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- Jun 22, 2012

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Determine the degree over [TEX] \mathbb{Q} [/TEX] of [TEX] \ 2 + \sqrt{3} [/TEX] and of [TEX] 1 + \sqrt[3]{2} + \sqrt[3]{2}[/TEX]

Peter

[This has also been posted on MHF]

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- Thread starter
- #1

- Jun 22, 2012

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Determine the degree over [TEX] \mathbb{Q} [/TEX] of [TEX] \ 2 + \sqrt{3} [/TEX] and of [TEX] 1 + \sqrt[3]{2} + \sqrt[3]{2}[/TEX]

Peter

[This has also been posted on MHF]

- Feb 15, 2012

- 1,967

Let $u = 2 + \sqrt{3}$. Then $(u - 2)^2 = 3$, so:

$(u - 2)^2 - 3 = 0$.

Multiplying this out, we get:

$u^2 - 4u + 1 = 0$, that is, $u$ is a root of the quadratic polynomial:

$x^2 - 4x + 1$.

If one is willing to accept that $u \not\in \Bbb Q$, this of course means that:

$x^2 - 4x + 1$ is irreducible over $\Bbb Q$. Can you continue?

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- Jun 22, 2012

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Thanks Deveno. Sorry for late reply - day job intervened in things!

Let $u = 2 + \sqrt{3}$. Then $(u - 2)^2 = 3$, so:

$(u - 2)^2 - 3 = 0$.

Multiplying this out, we get:

$u^2 - 4u + 1 = 0$, that is, $u$ is a root of the quadratic polynomial:

$x^2 - 4x + 1$.

If one is willing to accept that $u \not\in \Bbb Q$, this of course means that:

$x^2 - 4x + 1$ is irreducible over $\Bbb Q$. Can you continue?

I followed your post and assume that all we have to do now is quote Theorem 4 of Dummit and Foote section 13.1 which states the following:

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Theorem 4. Let \(\displaystyle p(x) \in F[x] \) be an irreducible polynomial of degree n over a field F and let K be the field \(\displaystyle F[x]/(p(x))\). Let \(\displaystyle \theta = x mod(p(x)) \in K \). Then the elements

\(\displaystyle 1, \theta, {\theta}^2, ... ... , {\theta}^{n-1} \)

are a basis for K as a vector space over F, so the degree of the extension is n i.e.

\(\displaystyle [K \ : \ F] = n \). ... ... ... etc etc

----------------------------------------------------------------------------------

In our situation we have

\(\displaystyle p(x) = x^2 - 4x - 2 \) where \(\displaystyle p(x) \in \mathbb{Q}[x] \).

p(x) has a root u in \(\displaystyle \mathbb{Q}[x]/(p(x)) \).

Specifically \(\displaystyle u = x \ mod(p(x)) \in K = \mathbb{Q}(2 + \sqrt{3}) \)

Then the elements 1 and u are a basis for \(\displaystyle K = \mathbb{Q}(2 + \sqrt{3})\).

Thus the degree of the extension \(\displaystyle \mathbb{Q}(2 + \sqrt{3}) \) over \(\displaystyle \mathbb{Q} \) is 2.

That is [K : F] = 2

Can someone please confirm that the above is correct?

Peter

- Feb 15, 2012

- 1,967

You have to realize that $\Bbb Q[x]/(p(x))$ doesn't EQUAL $\Bbb Q(u)$, but they are ISOMORPHIC.

So you need one small theorem from Linear Algebra:

Isomorphic vector spaces over a common field have the same dimension.

This is what allows us to conclude that $[\Bbb Q(u):\Bbb Q] = \text{deg}(p)$.

It may be instructive to convince yourself that an isomorphism between two extension fields of a base field $F$ that preserves the base field is an $F$-linear map (this will be helpful to you later on when you consider automorphisms of an extension field).

Another approach:

Show that: $\Bbb Q(2 + \sqrt{3}) = \Bbb Q(\sqrt{3})$.