Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

[Math Amateur]

Can someone help me get started on the following problem.

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]

 

[Deveno]

Let's try a naive approach:

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?

 

[Math Amateur]

Let's try a naive approach:

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?

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

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

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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

 

[Deveno]

More or less.

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

 

[blue_raver22]

To determine the degree over \mathbb{Q} of 2 + \sqrt{3}, we need to find the minimal polynomial of this element. We can do this by considering the elements 1, \sqrt{3}, and 2 + \sqrt{3} as a basis for the field extension \mathbb{Q}(\sqrt{3}). Now, we can express any element in this field as a linear combination of these basis elements. Therefore, we can write 2 + \sqrt{3} as a polynomial in \sqrt{3} with rational coefficients.

Let's call this polynomial f(x) = ax + b, where a and b are rational numbers. We can see that f(2 + \sqrt{3}) = 0, which means that 2 + \sqrt{3} is a root of f(x). This means that f(x) is a factor of the minimal polynomial of 2 + \sqrt{3}. Now, we can calculate f(x) by plugging in x = 2 + \sqrt{3} and setting it equal to 0. This gives us f(x) = 3 + 2\sqrt{3}, which is a polynomial of degree 1. Therefore, the degree over \mathbb{Q} of 2 + \sqrt{3} is 1.

To determine the degree over \mathbb{Q} of 1 + \sqrt[3]{2} + \sqrt[3]{2}, we can follow a similar approach. We consider the elements 1, \sqrt[3]{2}, and 1 + \sqrt[3]{2} + \sqrt[3]{2} as a basis for the field extension \mathbb{Q}(\sqrt[3]{2}). Now, we can express any element in this field as a linear combination of these basis elements. Therefore, we can write 1 + \sqrt[3]{2} + \sqrt[3]{2} as a polynomial in \sqrt[3]{2} with rational coefficients.

Let's call this polynomial g(x) = ax^2 + bx + c, where a, b, and c are rational numbers. We can see that g(1 + \sqrt[3]{2} + \sqrt[3]{2}) = 0, which means that 1 + \sqrt[3]{2} + \sqrt[3]{2