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

  1. Nov 11, 2008 #1
    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:


    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:


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


    with the ci in F.

  2. jcsd
  3. Nov 11, 2008 #2


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

    But a field also contains multiplicative inverses for all non-zero numbers. What is the multiplicative inverse of [itex]a\sqrt{2}+ b[/itex]? "Rationalize" the denominator of
    [tex]\frac{1}{a\sqrt{2}+ b}[/tex]
    by multiplying numerator and denominator by [itex]a\sqrt{2}- b[/itex]:
    [tex]\frac{1}{a\sqrt{2}+ b}\frac{a\sqrt{2}-b}{a\sqrt{2}-b}= \frac{a\sqrt{2}-b}{2a^2-b^2}[/tex]
    [tex]= \frac{a}{2a^2- b^2}\sqrt{2}+\frac{-b}{2a^2- b^2}[/itex]
    again a "rational number times [itex]\sqrt{2}[/itex] plus a rational number."
    (Note that [itex]2a^2- b^2[/itex] cannot be 0- if [itex]2a^2- b^2= 0[/itex], then [itex]2a^2= b^2[/itex] so [itex]b^2/a^2= 2[/itex] or [itex]b/a= \sqrt{2}[/itex], which is impossible as [itex]\sqrt{2}[/itex] is not a rational number.)

    In other words, every member of [itex]Q(\sqrt{2})[/itex] can be written in the form "[itex]a\sqrt{2}+ b[/itex]" for rational numbers a and b. Notice that is the same as saying that we can use "[itex]\sqrt{2}[/itex]" and "1" as "basis vectors" to write [itex]Q(\sqrt{2})[/itex] as a vector space of dimension 2 over the rational numbers.

    All of that clearly depends on the algebraic properties of "[itex]\sqrt{2}[/itex]". 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 "[itex]^9\sqrt{x}[/itex]" 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. [itex]\sqrt{2}[/itex] 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 "[itex]\sqrt{2}[/itex]". The number [itex]^9\sqrt{n}[/itex] 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 [itex]^9\sqrt{n}[/itex] 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 "[itex]\pi[/itex]" that are "transcendental", not algebraic of any degree. Members of Q(e) or [itex]Q(\pi)[/itex] cannot be written in that way at all.
  4. Nov 11, 2008 #3

    Okay that helped a lot. So, just to see if I'm getting this the members of [tex]Q(\pi)[/tex] would be everything in Q, [tex]\pi[/tex], and all of the other real numbers we'd need to make that a field.

    So, we'd also need [tex]r \pi[/tex] where r is rational. We'd need [tex]r_{1} \pi + r_{2}[/tex]... [tex](r_{1} \pi + r_{2})(r_{3} \pi + r_{4}) = r_{5} \pi^{2}+ r_{6} \pi + r_{7}[/tex]... so we also require the powers of [tex]\pi[/tex]... 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.
  5. Nov 11, 2008 #4


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    The required elements for the field are NOT always in the form of a polynomial. For [tex]Q(\pi)[/tex] you need [tex]\frac{1}{\pi}[/tex] 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 [tex]p(x) = a_0 + a_1x +... + a_nx^n[/tex] 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 [tex] 0 = a_0 + a_1a +... + a_na^n[/tex] so dividing both sides by a and rearranging we get

    [tex]\frac{1}{a} = -\frac{1}{a_0}*(a_1 + a_2a +... + a_na^{n-1})[/tex] (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
  6. Nov 11, 2008 #5
    Office_Shredder, I meant a polynomial where the powers could be any integers... that would cover [tex]\frac{1}{\pi}[/tex] 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?
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