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Subrings and Subfields

  1. Jan 21, 2009 #1
    I'm taking an independent study class over Groups. Rings, and Fields. It's been really confusing. On one page I understand everything completly and the next page I'm completly lost. I'm looking at a problem where I has to show that {x+y(cuberoot of 3) + z(cuberoot of 9) | x,y,z is in Q} is a subring. Now I think I got that...but then I want to prove that it's a subfield. Since I prove that it's a non-empty subset and closed under addition and multiplication by showing that it's a subring, then all I further have to show is that it's a field. (Because to show something is a subfield you just have to show that it's a subset and a field.)

    Then to show that the ring is a field you just have to prove it's closed under addition, subtraction, multiplication, division, and it must be cummutative....Right??? So in this case I just have to prove that it's closed under division and cummutative??

    I know that my denominator will look like: x^3 + 3y^3 +9z^3 - 9xyz but I really need help!!

    Last edited: Jan 21, 2009
  2. jcsd
  3. Jan 21, 2009 #2
    To show something is a field you need to show field axioms. Properties like division for example don't always make sense when talking about abstract fields.

    Have you looked at showing these properties hold:

  4. Jan 21, 2009 #3
    So if I can show that all 5 of those properties hold for x+y(cuberoot of 3) + z(cuberoot of 9) than I can prove that x+y(cuberoot of 3) + z(cuberoot of 9) | x,y,z is in Q} is a subfield?

    If that's the case can you maybe help me with the first one? Like associativity?

    I would really appreciate it. I think I'm making it harder than it has to be...and where does the denominator looking like: x^3 + 3y^3 +9z^3 - 9xyz come into play?
  5. Jan 21, 2009 #4
    Well you want associativity for both operations. You said you have addition and multiplication so for addition, you want to show

    [tex] (a + b \sqrt[3]{3} + c \sqrt[3]{9} + d + e \sqrt[3]{3} + f \sqrt[3]{9}) + g + h \sqrt[3]{3} + i \sqrt[3]{9} = a + b \sqrt[3]{3} + c \sqrt[3]{9} + (d + e \sqrt[3]{3} + f \sqrt[3]{9} + g + h \sqrt[3]{3} + i \sqrt[3]{9}) [/tex]

    Can you show that just because rationals are associative under addition?
  6. Jan 21, 2009 #5
    I'm not sure? I know that it's proven that addition and multiplication of rational numbers are associative and cummutative, but I don't know if thats all you have to show to prove it?

    Also...are you telling me that if I prove associativity, commutativity, distributity, identity, and inverses for both addition and multiplication for this equation that I have proved that it's a subfield? I thought I had to prove so much more!!
  7. Jan 21, 2009 #6
    I just don't see that...to prove something is a subring it says you have to show that

    1. S doesn't equal 0
    2. S is closed under subtraction
    3. S is closed under multiplication

    More precisely:
    1. S doesn't equal 0
    2. For all a,b in S we have a-b in S and a,b in S

    Then for subfield you must show that:
    1. S - {0} doesn't equal 0
    2. S is closed under subtraction
    3. S is closed under multiplication
    4. x is in S-{0} ==> x^(-1) is in S
  8. Jan 21, 2009 #7
    "To show something is a field you need to show field axioms" is not the only way, in fact, it is usually much easier to show that something is a subfield of something.
    lilcoley23@ho, your last post contains all you need. Now that you proved that A is a subring of B. You need to do two more things, show that 1!=0, and that the set A\{0} has multiplicative inverses!

    p.s What text are you using?
  9. Jan 21, 2009 #8
    There's a type in your last post though, in case you aren't aware of (but you probably are)
    It is For all a,b in S we have a-b in S and a.b in S
  10. Jan 21, 2009 #9
    So to show S was a subring of (R , + , .) I used the more precise version I talked about in my last post:

    So, = 0 is in S and so, S is a non empty subset of R

    Therefore, S is a subring of R.

    Now condition 2:

    Let say, x1+y1(cuberoot 3) + z1(cuberoot 9), x2+y2(cuberoot 3) + z2(cuberoot 9) are in S

    Then, x1, x2, y1, y2, z1, z2 are in Q

    So, (x1-x2), (y1-y2), (z1-z2) are in Q and (x1-x2)+(y1-y2)(cuberoot 3) + (z1-z2)(cuberoot 9) are in S

    Therefore, (x1+y1(cuberoot 3) + z1(cuberoot 9)) - (x2+y2(cuberoot 3) + z2(cuberoot 9)) are in S


    So: (x1+y1(cuberoot 3) + z1(cuberoot 9) * (x2+y2(cuberoot 3) + z2(cuberoot 9))

    ==> {x1x2 + 3y1z2 + 3y2z1} + {x1y2 + 3z1z2 + 3x2y1}(cuberoot 3) + {y1y2 + x1z2 + x2z1} (cuberoot 9)

    Which shomws that if x1+y1(cuberoot 3) + z1(cuberoot 9), x2+y2(cuberoot 3) + z2(cuberoot 9) are in S

    Therefore: (x1+y1(cuberoot 3) + z1(cuberoot 9)) * (x2+y2(cuberoot 3) + z2(cuberoot 9)) are in S

    So is that right? And if it is can you please tell me what else I have to show to show it's a subfield...

    I've never seen anything with "show that 1!=0, and that the set A\{0} has multiplicative inverses"

    And about my book...I'm not sure...I use mostly on-line. Really none of the problems I'm trying to solve are like any in the book!!

  11. Jan 21, 2009 #10
    Regarding "1!=0" and " the set A\{0} has multiplicative inverses", they'are just anyway of rephrasing the subfield test, so nevermind about them.

    To show it's a subfield, assume that an element exists and it doesn't equal to zero, now show that its inverse also exists.
  12. Jan 21, 2009 #11
    I think I haven't been quite explicit enough, sorry,

    The last question about the existence of a multiplicative inverse is: is it possible to write [tex]\frac{1}{ x+y\sqrt[3]{3} + z\sqrt[3]{9} }[/tex] in the form [tex]a+b\sqrt[3]{3} + c\sqrt[3]{9}[/tex]. What values of [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex] (for each x, y, and z) do we need then.
    Do they belong to [tex]\mathbf{Q}[/tex].
  13. Jan 28, 2009 #12
  14. Jan 28, 2009 #13
    well, so you need to show Z[cuberoot of 3] is a field.

    Proof 1.
    Think of the map Q[X] to Z[cuberoot of 3] that maps X to "cuberoot of 3". This is obviously a surjective ring-homomorphism. The kernel is the ideal generated by an irreducible polynomial X^3-3. Thus we have an isomorphism Q[X]/(X^3-3) onto Z[cuberoot of 3]. Since (X^3-3) is a maximal ideal, Q[X]/(X^3-3) is a field, so Z[cuberoot of 3] is also a field. ////
  15. Jan 28, 2009 #14
    I'm sorry,

    but I dont' understand why you're just looking at Z(cuberoot 3) instead of the whole equation? Is the x+y(cuberoot of 3) part of the equation irrelevant?

    Please help me understand!
  16. Jan 29, 2009 #15
    I denoted a ring generated by "cuberoot of 3" over Z by Z[cuberoot of 3].

    Let us denote the cube root of 3 by r for the sake of notation.

    Then the set in question is the same as Z[a] because a^3=3.

    By the way, in general, if K is a subfield of F and a is an element of F algebraic over K, F[a]=F(a). Here, F(a) denotes the smallest subfield of F that contains K and a.

    HERE IS ANOTHER PROOF (basically the same as the one I gave yesterday).

    Cuberoot of 3 is a root of an irredicuble polynomial X^3-3. Any nonzero element p of the ring in question is of a form f(cuberoot of 3), where f (=:f(X)) is a polynomial over Z. Since p is not zero, X^3-3 doesn't divide f(X). (If it does, f(p)=0.) As X^3-3 is irredicible, X^3-3 and f(X) are relatively prime. Hence by the Chinese remainder theorem there are nonzero polynomials h(X) and k(X). such that h(X)(X^3-3)+k(X)f(X)=1. Substitute "cuberoot of 3" for X. Then we obtain k(cuberoot of 3)p=1. (f(cuberoot of3)=p by def.) So p has an inverse. ////
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