Isomorphic Math Help: Proving Quotient Ring and Polynomial Irreducibility

[Oxymoron]

Questions

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

2) Show that the polynomial x^2+x+1 is irreducible over \mathbb{Z}_2.

 

[Hurkyl]

Yep, they look like questions, all right.

 

[Oxymoron]

Solution

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

That is

a + bx \rightarrow a + b\lambda

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

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

(a+bx)(c+dx) \rightarrow (a+b\lambda)(c+d\lambda)

(a+bx)(c+dx) = (ac + (bc+ad)x + bdx^2)

(a+b\lambda)(c+d\lambda) = ac + (bc+ad)\lambda +bd\lambda^2

 

[Hurkyl]

What's the kernel of the map R[x] --> C that maps x to λ? Is it surjective? Know any theorems about homomorphisms?

 

[Oxymoron]

The polynomial x^2-x+1 is irreducible over \mathbb{R} since if the polynomial was reducible it would have a linear factor in \mathbb{R}[x] and hence a zero in \mathbb{R}. But the polynomial has no zeroes hence factorization is impossible.

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

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

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

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

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

\phi(a+b) = \phi(a') + \phi(b')
\phi(ab) = \phi(a')\phi(b')

Where a,b are elements of the quotient ring and a',b' are elements of the complex numbers.

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

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 \phi(ab) = \phi(a')\phi(b').

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

And

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

And so \phi(ab) = \phi(a')\phi(b')



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

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

Not sure what all this means though.

 

[Oxymoron]

Solution 2

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

0^2+0+1 = 1 \neq 0

1^2+1+1 = 3 \equiv 1 \neq 0

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

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

is a field.

This field has 2^2=4 elements, namely

0 + 0x = 0
1 + 0x = 1
0 + 1x = x
1 + 1x = 1+x

For example,

(1+x)(1+x) = x^2 + 2x + 1 \equiv (x+1) + 0x + 1 = x + 2 \equiv x

 

[Oxymoron]

Does anyone know if I have done these correctly?