Proof of closure under addition and multiplication in a field


by karnten07
Tags: addition, closure, field, multiplication, proof
karnten07
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#1
Feb24-08, 09:08 PM
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1. The problem statement, all variables and given/known data


Does anyone know a generic way of showing that a field is closed under multiplication and addition? Please, thanks
2. Relevant equations



3. The attempt at a solution

Just need to prove that a+b and ab are in the field that each element a and b are from. Any ideas??
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Mystic998
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#2
Feb24-08, 09:14 PM
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There's not really a generic way to do it. It all depends on what set you're claiming is a field. Incidentally, calling it a field before you show it's closed under the operations is bad form.
karnten07
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Feb24-08, 09:33 PM
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Quote Quote by Mystic998 View Post
There's not really a generic way to do it. It all depends on what set you're claiming is a field. Incidentally, calling it a field before you show it's closed under the operations is bad form.
The field im trying to show is a field is of the form Q[tex]\sqrt{}[/tex]d where Q is the set of rational numbers and d is in the set of complex numbers, C.

Hurkyl
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Feb24-08, 10:25 PM
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Proof of closure under addition and multiplication in a field


Quote Quote by karnten07 View Post
Does anyone know a generic way of showing that a field is closed under multiplication and addition? Please, thanks
Yes -- the definition of "field" mandates that it is closed under multiplication and addition.

I suspect you meant to ask something else -- judging by your wording, did you mean to ask about showing whether a subset of a field (with the induced arithmetic operations) is a subfield?
Mystic998
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#5
Feb25-08, 12:53 AM
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Okay, so what you want to do is take arbitrary elements of the field, namely elements of the form [itex]a + b\sqrt{d}[/itex] and [itex]c + e\sqrt{d}[/itex], with [itex]a, b, c, e \in \mathbb{Q}[/itex], and show that you get something of that form back when you multiply or add them.
HallsofIvy
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Feb25-08, 05:34 AM
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As Mystic988 said in the first post, how you show a set if closed under operations (and so is a field) depends on how the set and operations are defined. How you would show that "a field" is of the form Q([itex]\sqrt{d}[/itex]) where d is a complex number depends upon exactly how "a field" is defined!

Exactly how is your field defined? Mystic988's suggestion is assuming you already know d and want to show that the set of numbers of the form a+ b[itex]\sqrt{d}[/itex] is a field. That is quite correct if d is an integer but if, for example, d is a trancendental number, it is not at all correct.

Again, exactly what is the problem you are working on?
karnten07
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#7
Feb25-08, 10:34 AM
P: 208
Quote Quote by HallsofIvy View Post
As Mystic988 said in the first post, how you show a set if closed under operations (and so is a field) depends on how the set and operations are defined. How you would show that "a field" is of the form Q([itex]\sqrt{d}[/itex]) where d is a complex number depends upon exactly how "a field" is defined!

Exactly how is your field defined? Mystic988's suggestion is assuming you already know d and want to show that the set of numbers of the form a+ b[itex]\sqrt{d}[/itex] is a field. That is quite correct if d is an integer but if, for example, d is a trancendental number, it is not at all correct.

Again, exactly what is the problem you are working on?
It's okay, i did the assignment now, i apologise for my poor wording of the question. But i understand how to do the question now, thanks guys
Mystic998
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#8
Feb25-08, 05:26 PM
P: 206
Oh, yeah, I was assuming we were talking about quadratic rationals (I think that's the right name. My brain is not functioning fully at this very moment), not more exotic field extensions. Sorry about that.
HallsofIvy
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Feb25-08, 08:54 PM
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I started to answer the question in exactly the same way!


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