correct and complete version of your question and the solution:

__Problem__: let be a domain and its field of fractions. show that if is algebraic over , then the set is a non-zero ideal of .

__Proof__: is obviously closed under addition because if then are integral over i.e. are in the integral closure of in since is a ring, we have

thus now suppose and we must show that : so and thus i.e. this proves that is an ideal

of the only thing left is to show that : since is algebraic over , we will have: for some integer and hence

so is integral over i.e.

__something for you to think about__:

did we use the assumption that is a domain and is its fraction field? is the claim in the problem true for any (unitary) commutative ring R and any ring extension L of R?