Proof of closure under addition and multiplication in a fieldby karnten07 Tags: addition, closure, field, multiplication, proof 

#1
Feb2408, 09:08 PM

P: 208

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?? 



#2
Feb2408, 09:14 PM

P: 206

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.




#3
Feb2408, 09:33 PM

P: 208





#4
Feb2408, 10:25 PM

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Proof of closure under addition and multiplication in a fieldI 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? 



#5
Feb2508, 12:53 AM

P: 206

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.




#6
Feb2508, 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? 



#7
Feb2508, 10:34 AM

P: 208





#8
Feb2508, 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.




#9
Feb2508, 08:54 PM

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I started to answer the question in exactly the same way!



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