1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook