proof that a gcd domain is integrally closed
Proposition 1.
Every gcd domain is integrally closed.
Proof.
Let be a gcd domain. For any , let be the collection of all gcd’s of and . For this proof, we need two facts:
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1.
.
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2.
If and , then .
The proof of the two properties above can be found here (http://planetmath.org/PropertiesOfAGcdDomain). For convenience, we let be any one of the representatives in .
Let be the field of fraction of , and ( and ) is a root of a monic polynomial . We may, from property (1) above, assume that .
Write
So we have
Multiply the equation by then rearrange, and we get
Therefore, . Since , , by repeated applications of property (2), and one application of property (1) above. Therefore is an associate of 1, hence a unit and we have .
∎
Together with the additional property (call it property 3)
if and , then (proof found here (http://planetmath.org/PropertiesOfAGcdDomain)),
we have the following
Proposition 2.
Every gcd domain is a Schreier domain.
Proof.
That a gcd domain is integrally closed is clear from the previous paragraph. We need to show that is pre-Schreier, that is, every non-zero element is primal. Suppose is non-zero in , and with . Let and , . Then by property (1) above. Next, since , write so that . This implies that . So together with show that by property (3). So we have just shown the existence of with , and . Therefore, is primal and is a Schreier domain.
∎
Title | proof that a gcd domain is integrally closed |
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Canonical name | ProofThatAGcdDomainIsIntegrallyClosed |
Date of creation | 2013-03-22 18:19:27 |
Last modified on | 2013-03-22 18:19:27 |
Owner | CWoo (3771) |
Last modified by | CWoo (3771) |
Numerical id | 6 |
Author | CWoo (3771) |
Entry type | Derivation |
Classification | msc 13G05 |