Quotient ring isomorphism to C and irreducibility over Z_2

[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?