I am trying to construct a field with 27 elements. So far I have found an irreducible polynomial of degree 3, $2x^3 + x + 2$ in $\mathbb{Z}_3$ and thus $\langle 2x^3 + x + 2 \rangle$ is maximal in $\mathbb{Z}_3$. Now all that remains is for me to prove the field $\mathbb{Z}_3/ \langle 2x^3 + x + 2 \rangle$ has 27 elements. I know the elements in this field look like $ax^2 + bx + c$ + $f(x)$ with 3 choices for each constant term in the left coset. All that remains is for me to provide justification that The example I provided indeed has 27 elements. Any help would be appreciated on how to do so.
[Math] Field with 27 Elements
abstract-algebrafield-theoryirreducible-polynomialsmaximal-and-prime-ideals
Related Solutions
Lemma 1
If $F$ is a field and if $p$ is an irreducible polynomial with coefficients in $F$, then $F[x]/(p)$ is a field.
Proof
A polynomial $p\in F[x]$ is irreducible if and only if $(p)$ is a maximal ideal in $F[x]$. A quotient $R/I$ of a commutative ring $R$ by an ideal $I$ is a field if and only if $I$ is a maximal ideal in $R$. Q.E.D.
Lemma 2
In the context of Lemma 1, $F\subseteq F[x]/(p)$ so $F[x]/(p)$ is a vector space over $F$. The dimension of this $F$-vector space is $\text{deg}(p)$.
Proof
Firstly, technically $F$ is not a subset of $F[x]/(p)$. However, the canonical homomorphism $F[x]\to F[x]/(p)$ restricts to a homomorphism of $F\to F[x]/(p)$ (since $F$ is a subset of $F[x]/(p)$). Furthermore, this map $F\to F[x]/(p)$ is injective (an element of $F[x]$ is sent to $0$ under $F[x]\to F[x]/(p)$ if and only if its divisible by $p$; clearly, no constant polynomial is divisible by $p$ except $0$). Therefore, we can view $F$ as a subset (in fact, a subfield) of $F[x]/(p)$.
We claim that the set ${\cal B}=\{1+(p),x+(p),x^2+(p),\dots,x^{\text{deg}(p)-1}+(p)\}$ is a basis for $F[x]/(p)$ as an $F$-vector space. Firstly, if $q+(p)\in F[x]/(p)$ (for $q\in F[x]$), then the division algorithm implies that $q=pa+b$ for $a,b\in F[x]$ and $b=0$ or $\text{deg}(b)<\text{deg}(p)$. Since $q+(p)=b+(p)$ and $b+(p)$ is clearly in the span of ${\cal B}$, it follows that ${\cal B}$ spans $F[x]/(p)$ as an $F$-vector space.
An equation of linear dependence for ${\cal B}$ is equivalent to a polynomial $q$ of degree less than that of $p$ equalling zero in $F[x]/(p)$. If the coefficients of $q$ were non-zero, then this would be a contradiction as $p$ cannot divide any non-zero polynomial of smaller degree. Therefore, $q=0$ and ${\cal B}$ is linearly independent.
Therefore, $F[x]/(p)$ is $\text{deg}(p)$-dimensional as an $F$-vector space. Q.E.D.
Lemma 3
If $V$ is a vector space of dimension $n$ over a finite field $F$ with $k$ elements, then $V$ has $k^n$ elements.
Proof
Exercise! Q.E.D.
Exercise: What is the number of elements in $\mathbb{Z}_2[x]/(x^3+x^2+1)$?
I hope this helps!
If a cubic polynomial of $\mathbb{F}_5[x]$ is reducible, then it splits into a linear factor and a quadratic factor or into the product of three linear factors. Linear factors are very easy to test for, as $(x-a)$ is a factor of $f$ if and only if $f(a) = 0$.
So you might choose a random degree $3$ polynomial and test for the five possible roots.
For instance, I'm tempted to try $f(x) = x^3 + 2x + 1$. Then $$\begin{align} &f(0) = 1, \\ &f(1) = 4, \\ &f(2) = 8 + 4 + 1 \equiv 3, \\ &f(3) = 27 + 6 + 1 \equiv 4, \\ &f(4) = 64 + 8 + 1 \equiv 3. \end{align}$$
As none of these are zero, $f(x)$ is an irreducible cubic.
Note that testing for roots explicitly doesn't work for higher degree polynomials. It is possible that a reducible quartic factors into the product of two quadratics. So though there are no roots, it might still be reducible. However since we know that a reducible cubic has a linear factor, we can test just for the linear factor.
Best Answer
If every element in the field is of the form $ax^2+bx+c + <f(x)>$ and there are three choices for $a$, three choices for $b$, and three choices for $c$, and different choices give rise to different elements, then by some very elementary combinatorics there are $3 \times 3 \times 3=27$ possible choices.