# Minimum polynomial over a field

1. Jan 17, 2009

### math8

The question is to determine the dimension over Q(rationals)of the extension Q(sqrt(3+2sqrt2)).
So we need to find [Q(sqrt(3+2sqrt2)): Q].

All I can say is that (3+2sqrt2) = (1+sqrt2)^2.
I also know that we're trying to find the degree of the minimum polynomial over Q that has sqrt(3+2sqrt2) as a root.

But I don't know how to proceed.

2. Jan 17, 2009

### Hurkyl

Staff Emeritus
I'm not really sure how to explain it, but it seems obvious to me how to go about constructing that particular field extension out of other field extensions that are very easy to understand. (And your observation makes this even more obvious)

You have a problem: you want to compute [Q(sqrt(3+2sqrt2)): Q].
You can simplify this problem.
Do so.

3. Jan 17, 2009

### math8

I would say that this degree is 2, and the minimum polynomial is (x^2)-2x-1.

4. Jan 17, 2009

### math8

My question is, are there some other details that need to be specified other than showing that (x^2)-2x-1 is irreducible?

5. Jan 17, 2009

### Hurkyl

Staff Emeritus
You don't actually need to find the minimal polynomial of that element. You just have to find the degree of the field extension. And you already know the degree of Q(sqrt(3+2sqrt2)); it would be clear if you wrote the extension differently....

If you're not sure about the details, you can always look at the definitions and theorems! It would certainly be faster than asking for help over the internet.

But yes, the relevant theorem is:
Theorem: If f is an irreducible integer polynomial of degree d, and f(a) = 0, then [Q(a) : Q] = d.

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