How do you find the elements of $\mathbb{F} _5 [x] / (x^2 + 2)$?

Question

I am new to the field $\mathbb{F} _5 [x] / (x^2 + 2)$. How would I find all the elements present in this field?

Additionally, I know that the order of the element $x$ is 8 and the order of element $(1+x)$ is 1 (it is the generator), but how would I prove this?

Edit: the order of $(1+x)$ is the same as the order of the group as it is the generator. How would I prove this?

Answer 1

Hint to get you started.

For example, to "find" all the elements of $\mathbb{Z}/(22)$ you can just write the numbers $0, 1, \ldots. 21$ since that gives you one representative of each equivalence class. Those are all the possible remainders when you divide an integer by $22$.

By analogy, write down all the polynomials that can be the remainders of polynomials in $\mathbb{Z}_5[x]$ when you divide by $x^2 -2$. (They will be of degree less than $2$.) That will give you a list of all the elements of that ring. Then you can do all the calculations you need to.

Answer 2

$\mathbb{F} _5 [x] / (x^2 + 2) \cong \mathbb{F} _5 [u]$ with $u^2 =-2$.

$\mathbb{F} _5 $ is the set of all polynomials expressions in $u$ with coefficients in $\mathbb{F} _5$. Since $u^2 =-2$, all polynomials expressions can be reduced to the for $a+bu$ with $a,b \in \mathbb{F} _5$.

Answer 3

To add to @Ethan Bolker's answer, I want to address your question about the order of $x$ and $(x+1)$. You are correct about the order of $x$, since

$$x^8=(x^2)^4=(-2)^4=16=1.$$

However, the order of $x+1$ is not one. This would imply that $(x+1)^2=(x+1)$, which is not true. It should be noted that the order of a generator of a nontrivial group is never 1. The order of $(x+1)$ is 12. You should check this is true.

Answer 4

Yet another way of doing it:

Just as element of $\Bbb C$ may be written $a+bi$, where $a$ and $b$ are allowed to be any real number, and $i^2+1=0$, so in your field you have adjoined the square root of $3\equiv-2\pmod5$ and you can mimic the construction of the complex numbers.

Thus, if you like, you can write the elements of your field as $a+b\sqrt3$, where $a$ and $b$ are allowed to be any elements of $\Bbb F_5$, and “$\sqrt3$” may be taken as a symbol whose square is $3$. Then multiplication is easy: $(a_1+b_1\sqrt3)(a_2+b_2\sqrt3)=a_1a_2+3b_1b_2+(a_1b_2+a_2b_1)\sqrt3$. One way of finding the reciprocal of any quantity here is by the technique of “rationalizing the denominator” that you learned in high-school.

Maybe I can say one more thing, since I’m on my soapbox. I’ve done a lot of hand computation in finite fields $\kappa$ of relatively small cardinality, and I’ve found that the best way to work, if you have a lot of multiplications or divisions to perform, is to write down a list of the powers of a generator of the (cyclic) multiplicative group $\kappa^\times$ of $\kappa$. Then you have what amounts to an (anti)logarithm table, and by making table-lookups, you can do multiplications and divisions with a flick of your wrist instead of a relatively tiresome pencil-and-paper calculation. It seems, in your case, that $1+\sqrt3$ is such a generator (“primitive element”), and to implement my suggestion, of course you’d have to write down all $23$ powers of it, but once that’s done, you have smooth sailing.