degree of [$\mathbb Z_2[x]/f\mathbb Z_2[x]: \mathbb Z_2$]

Question

I got maybe easy problem. I am not sure if it is true that [$\mathbb Z_2[x]/f\mathbb Z_2[x]: \mathbb Z_2$]=deg $f$ where $f \in \mathbb Z_2[x]$ irreducible. Can anybody help me ? Thanks

Answer 1

Let $\langle f\rangle=f\Bbb{Z}_2[x]$. Now, consider a polynomial $p(x)+\langle f\rangle \in \Bbb{Z}_2[x]/\langle f\rangle$. By the Division Theorem for polynomials, $p(x)=q(x)f(x)+r(x)$ where $r(x)$ has degree less than that of $f(x)$. Therefore, we have the following:

$$p(x)+\langle f\rangle=q(x)f(x)+r(x)+\langle f\rangle=r(x)+\langle f\rangle$$

Thus, all of the elements in $\Bbb{Z}_2[x]/\langle f\rangle$ can be written as polynomials with degree less than $\deg f$. Thus, the following elements span the field:

$$1, x, x^2, [...], x^{\deg f-1}$$

This is because all polynomials with degree less than that of $\deg f$ can be written using exponents of $x$ less than $\deg f$. Thus, we have found a set of $\deg f$ elements that span $\Bbb{Z}_2[x]/\langle f \rangle$. Now, we need to show that these elements are linearly independent.

Let's say we have a relation of linear dependence as follows:

$$a_0+a_1x+a_2x^2+[...]+a_{\deg f-1}x^{\deg f-1}+\langle f\rangle=\langle f\rangle$$

Since $\langle f\rangle$ represents a multiply of $f(x)$, we can rewrite this in terms of regular polynomials as the following ($q(x)$ is simply some arbitrary polynomial):

$$a_0+a_1x+a_2x^2+[...]+a_{\deg f-1}x^{\deg f-1}=q(x)f(x)$$

The left-hand side can be, at most, of degree $\deg f-1$ because that is its biggest exponent of $x$. The right-hand side, however, must be either $0$ or have degree $\deg f$ because $f(x)$ times any polynomial can only have a bigger, not smaller degree. Since the left-hand side can never reach $\deg f$, the right-hand side must be $0$, so both sides are simply $0=0$. Thus, all of the $a_i$ must be $0$ because for a polynomial to be $0$, all of its coefficients must be $0$. Thus, the only relation of linear dependence between these elements is the trivial one, meaning all of these elements are linearly independent.

We have now found a linearly independent basis of $\Bbb{Z}_2[x]/\langle f\rangle$ that has $\deg f$ elements in it. This means that $\Bbb{Z}_2[x]/\langle f\rangle$ is a vector space of dimensionality $\deg f$ over $\Bbb{Z}_2[x]$, concluding the proof.

(Notice that we did not use the fact that $f(x)$ is irreducible. However, $f(x)$ must be irreducible because otherwise, $\Bbb{Z}_2[x]\langle f\rangle$ will not be a field. If you want me to prove this, I can, but otherwise, this is a valid proof.)

Answer 2

As a $\mathbf Z_2$ vector-space, it is isomorphic to the vector space $P_n$ of polynomials of degree at most $n$ (including $0$), via the linear map: \begin{align*} \mathbf Z_2[x]/f\mathbf Z_2[x]&\longrightarrow P_n,\\ p(x)+f\mathbf Z_2[x]&\longmapsto p(x)\bmod f(x). \end{align*}