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Math Help - Polynomial Ring Z[x]/(x^2)

  1. #1
    Super Member Bernhard's Avatar
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    Polynomial Ring Z[x]/(x^2)

    Describe the ring structure of  \mathbb{Z} [x]/(x^2)

    Also, indicate how the ring structure differs from  \mathbb{Z} [x]/(x^2 + 1)

    Would appreciate help with this exercise

    Peter
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  2. #2
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    Re: Polynomial Ring Z[x]/(x^2)

    Quote Originally Posted by Bernhard View Post
    Describe the ring structure of  \mathbb{Z} [x]/(x^2)

    Also, indicate how the ring structure differs from  \mathbb{Z} [x]/(x^2 + 1)

    Would appreciate help with this exercise

    Peter
    Can you see that:

    The former is isomorphic to the ring  \mathbb{Z}[\gamma]=\{a+b\gamma | a,b\in \mathbb{Z}, \gamma^2=0\}

    The latter is isomorphic to the ring of Gaussian integers \mathbb{Z}[i]=\{a+bi|a,b\in \mathbb{Z}, i^2=-1\}
    Thanks from Bernhard
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Polynomial Ring Z[x]/(x^2)

    Thanks ... but I still need some clarification ...

    My thinking ... and then some of my issues/problems follow ...

    Following an example I found in Gallian (page 257), first consider  \mathbb{R}[x]/<x^2> where   \mathbb{R}[x] is the ring of polynomials with real co-efficients.

    Then  \mathbb{R}[x]/<x^2> = \{ g(x) + <x^2> | g(x) \in \mathbb{R}[x] \}

    But   \mathbb{R}[x] is a Euclidean Domain and hence possesses a Division Algorithm, so we may write:

      g(x) = q(x)(x^2) + r(x) where r(x) = 0 or r(x) has degree less than 2.

    so we can write r(x) = ax + b where a, b  \in \mathbb{R}

    Thus  g(x) + <x^2> = q(x)(x^2) + r(x) + <x^2>

    =  r(x) + <x^2>  since the ideal  <x^2> absorbs the term  q(x)(x^2) + r(x)

    =  ax + b + <x^2>

    Thus  \mathbb{R}[x]/<x^2>  = \{ ax + b + <x^2> | a, b \in \mathbb{R} \}

    Now, by a similar argument we can demonstrate that

     \mathbb{R}[x]/<x^2 +1>  = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{R} \}

    which makes the two rings  \mathbb{R}[x]/<x^2> and   \mathbb{R}[x]/<x^2 +1> look to have the same structure?

    One of my questions is how exactly are these two ring structures different?

    A second worry is that the above demonstration works because   \mathbb{R}[x] is a Euclidean Domain ... so the same argument as above does not apply to

      \mathbb{Z}[x] because   \mathbb{Z}[x] is not a Euclidean Domain and hence we cannot use the Division algorithm.

    How do we rigorously demonstrate that

     \mathbb{Z}[x]/<x^2>  = \{ ax + b + <x^2> | a, b \in \mathbb{Z} \} and


     \mathbb{Z}[x]/<x^2 +1>  = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{Z}  \}

    Can someone please help clarify the above problems and issues?

    Peter
    Last edited by Bernhard; April 4th 2013 at 09:49 PM.
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    Re: Polynomial Ring Z[x]/(x^2)

    Quote Originally Posted by Bernhard View Post

    How do we rigorously demonstrate that

     \mathbb{Z}[x]/<x^2>  = \{ ax + b + <x^2> | a, b \in \mathbb{Z} \} and


     \mathbb{Z}[x]/<x^2 +1>  = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{Z}  \}
    Write a surjective homomorphism \mathbb{Z}[x] \to \mathbb{Z}[i] with kernel \langle x^2 +1 \rangle. The first isomorphism theorem for rings gives an isomorphism between \mathbb{Z}[x]/\langle x^2+1\rangle and \mathbb{Z}[i].

    As for how the structures are different: let's look at the \mathbb{R}/\langle x^2 \rangle and \mathbb{R}/\langle x^2+1 \rangle. The first one introduces a non-trivial element \gamma which squares to zero. There is no such element in the second one, which introduces a square root for -1. It is easy to see that this second field is actually \mathbb{C}. If you try to write an ring isomorphism from the first ring to the second ring, you need to send this \gamma to a nilpotent element in \mathbb{C}, but there is no such element. As such, the rings are completely different.
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  5. #5
    Super Member Bernhard's Avatar
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    Re: Polynomial Ring Z[x]/(x^2)

    Thanks Gusbob

    But I am still not sure how you rigorously demonstrate that

     \mathbb{Z}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{Z} \}

    Can you elaborate?

    Peter
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    Re: Polynomial Ring Z[x]/(x^2)

    Quote Originally Posted by Bernhard View Post
    Thanks Gusbob

    But I am still not sure how you rigorously demonstrate that

     \mathbb{Z}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{Z} \}

    Can you elaborate?

    Peter
    Basically the same as the one I showed you.

    f:\mathbb{Z}[x] \to \mathbb{Z}[\gamma] defined by a\mapsto a \quad \forall a\in \mathbb{Z}, x\mapsto \gamma is surjective with kernel \langle x^2 \rangle since \gamma^2=0
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  7. #7
    Super Member Bernhard's Avatar
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    Re: Polynomial Ring Z[x]/(x^2)

    Sorry to be slow but what is  \gamma exactly?

    Further you say that "one introduces a non-trivial element \gamma which squares to zero" - but the elements of  \mathbb{Z}[x]/<x^2> are cosets.

    Do you mean \gamma is the coset  \{ 0 + <x^2> \} ? Then you would bbe saying that every element of this like, for example  x^4 + x^2 is nilpotent?

    Please correct me if necessary.

    Peter
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    Re: Polynomial Ring Z[x]/(x^2)

    Quote Originally Posted by Bernhard View Post
    Sorry to be slow but what is  \gamma exactly?

    Further you say that "one introduces a non-trivial element \gamma which squares to zero" - but the elements of  \mathbb{Z}[x]/<x^2> are cosets.

    Do you mean \gamma is the coset  \{ 0 + <x^2> \} ? Then you would bbe saying that every element of this like, for example  x^4 + x^2 is nilpotent?

    Please correct me if necessary.

    Peter
    Sorry for the confusion. I was referring to my previous notation \mathbb{Z}[\gamma]= \{a+b\gamma |a,b\in \mathbb{Z}, \gamma^2=0\}. You can check that this is a ring by itself. Here, \gamma is meant to be a variable used to distinguish it from x. When I say we introduce a non-trivial element, what I actually meant was adding an element to \mathbb{Z}. I apologise for the ambiguity. In general, when we want to add an element satisfying some polynomial property to a ring R, we add it by forming the associated polynomial ring, then quotient out by the relation it satisfies. For example, to add i satisfying i^2+1=0, we write \mathbb{R}[x]/(x^2+1)\cong \mathbb{R}[i]\cong  \mathbb{C}

    It is true that every element of  \mathbb{Z}[x]/\langle x^2\rangle are cosets, but I was writing a ring homomorphism from that ring to \mathbb{Z}[\gamma]. If you wish, \gamma does correspond to  \{0+\langle x^2 \rangle\}, in which case your third sentence is correct.
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