Basis of field - polynomial

Question

Going through some old exams in my abstract algebra course, and was a bit curious to how I should neatly approach this problem.

Let $F=\mathbb{Z_5}[x]/\langle x^3+x^2+1\rangle$

a) Give a basis of F over $\mathbb{Z_5}$

b) Express $\alpha^4,\alpha^5,1/(\alpha+1)$ in the basis you found in a).

Obviously $x^3+x^2+1$ is irreducible in $\mathbb{Z_5}$ and thus gives us a field. Also if I'm not wrong the number of elements in this field is $5^3 = 125$ and since the polynomial is of degree 3, I'm assuming the basis will be on the form $\{1,\alpha,\alpha^2\}$ but I'm not completely sure about this.

If I'm correct about this, we would end up with something like $\alpha = x + \langle x^3 + x^2 + 1\rangle$

$\alpha^2 = x^2 + \langle x^3 + x^2 + 1\rangle$

and this is where I'm sort of stuck. I don't see how this will lead me anywhere.

Would anyone try to explain this to me, and I'm really interested to see how this will lead me expressing $1/(\alpha+1)$ etc.

Appreciate any help.

Answer 1

You are perfectly right - the basis is $\{1,\alpha,\alpha^2\}$, as a vector space over $\mathbb{F}_5$, where $\alpha$ satisfies $\alpha^3+\alpha^2+1=0$. But then your notation gets in the way: every element of this field of 125 elements can be expressed as $a+b\alpha+c\alpha^2$, with $a,b,c \in \mathbb{F}_5$. For example: $\alpha^4=-\alpha^3-\alpha=\alpha^2+1-\alpha$. And since $\alpha^3+\alpha^2+1=0$, we have $1/\alpha=-\alpha^2-\alpha$. Also $\alpha^2(\alpha+1)+1=0$, so $1/(\alpha+1)=-\alpha^2$, etc.. Can you take it from here?

Answer 2

You're on the right track. You've done a) so I'll give you some hints for b).

Let's determine $\alpha^3$. As $\alpha = x + \langle x^3 + x^2 + 1\rangle$, $\alpha^3 = x^3 + \langle x^3 + x^2 + 1\rangle$. However, $\alpha^3$ is an element of $F$ which is a three-dimensional vector space over $\mathbb{Z}_5$ with basis $\{1, \alpha, \alpha^2\}$, so we must be able to write $\alpha^3$ as a $\mathbb{Z}_5$ linear combination of $\{1, \alpha, \alpha^2\}$. To do this, note that $\alpha^3 + \alpha^2 + 1 = x^3 + x^2 + 1 + \langle x^3 + x^2 + 1\rangle = 0 + \langle x^3 + x^2 + 1\rangle = 0$; I'll leave this to you. Once you have $\alpha^3$ in terms of the basis vectors, you can obtain $\alpha^4, \alpha^5$ by multiplying with $\alpha$, $\alpha^2$ respectively.

As $\alpha, 1 \in F$, and $\alpha \neq -1$, $\alpha + 1 \neq 0$, so it has a multiplicative inverse $\frac{1}{\alpha + 1}$. One way to find this explicitly is to realise that $\frac{1}{\alpha} \in F$ so $\frac{1}{\alpha + 1} = x + y\alpha + z\alpha^2$ for some $x, y, z \in \mathbb{Z}_5$. To determine $x, y, z \in \mathbb{Z}_5$, consider the equation $\frac{1}{\alpha + 1}(\alpha + 1) = 1$ and use the fact that $\{1, \alpha, \alpha^2\}$ is a basis for $F$.

Answer 3

You're correct so far. If $\alpha$ is a root of that polynomial, then the basis of the resulting field will be $\{1, \alpha, \alpha^{2}\}$. Now remember, when you're doing calculations with $\alpha$, it's a root of the polynomial $x^{3} + x^{2} + 1$. So that means that $\alpha^{3} + \alpha^{2} + 1 = 0$, or $\alpha^{3} = -\alpha^{2}-1$.

So, for example, $\alpha^{4} = \alpha^{3} \cdot \alpha = (-\alpha^{2}-1) \cdot \alpha = -\alpha^{3} - \alpha = \alpha^{2} - \alpha + 1$. (In the last expression, you substitute for $\alpha^{3}$ one more time.) $\alpha^{5}$ should be similar.

Now, for $\frac{1}{\alpha+1}$, you really want a solution of $(\alpha + 1)x = 1$. But you know that $x$ is of the form $a + b\alpha + c\alpha^{2}$ for some $a,b,c \in \mathbb{Z}_{5}$. Using that expression for $x$ with give you a set of linear equations to solve to find those coefficients.