Quick question regarding isomorphic groups?

  • #1
139
0
Let F be a field. R is an element of Mat(2,2)
[a -b
b a]
for a, b in F with matrix operations.
a. Show that R is a commutative ring with 1 and the set of diagonal matrices are
naturally isomorphic to F .
b. For which of the fields Q , R , C , F5
, F7
is R a field?
c. Characterize which elements of R have a multiplicative inverse.
d. Characterize the fields F for which R will be a fi eld.
e. For which Fp
( p prime) is R a field?
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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That looks like an entire homework problem.
So where do you get stuck? What have you tried already?
 
  • #3
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0
Just wanted to confirm my answer.only F2 would be a prime subfield such that R is a field right?

There's just one problem I can't figure out. Give examples of fields F such that
f is isomorphic to f(t) where f(t) is the set of formal fractions.

And just one more thing. How would you show Q[sqrt2 sqrt3] is a field? I'm having trouble showing multiplicative inverse.

Thanks
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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If you want to "confirm an answer", then tell us what your answer is!
 
  • #5
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34
inverse element for field a + bsqrt(2) + c(sqrt(3)

Yes there is an inverse element, you just have to work at it a little. Let's say your element is a + b√2 + c√3 and you want to show that 1/( a + b√2 + c√3) is in the field. As a first step multiply top and bottom of that fraction by a + b√2 - c√3 giving you

( a + b√2 + c√3)/((a + b√2)[itex]^2[/itex] +3c[itex]^{2}[/itex])

the denominator becomes a[itex]^{2}[/itex] + 2b[itex]^{2}[/itex] +3c[itex]^{2}[/itex] + ab√2.

Now multiply numerator and denominator by
a[itex]^{2}[/itex] + 2b[itex]^{2}[/itex] +3c[itex]^{2}[/itex] - ab√2

Your denominator will be all real and the numerator will have various terms that can be simplified down to something in your field.
 
Last edited:
  • #6
329
34
For some reason the answer up above got placed in this topic. But it was the answer to another question. I wonder if I managed to actually do this, or is there a bug somewhere?
 

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