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Subfield criterion

I am trying to show that
F := {a + b∛7+ c(∛7)^2 : a,b,c ∈Q} is a subfield of the reals using the subfield criterion. I have done all of it apart from showing that is as multipication inverses also in F and wondered if anyone could help?

Thank you in advance.
Reply 1
Avatar for RichE
RichE
15
Original post by mollyjordansmith
I am trying to show that
F := {a + b∛7+ c(∛7)^2 : a,b,c ∈Q} is a subfield of the reals using the subfield criterion. I have done all of it apart from showing that is as multipication inverses also in F and wondered if anyone could help?

Thank you in advance.


Try writing

1/(a + b∛7+ c(∛7)^2) = A + B∛7+ C(∛7)^2

Multiply up and then compare coefficients.
Reply 2
Avatar for RichE
RichE
15
Original post by mollyjordansmith
Thank you!
From doing this I have got:
Aa+7Bc+7Cb=1
Ab+Ba+7Cc=0
Ac+Bb+Ca=0

and I am unsure how to solve this system for A,B and C


This is a linear system in A,B,C - can you show it's uniquely solvable?
Alternatively, consider F as a vector space over Q. Given non-zero x in F, multiplication by x defines a linear map on the vector space. What's the kernel of x? So, by rank-nullity, what's the image? Deduce 1 is in the image and so x has an inverse in F.
Original post by mollyjordansmith
Just to confirm, I would have a linear map going from F\{0} to F\{0} and will have that the kernel of the map is empty?
F \ {0} isn't a vector space.

It's a linear map from F to F. The kernel won't *quite* be empty.

(You may need to review your vector space material if this is confusing - it's all very standard stuff).
Reply 5
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RichE
15
Original post by DFranklin
Alternatively, consider F as a vector space over Q. Given non-zero x in F, multiplication by x defines a linear map on the vector space. What's the kernel of x? So, by rank-nullity, what's the image? Deduce 1 is in the image and so x has an inverse in F.


Yes, this is a much smarter was of doing this having had a look at the determinant of the linear system I suggested.That there are no zero-divisors means you can work out this kernel.
Original post by DFranklin
F \ {0} isn't a vector space.

It's a linear map from F to F. The kernel won't *quite* be empty.

(You may need to review your vector space material if this is confusing - it's all very standard stuff).


Yes sorry a bit of confusion there.
But I can conclude that dim(Ker)=0?
Original post by mollyjordansmith
Yes sorry a bit of confusion there.
But I can conclude that dim(Ker)=0?
Yes, but you should have at least a few words of justification IMHO.
Original post by DFranklin
...

I might be missing something but how is showing that the kernel of the multiplication map is only the zero vector any easier than what was initially asked for? As far as I can see it seems to boil down to the same set of equations but with all zeroes instead of a one
(edited 6 years ago)
Original post by silentshadows
I might be missing something but how is showing that the kernel of the multiplication map is only the zero vector any easier than what was initially asked for? As far as I can see it seems to boil down to the same set of equations but with all zeroes instead of a one
Because everything embeds in the reals, if X is non-zero, it's invertible (in the reals), so can't be a zero divider.
Hi I am currently in A levels and a friend of mine who lives in Canada sent me this. Can you help solving these please?

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