# Homework Help: In algebra I'm having trouble with this definition: simple extension

1. Nov 11, 2008

### futurebird

In algebra I'm having trouble with this definition:

The extensions K of F is a simple extension of F if F = F(a) for some a in K.

What I understand so far:

F is a set. But the notation F(a) has me lost-- what is it? Is is the values you get when you plug a in to any polynomial over th field F? So, one value in the set F(a) would be:

$$b_{1}(a)+b_{2}(a)^{3}+b_{3}(a)^{9}+b_{4}$$

Where the bi are from F ? And I can make any polynomial like that then fin it's value for a?

This is hard to think about.

Also... is the meaning for F(a,b) ... that I can plug in a or b? or can I mix and match like this:

$$c_{1}(a)-c_{2}(b)^{5}+c_{3}(b)^{9}+c_{4}$$

Is that in F(a, b)... or is it just stuff like:

$$c_{1}(a)+c_{2}(a)^{5}-c_{3}(a)^{9}+c_{4}$$
$$c_{1}(b)+c_{2}(b)^{5}+c_{3}(b)^{9}+c_{4}(b)^{9}+c_{5}$$

with the ci in F.

Help!

2. Nov 11, 2008

### HallsofIvy

What you "understand" apparently is not correct. F is NOT just a set, it is a subFIELD of K. If F is a subfield of K and a is a member of K, then F(a) is the smallest subfield of K that contains all member of F and a. If, for example, a is already in F, then F(a)= F. If a is not in F then F(a) is some larger field, the "extension" field.

Here's an example. Let F be the field of real numbers and R be the field of rational numbers. $\sqrt{2}$ is a real number but not a rational number. What is the smallest field that contains all rational numbers and $\sqrt{2}$. Since a field must be closed under multiplication, it certainly must contain any number of the form $a\sqrt{2}$ where a is any rational number. Since a field is closed under addition, it must contain any number of the form $a\sqrt{2}+ b$ where a and b are rational numbers. We don't have to specifically mention numbers of the form $a\sqrt{2}+ b\sqrt{2}$ because that is equal to $(a+b)\sqrt{2}$, already a "rational number times $\sqrt{2}$". But, again because a field is closed under multiplication, it must contain things like $(a\sqrt{2}+ b)(c\sqrt{2}+ d)$. Multiplying that out, we get $ac(2)+ (ad+bc)\sqrt{2}+ bd= (2ac+bd)+ (ad+bc)\sqrt{2}$, already of the form "a rational number times $\sqrt{2}$ plus a rational number".

But a field also contains multiplicative inverses for all non-zero numbers. What is the multiplicative inverse of $a\sqrt{2}+ b$? "Rationalize" the denominator of
$$\frac{1}{a\sqrt{2}+ b}$$
by multiplying numerator and denominator by $a\sqrt{2}- b$:
$$\frac{1}{a\sqrt{2}+ b}\frac{a\sqrt{2}-b}{a\sqrt{2}-b}= \frac{a\sqrt{2}-b}{2a^2-b^2}$$
$$= \frac{a}{2a^2- b^2}\sqrt{2}+\frac{-b}{2a^2- b^2}[/itex] again a "rational number times $\sqrt{2}$ plus a rational number." (Note that $2a^2- b^2$ cannot be 0- if $2a^2- b^2= 0$, then $2a^2= b^2$ so $b^2/a^2= 2$ or $b/a= \sqrt{2}$, which is impossible as $\sqrt{2}$ is not a rational number.) In other words, every member of $Q(\sqrt{2})$ can be written in the form "$a\sqrt{2}+ b$" for rational numbers a and b. Notice that is the same as saying that we can use "$\sqrt{2}$" and "1" as "basis vectors" to write $Q(\sqrt{2})$ as a vector space of dimension 2 over the rational numbers. All of that clearly depends on the algebraic properties of "$\sqrt{2}$". I don't know where you got the exponents "3" and "9". Now, I don't know where you got the exponents 3 and 6 but they would have to depend on the specific value of "a". Is it something like "$^9\sqrt{x}$" for some rational number x? What can be shown is "If a is algebraic of degree n, then Q(a) can be written as a vector space of dimension n over the rational numbers." "Algebraic of degree n" means "satisfies a polynomial equation with rational coefficients but no such polynomial equation of lower degree". "A vector space of dimension n over the rational numbers" means we can pick out n numbers in the set and write all the others as sums of rational numbers times those n numbers. $\sqrt{2}$ is "algebraic of order 2" because it satisfies x2= 2 but no linear equation. The two "basis" vectors, as I said above, are "1" and "$\sqrt{2}$". The number $^9\sqrt{n}$ where n is, say, an integer, not a perfect 9th power,or cube, is algebraic of degree 9. If it happens to be a perfect cube, then $^9\sqrt{n}$ is algebraic of order 3 which looks like the case here. But, again, the way you can write members of F(a) depends strongly on what both F and a are! In fact, there exist numbers, like "e" and "$\pi$" that are "transcendental", not algebraic of any degree. Members of Q(e) or $Q(\pi)$ cannot be written in that way at all. 3. Nov 11, 2008 ### futurebird Thanks!!! Okay that helped a lot. So, just to see if I'm getting this the members of [tex]Q(\pi)$$ would be everything in Q, $$\pi$$, and all of the other real numbers we'd need to make that a field.

So, we'd also need $$r \pi$$ where r is rational. We'd need $$r_{1} \pi + r_{2}$$... $$(r_{1} \pi + r_{2})(r_{3} \pi + r_{4}) = r_{5} \pi^{2}+ r_{6} \pi + r_{7}$$... so we also require the powers of $$\pi$$... so any polynomial with rational coefficient ... and then you can plug in pi.

A field needs multiplicative inverses, if we let the powers of the polynomial be integers we have those... But the thing is this only works for Q-- because in Q we are able to show that the required elements for the field are in the form of the polynomial ... OK. I think that makes sense--- But, if we don't know what the field is we don't know that it will take on the polynomial form.

4. Nov 11, 2008

### Office_Shredder

Staff Emeritus
The required elements for the field are NOT always in the form of a polynomial. For $$Q(\pi)$$ you need $$\frac{1}{\pi}$$ which in this case can't be written as a polynomial in terms of pi (if it could, then pi would be algebraic)

Often, the element you're extending the field by is the solution to a polynomial $$p(x) = a_0 + a_1x +... + a_nx^n$$ where the ai's are all in your base field. In cases like these the field extension behaves nicely... for example, if your new element is a solution and we call it 'a', p(a) = 0 gives us $$0 = a_0 + a_1a +... + a_na^n$$ so dividing both sides by a and rearranging we get

$$\frac{1}{a} = -\frac{1}{a_0}*(a_1 + a_2a +... + a_na^{n-1})$$ (if the constant term is zero, you just need to divide by a twice, or three times, or however many times to get a 1/a term).

In fact, it turns out that if you're extending by an algebraic element (some a s.t. p(a) = 0 for a polynomial in the base field) then your field extension is always a finite dimensional vector space over the base field. If the element isn't algebraic, then your field extension is never a finite dimensional vector space over the base field

5. Nov 11, 2008

### futurebird

Office_Shredder, I meant a polynomial where the powers could be any integers... that would cover $$\frac{1}{\pi}$$ right?

This comment is very helpful-- I think I'm starting to get this a little more. I've been confused because of something my prof. said-- I might have heard him wrong, but he was explaining the difference between:

F[x] and F(x)

He said that F[x] is the set of all polynomials in x with coefficients from the field F. And F(x) is the set of polynomials and quotients of polynomials. Is this true for all kinds of fields?