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Proving a subring

  1. Jan 25, 2010 #1
    Prove that Q[22/3]={a+b22/3+c42/3} is a subring of the R. Here Q denotes rationals and R denotes reals (I'm not sure how to do the boldface).

    In my algebra class we haven't gotten to the point in which we can just show that it is a subring through subtraction and multiplication.

    This is still a abelian group under addition:
    - (a+b22/3+c42/3)+(c+d22/3+f42/3)=(a+c)+(b22/3+d22/3)+(c42/3+f42/3); closure
    - It has additive identity 0, 0 is in C
    - Additionally it has additive inverses

    For multiplication:
    - one can multiply two elements and still get an element in the ring

    Is this sufficient?
     
  2. jcsd
  3. Jan 25, 2010 #2

    Hurkyl

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    You forgot the multiplicative identity, and the rest of the ring axioms.


    But there is a problem: this would merely show it to be a subset and a ring, but it's not enough to be a subring! You have to remark that 0,1,+,-,* for the larger ring have to, when applied to the smaller ring, give the same values as those operations on the smaller ring.


    There may be another problem -- what precisely do you mean by {a+b22/3+c42/3}?
     
  4. Jan 25, 2010 #3
    The larger ring is C (the complex number), we just have to show that this subset is in fact a subring. I thought a subring needed to suffice 4 axioms: closure under addition, additive inverse, additive identity, and closure under multiplication

    Additionally, I forgot to answer this. It's the cube root, I misinterpreted it.

    a+b[tex]^3\sqrt[/tex]2+c[tex]^3\sqrt[/tex]4
     
    Last edited: Jan 25, 2010
  5. Jan 26, 2010 #4

    Hurkyl

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    I was more wondering if a+b22/3+c42/3 was supposed to be a formal linear combination, or if it was supposed to be the real number named by that linear combination.


    In the general context I'm used to, the phrase "A is a subXXX of B" is defined to mean "A is an XXX, B is an XXX, the underlying set of A is a subset of the underlying set of B, and the inclusion function on those sets defines a homomorphism of XXX's". So I was speaking to that definition.


    But if you are using a different definition -- and there are lots of equivalent ways to define things -- then what you need to prove will need to be geared for that definition instead.

    As an FYI, I feel like making something explicit: something implicit in your definition is that it's not saying "A is a subring", but instead "A, together with the 0, 1, +, -, * formed by taking the operations of B and restricting them to A, is a subring of B"
    _____________________________________________________

    All that said... are you really sure your definition of "subring" doesn't say anything about 1?

    I've been assuming all along that "ring"s have to include the multiplicative identity -- and thus to show something is a subring, it has to include it.

    However, there is another convention that defines "ring" without any reference to multiplicative identity. In that case, it would have nothing to do with whether something is a subring.
     
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