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Degrees of a Field Extension

shmounal

New member
Feb 14, 2012
3
Hi,

I've been asked to find $[ \mathbb{Q}(\sqrt{2}, i ): \mathbb{Q} ]$, and to write down a basis.

Now I know that $[ \mathbb{Q}(\sqrt{2}, i ): \mathbb{Q} ] = 4$, and that a basis could be $ \left \{ 1, i, \sqrt{2}, i\sqrt{2}\right \} $ it is whether the way I am explaining how I arrived here is satisfactory explanation.

Utilising the tower law I want to find both $[ \mathbb{Q}(\sqrt{2}, i ): \mathbb{Q(\sqrt{2})} ]$ and $[ \mathbb{Q}(\sqrt{2}): \mathbb{Q} ]$ and multiply.

For $[ \mathbb{Q}(\sqrt{2}): \mathbb{Q} ] = 2$ as $\mathbb{Q}(\sqrt{2}) = [a+b\sqrt{2} | a,b \in \mathbb{Q}]$ so $\mathbb{Q}(\sqrt{2})$ is a 2d vector space over $\mathbb{Q}$ with basis $\left \{1, \sqrt{2}\right \}$

I then use a similar argument for $[ \mathbb{Q}(\sqrt{2}, i ): \mathbb{Q(\sqrt{2})} ] = 2$ in that $\mathbb{Q}(i)(\sqrt{2}) = [a+bi | a,b \in \mathbb{Q(\sqrt{2})}]$

and the rest follows. I'm not sure if this is enough to show it rigidly. Some of my friends have been using minimal polynomials to show the different degrees and I don't entirely follow the logic!

Any help appreciated!

xx
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Some of my friends have been using minimal polynomials to show the different degrees and I don't entirely follow the logic!

Certainly you should justify why $[\mathbb{Q}(\sqrt{2}): \mathbb{Q}]=2$ . As $\sqrt{2}\not\in \mathbb{Q}$ , then $[\mathbb{Q}(\sqrt{2}): \mathbb{Q}]>1$ , but $p(x)=x^2-2\in\mathbb{Q}[x]$ satisfies $p(\sqrt{2})=0$ , so $p(x)$ is the minimal polynomial of $\sqrt{2}$ and a basis of $\mathbb{Q}(\sqrt{2})$ over the field $\mathbb{Q}$ is $\{(\sqrt{2})^0,(\sqrt{2})^1\}=\{1,\sqrt{2}\}$ . In the same way , $i\not\in\mathbb{Q}(\sqrt{2})$ , then ...
 
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shmounal

New member
Feb 14, 2012
3
Thanks for the help!! Think I understand now.

What I was doing only showed that the degree was 2 or more while finding the min polynomial says it must be exactly 2?

i.e. $p(x) = x^2 + 1 \in\mathbb{Q} (\sqrt{2})[x]$ satisfies $p(i) = 0$, ....
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
What I was doing only showed that the degree was 2 or more while finding the min polynomial says it must be exactly 2?
Right, that is the question.

i.e. $p(x) = x^2 + 1 \in\mathbb{Q} (\sqrt{2})[x]$ satisfies $p(i) = 0$, ....
Also right.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
another way to show that [Q(√2):Q] = 2 is to show that {1,√2} generate Q(√2), to do this, you need to prove that every element of Q(√2) is of the form a+b√2 (i.e. that the set of all such elements with a,b in Q does indeed form a field, and this field is contained in any field that contains Q and √2).

finding an irreducible polynomial of minimal degree is often easier, because of the isomorphism Q(√2) ≅ Q[x]/(x^2 - 2). note that proving x^2 - 2 is irreducible over Q is the same as showing √2 is irrational, something often glossed over as "obvious" and stated without proof. a similar consideration holds for i, and to be really nit-picky, showing x^2+1 is irreducible over Q(√2)[x] one ought to appeal to the fact that Q(√2) is a sub-field of the reals, which has no square roots of negative numbers (a consequence of being a totally ordered field).

when one has a cubic (or higher) extension, the polynomial approach really shines, because then one knows that successive powers of a root (up to one less than the degree of the minimal polynomial, in this case: 2, since we have a cubic) are linearly independent (or else the root would have a minimal polynomial of lesser degree). so the polynomial itself provides the basis.