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Math Help - The subring Z[p/q] of Q

  1. #1
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    The subring Z[p/q] of Q

    Hello. I am working with the subring Z[p/q] of the ring of integers Q. Among other things, I want to show that Z[1/2], as a Z-module, cannot be generated by an element a + b/2 in Z[1/2]. Otherwise, if we take the element 1/2 in Z[1/2] then for some m in Z we can write

    1/2 = ma +mb/2.

    So this implies a = 0 and b = 1 so Z[1/2] is generated by 1/2. But then, for 1 + 1/2 in Z[1/2],

    1 + 1/2 = m/2

    for some m in Z is impossible. This is where I am starting to get confused. Couldn't we just add the left hand side of the last equation and then say m had to of been 3?



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  2. #2
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    Quote Originally Posted by Jumping View Post
    Hello. I am working with the subring Z[p/q] of the ring of integers Q. Among other things, I want to show that Z[1/2], as a Z-module, cannot be generated by an element a + b/2 in Z[1/2].


    Do you mean here \displaystyle{a+\frac{b}{2}\,,\,\,a,b\in\mathbb{Z}  } ?



    Otherwise, if we take the element 1/2 in Z[1/2] then for some m in Z we can write

    1/2 = ma +mb/2.

    So this implies a = 0 and b = 1


    If the answer to my question above is yes then no: it doesn't imply this. For example, take a=2\,,\,b=-3 , so

    the question is: what do you really mean here?

    Tonio



    so Z[1/2] is generated by 1/2. But then, for 1 + 1/2 in Z[1/2],

    1 + 1/2 = m/2

    for some m in Z is impossible. This is where I am starting to get confused. Couldn't we just add the left hand side of the last equation and then say m had to of been 3?



    .
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  3. #3
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    Thank you for the help Tonio. My goal is to show that the subring Z[1/2] of Q, when considered as a Z-module, cannot be finitely generated. I figure I ought to first show that Z[1/2] cannot be generated by a single element from Z[1/2].

    I think my confusion starts when I was writing on my scratch paper that

    Z[1/2] = {a + b*(1/2) : a, b are in Z}.

    But, Z[1/2] is a subring of Q and so it also has a multiplicative structure. For example 1/2 is in Z[1/2] and so (1/2)(1/2) = 1/4 is also in Z[1/2]. But I don't see how I can produce an element 1/4 in Z[1/2] by my above description of Z[1/2].

    So I think what I am really trying to do is show that if a is any element of Z[1/2], then I can find some element b in Z[1/2] so that it is impossible to write

    b = m*a

    for any m in Z. This would then show that the Z-module Z[1/2] cannot be generated by any single element a in Z[1/2].

    I will keep working on this. Thanks again Tonio.
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    Quote Originally Posted by Jumping View Post
    Thank you for the help Tonio. My goal is to show that the subring Z[1/2] of Q, when considered as a Z-module, cannot be finitely generated. I figure I ought to first show that Z[1/2] cannot be generated by a single element from Z[1/2].

    I think my confusion starts when I was writing on my scratch paper that

    Z[1/2] = {a + b*(1/2) : a, b are in Z}.


    This, of course, is false. We can define \displaystyle{\mathbb{Z}\left[\frac{1}{2}\right]:=\{f(1/2)\,;\,\,f(x)\in\mathbb{Z}[x]\} , so for example, with f(x)=x^4 , we get

    \displaystyle{f\left(\frac{1}{2}\right)=\frac{1}{1  6}\in\mathbb{Z}\left[\frac{1}{2}\right]-\left\{a+b\cdot\frac{1}{2}\,;\,\,a,b\in\mathbb{Z}\  right\}}



    But, Z[1/2] is a subring of Q and so it also has a multiplicative structure. For example 1/2 is in Z[1/2] and so (1/2)(1/2) = 1/4 is also in Z[1/2]. But I don't see how I can produce an element 1/4 in Z[1/2] by my above description of Z[1/2].

    So I think what I am really trying to do is show that if a is any element of Z[1/2], then I can find some element b in Z[1/2] so that it is impossible to write

    b = m*a

    for any m in Z. This would then show that the Z-module Z[1/2] cannot be generated by any single element a in Z[1/2].

    I will keep working on this. Thanks again Tonio.

    Do you know what an integral element is? And do you know that \mathbb{Z} is an integrally

    closed domain?

    If you know the above then it is inmediate that \mathbb{Z}\left[\frac{1}{2}\right] cannot be a f.g. \mathbb{Z}- module,

    lest \frac{1}{2} is an integral non-integer rational number, which of course is absurd.

    If you still don't know the above, then you'll have to work (slightly) harder: suppose the above ring is a f.g.

    module over the integers, so that \displaystyle{\mathbb{Z}\left[\frac{1}{2}\right]=\left\{\sum\limits^r_{k=1}m_kf_k(1/2)\,,\,\,m_k\in\mathbb{Z}\,,\,\,f_k(x)\in\mathbb{Z  }[x]\right\}} , and let

    n:=\max\{\deg f_1,\,\deg f_2,\ldots,\deg f_r\} . Well, now I let you to prove the

    easy fact that \displaystyle{\frac{1}{2^{n+1}}} cannot be generated by the above "generators"...

    Tonio
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