Thread: Extension Fields

Prove that
( , ) is isomorphic to ( + )

ATTEMPT:

Well we have been learning about extension fields. I know that if F is a field and a is a zero of a polynomial p(x) in F[x] then F(a) is isomorphic to F[x]/<p(x)>. So if i can find a polynomial such that , and + are all zero that would be great.

I really have no idea what to do can you guys give me hints?

Originally Posted by mulaosmanovicben

Prove that
( , ) is isomorphic to ( + )

ATTEMPT:

Well we have been learning about extension fields. I know that if F is a field and a is a zero of a polynomial p(x) in F[x] then F(a) is isomorphic to F[x]/<p(x)>. So if i can find a polynomial such that , and + are all zero that would be great.

I really have no idea what to do can you guys give me hints?

I figured it out:

We will show each set is a subset of the other:

elements of ( , ) have the form (a+b ) + (c+d )

let q be an element of ( + ) then q has the form:

q = a + b( + ) = a + b +b which is of the form mentioned above with a=a, b=b, c=0, d=b. Therefore ( + ) is a subset of ( , )


Now consider ( + )^3 = 11 + 9 ( I did the math)

subtract 9( + ) from both sides and you get:

2 = ( + )^3 - 9( + ) so is an element of ( + ) similar procedure can show that is also an element of that set and therefore ( , ) is a subset of ( + )

So the sets are the same

your conclusion is correct, but your reasoning is slightly off.

q has the form a + b√2 + c√3 + d√6, so √2 is the vector (0,1,0,0) and √3 is the vector (0,0,1,0) relative to the basis {1,√2,√3,√6}.

the implication the other way is fine.

a more in-depth look at Q(√p,√q) for p,q distinct primes was posted here: http://www.mathhelpforum.com/math-he...ml?pagenumber=