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Isomorphic math help

  1. May 25, 2005 #1

    1) Let [itex]x^2-x+1[/itex] be the ideal for [itex]\mathbb{R}[x][/itex] generated by the polynomial [itex]x^2-x+1[/itex]. Show that the quotient ring [itex]\mathbb{R}[x]/(x^2-x+1)[/itex] is isomorphic to the field [itex]\mathbb{C}[/itex] of complex numbers.

    2) Show that the polynomial [itex]x^2+x+1[/itex] is irreducible over [itex]\mathbb{Z}_2[/itex].
  2. jcsd
  3. May 25, 2005 #2


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    Yep, they look like questions, all right.
  4. May 25, 2005 #3

    The isomorphism is going to map solutions [itex]x[/itex] of the polynomial [itex]x^2-x+1[/itex] to a corresponding complex number [itex]\lambda[/itex] which is also a solution of the equation [itex]\lambda^2 - \lambda +1[/itex].

    That is

    [tex]a + bx \rightarrow a + b\lambda[/tex]

    where [itex]a+bx[/itex] is a linear factor of [itex]x^2-x+1[/itex] and [itex]a, b \in \mathbb{R}[/itex].

    So take [itex]x \in \mathbb{R}[x]/(x^2-x+1) [/itex]. Then [itex]x[/itex] is a solution of [itex]x^2-x+1 = 0[/itex]. Now we take a [itex]\lambda \in \mathbb{C}[/itex] such that [itex]\lambda[/itex] solves [itex]\lambda^2 -\lambda + 1[/itex]. To prove that this is an isomorphism we need to show

    [tex](a+bx)(c+dx) \rightarrow (a+b\lambda)(c+d\lambda)[/tex]

    [tex](a+bx)(c+dx) = (ac + (bc+ad)x + bdx^2) [/tex]

    [tex](a+b\lambda)(c+d\lambda) = ac + (bc+ad)\lambda +bd\lambda^2[/tex]
  5. May 25, 2005 #4


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    What's the kernel of the map R[x] --> C that maps x to λ? Is it surjective? Know any theorems about homomorphisms?
  6. May 25, 2005 #5
    The polynomial [itex]x^2-x+1[/itex] is irreducible over [itex]\mathbb{R}[/itex] since if the polynomial was reducible it would have a linear factor in [itex]\mathbb{R}[x][/itex] and hence a zero in [itex]\mathbb{R}[/itex]. But the polynomial has no zeroes hence factorization is impossible.

    Because of this fact, the quotient ring [itex]\mathbb{R}[x]/(x^2-x+1)[/itex] is a field.

    I now look for isomorphisms that take elements in the quotient ring to elements of the complex numbers.

    [tex]\phi : \mathbb{R}[x]/(x^2-x+1) \rightarrow \mathbb{C}[/tex]

    An element in the quotient ring will be of the form [itex](a+bx)[/itex], where [itex]x[/itex] is a solution to [itex]x^2-x+1=0[/itex]. And an element of the complex numbers will be of the form [itex](a+b\lambda)[/itex] where [itex]\lambda[/itex] is a solution to [itex]\lambda^2-\lambda+1=0[/itex].

    Now [itex]\phi[/itex] is an isomorphism if it is a homomorphism with respect to addition and multiplication. That is

    [tex]\phi(a+b) = \phi(a') + \phi(b')[/tex]
    [tex]\phi(ab) = \phi(a')\phi(b')[/tex]

    Where [itex]a,b[/itex] are elements of the quotient ring and [itex]a',b'[/itex] are elements of the complex numbers.

    The kernel of this map is that element of the quotient ring which maps to [itex]0 \in \mathbb{C}[/itex]. That is, [itex]\ker{\phi} = x[/itex] where [itex]x[/itex] is the solution to [itex]x^2-x+1=0[/itex].

    Im not sure if all this so far is necessary to show an isomorphism exists, but I wrote this just to make sure my reasoning is correct. It probably isn't, but someone can point that one out.

    Now it suffices to show that [itex]\phi(ab) = \phi(a')\phi(b')[/itex].

    [tex]\phi(ab) = \phi((a+bx)(c+dx)) [/tex]
    [tex]= \phi(ac + (bc+ad)x + bdx^2)[/tex]
    [tex]= \phi(ac + (bc+ad)x + bd(x-1))[/tex]
    [tex]= \phi(ac + (bc+ad)x + bdx - bd)[/tex]
    [tex]= \phi(ac + (bc+ad-bd)x + bd)[/tex]


    [tex]\phi(a')\phi(b') = \phi(a'+b'\lambda)\phi(c'+d'\lambda)[/tex]
    [tex]= \phi(a'c' + (b'c' + a'd')\lambda + b'd'\lambda^2)[/tex]
    [tex]= \phi(a'c' + (b'c' + a'd')\lambda + b'd'(\lambda -1))[/tex]
    [tex]= \phi(a'c' + (b'c' + a'd' - b'd')\lambda + b'd')[/tex]

    And so [itex]\phi(ab) = \phi(a')\phi(b')[/itex]

    Also note that if we divide [itex]bdx^2 + (bc+ad)x + ac[/itex] by [itex]x^2-x+1[/itex] using long division we obtain

    [tex]ac + (bc+ad-bd)x + bd = \phi^{-1}\phi(ab)[/tex]

    Not sure what all this means though.
    Last edited: May 25, 2005
  7. May 25, 2005 #6
    Solution 2

    We are considering the polynomial [itex]x^2+x+1[/itex] in [itex]\mathbb{Z}_2[/itex]. It suffices to show that it has no roots: [itex]\mathbb{Z}_2 = \{0,1\}[/itex]

    [tex]0^2+0+1 = 1 \neq 0[/tex]

    [tex]1^2+1+1 = 3 \equiv 1 \neq 0[/tex]

    Hence it cannot be factored non-trivially. This means that

    [tex]\mathbb{Z}_2[x]/(x^2+x+1) = \{a+bx | a,b \in \mathbb{Z}_2\}[/tex]

    is a field.

    This field has [itex]2^2=4[/itex] elements, namely

    [tex] 0 + 0x = 0[/tex]
    [tex] 1 + 0x = 1[/tex]
    [tex] 0 + 1x = x[/tex]
    [tex] 1 + 1x = 1+x[/tex]

    For example,

    [itex](1+x)(1+x) = x^2 + 2x + 1 \equiv (x+1) + 0x + 1 = x + 2 \equiv x[/itex]
    Last edited: May 25, 2005
  8. May 26, 2005 #7
    Does anyone know if I have done these correctly?
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