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Math Help - Desperate need of help

  1. #1
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    Desperate need of help

    Let p be a prime integer.

    1. Prove that Z_p is an integral domain.

    2. Prove that Z_p[x] is an integral domain.

    I know that if a ring is a ED it is a PID and if it is a PID it is a UFD and if it is a UFD it is an integral domain. So this can be proven by proving ED, PID or UFD. However, I'm stuck. I need a clear explanation/proof.
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  2. #2
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    Quote Originally Posted by pleasehelpme1 View Post
    I know that if a ring is a ED it is a PID and if it is a PID it is a UFD and if it is a UFD it is an integral domain. So this can be proven by proving ED, PID or UFD. .
    You don't really have to do all of that. To show that a ring is an integral domain, you just need to show that it's commutative, have unity, and have no zero divisors.

    1) Let a,b \in \mathbb{Z}_p and ab=0. Then ab=pk for some k \in \mathbb{Z}, which means that p|a or p|b. Therefore, in \mathbb{Z}_p, a=0 or b=0. \Box



    2) \mathbb{Z}_p[x] is a commutative ring with unity; therefore, we need to show that there are no zero divisors.

    Let f(x), g(x) \in \mathbb{Z}_p[x], where

    f(x)=a_mx^m+a_{m-1}x^{m-1}+\dots+a_0
    g(x)=b_nx^n+b_{n-1}x^{n-1}+\dots+b_0,

    and a_m \not=0 and b_n \not=0. Then f(x)g(x) has leading coefficient a_mb_n, but since \mathbb{Z}_p is an integral domain, a_mb_n \not=0. \Box
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  3. #3
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    ....

    can you help me prove that they are both unique factorization domains?
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  4. #4
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    For that one, I think it would be easier to prove that they are both PID's.


    1) \mathbb{Z}_p is a field, and the only ideals in a field is {0} and the field itself. Since {0} =(0) and \mathbb{Z}_p=(1) , \mathbb{Z}_p is a PID. \Box

    2) Let's look at the stronger case: if K is a field, then K[x] is a PID.

    Let I be a nontrivial ideal in K[x]. Let g(x) be an element in I with minimum degree. We claim that I=(g(x)).

    Clearly, (g(x))\subseteq I. Let f(x) \in I. We can write f(x) as f(x)=g(x)q(x)+r(x), where r(x)=0 or deg r(x)< deg g(x).
    Solving for r(x) gives us r(x)=f(x)-g(x)q(x) \in I \Longrightarrow r(x)=0 (minimality of g(x)). Therefore, f(x)=g(x)q(x) \in (g(x)) \Longrightarrow I \subseteq (g(x)). \Box
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  5. #5
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    ??

    how do you know the only ideal are {0} and the field itself? and how does this mean that every ideal can be written as <a> for some a in the field?
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  6. #6
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    Quote Originally Posted by pleasehelpme1 View Post
    how do you know the only ideal are {0} and the field itself?
    Well, you use the fact that if an ideal of a ring contains unity, then the ideal is equal to the ring. Every nonzero element in a field is a unit, so it pretty much follows from there.
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  7. #7
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    ??

    you are completely losing me here. can you explain this in a simpler way? isn't a field a PID only if every ideal is generated by only one ideal in the field? if so how is {0} and the field itself generated by the same ideal?
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  8. #8
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    No, {0} and the field isn't generated by the same element.

    Ok, let's say we have a field F and a nontrivial ideal I of F. Let a be an element in I. Since a is also in F, there exists an element b in F such that ab=1 (every nonzero element in a field is a unit). Then, by the definition of an ideal, 1=ab \in I. Since I contains unity, I=F. Therefore, the only ideals in the field F are {0} and F.

    In the case of \mathbb{Z}_p, 1 is a generator for \mathbb{Z}_p, so \mathbb{Z}_p=(1), and clearly {0} =(0).
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  9. #9
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    so the its that each ideal is generated by a single element? not that its the same elements that generates all ideals? also how do we know that all nonzero elements in a field are units?
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  10. #10
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    ???

    forget about my last question, but does this mean that the only ideals of all fields are {0} and the field itself?
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