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Math Help - Isomorphism

  1. #1
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    Exclamation Isomorphism

    Does anyone know how to do this question?

    In each of the following cases, decide whether or not the two rings are isomorphic.
    (a) Z and Q.
    (b) C and H.
    (c) Z6 R and R Z6.
    (d) Q and Q Z2.
    (e) R and C.


    Thanks very much.
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  2. #2
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    Quote Originally Posted by suedenation
    Does anyone know how to do this question?

    In each of the following cases, decide whether or not the two rings are isomorphic.
    (a) Z and Q.
    (b) C and H.
    (c) Z6 R and R Z6.
    (d) Q and Q Z2.
    (e) R and C.


    Thanks very much.
    (a) No. The integers are not dense. And the rationals are dense.

    (b)No. Because \mathbb{C} is a commutative ring, while \mathbb{H} the quaternion numbers are not. Because for example, i\times j=k while j\times i=-k

    (c)Yes. I really do not see how order is important in a direct product between rings.

    (e)No. The reals are ordered but the complex numbers cannot be ordered.

    I am thinking about (d)
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  3. #3
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    (d) Nope.

    Explanation
    -------------
    It is known that the direct product between two groups is a group. And also the direct product between two rings is a ring. But the direct product between two fields IS NOT a field EVER.

    Proof: Let F_1 and F_2 be fields. Then, by definition a field is a commutative division ring. Meaning that every non-zero element in a field has a multiplicative inverse. Thus, we have (a,b)\in F_1\times F_2. Now, (0,0')\in F_1\times F_2 acts like a zero. Thus, we need to show that for all (a,b)\in F_1\times F_2 has an inverse (a^{-1},b^{-1}) where not both a,b are zero. BUT (0,a) is a nonzero element where a\not= 0 has no multiplicative inverse. Because (0,a)(x,y)=(0,ay) thus, there is no way how it can produce (1,1') which acts like a multiplicative inverse in F_1\times F_2

    Thus, \mathbb{Q} is a field. And \mathbb{Z}_2 is also a field. But \mathbb{Q}\times \mathbb{Z}_2 cannot be field. Thus, there is no isomorphism between \mathbb{Q} and \mathbb{Q}\times\mathbb{Z}_2 because it is not a field.

    Note: By 0,0' I mean that 0\in F_1 which is the additive identity and 0'\in F_2 which is the additive identity. Similary 1,1'.
    Q.E.D.
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