Confusing question on $\mathbb{Z}_4$ and the polynomial $x^2+x+1$

Question

I saw on my papers that

Suppose $\mathrm{F}$ is a finite field with ${\rm char}(\mathrm{F})=p$, where is a prime number then cardinality of $F$ is exactly $p^n$ for some $n$.

But, indeed I considered $\mathbb{Z}_4$ (I know this is not a field since it is not even an integral domain) I take $f(x)=x^2+x+1$. Since $\deg(f)=2$ we can say if it has no root in $\mathbb{Z}_4$, then we can say it is irreducible by theorem. I checked for $$f(0)=1, \;f(1)=3,\;f(2)=3,\;f(3)=1$$

And then I said that let $\alpha$ be a root then I got:

$$(x-\alpha)(x+(1+\alpha)) = x^2+x+1$$

Then I observed that $\mathbb{Z}_4$ is a splitting field (how is that possible? How to figure out this? It is not even a field!) If we go as we have no such problems we have a field extension(!) $\mathbb{Z}_4(\alpha)$ over $\mathbb{Z}_4$ and for minimal polynomial of $\alpha$ we have $f(x)$ and thus $\mathbb{Z}_4(\alpha)=\lbrace a+b\alpha:a,b\in\mathbb{Z}_4\rbrace$ thus the cardinality 16?

Can someone help me? There is some absurdum.

Answer 1

You are correct that $\Bbb{Z}_{4}$ is not a field. But the finite field of order $4$ is very different from $\Bbb{Z}_{4}$.

Any finite field call it $F$(an integral domain) must have a prime characteristic say $p$. Thus you can embedd $\Bbb{Z}_{p}$ inside $F$ by the homomorphism $\bar{n}\mapsto n\cdot 1 $ where $1\in F$ is the multiplicative identity. Hence $\Bbb{Z}_{p}$ which we will henceforth denote by $\Bbb{F}_{p}$(A standard notation for a finite field with $n$ elements is $\Bbb{F}_{n}$) sits inside a finite field where $p$ is the characteristic of the finite field $F$.

Now it is a elementary result in Linear Algebra that any field forms a vector space over one of it's subfield.(Try and prove it as an easy excercise)

So we consider the vector space $F$ over it's subfield $\Bbb{F}_{p}$ . Now this vector space must have a basis and say it is $\{a_{1},...,a_{n}\}$ .

Hence any element in $F$ can be written as $c_{1}a_{1}+...+c_{n}a_{n}\,, c_{i}\in \Bbb{F}_{p}$ . But there are precisely $p$ choices for each $c_{i}$ as they are from $\Bbb{Z}_{p}$. So there are just $p^{n}$ elements in $F$.

You will learn in field theory that there is only one finite field of a particular order upto isomorphism. And it indeed occurs as a splitting field . So $\Bbb{F}_{4}$ is indeed a splitting field of $x^{4}-x$.

So to summerize. Any finite field of characteristic $p$ say has order $p^{n}$ for some $n$. And it occurs as a splitting field of the polynomial $x^{p^{n}}-x\in \Bbb{F}_{p}[x]$ .

Now if you are looking to construct such a field of order $p^{n}$ , simply take $\Bbb{F}_{p}[x]$ and find an irreducible polynomial $f(x)$ of order $n$ and quotient by the maximal ideal $(f(x))$ to get $\frac{\Bbb{F}_{p}[x]}{(f(x))}$ .

Answer 2

The polynomial $x^2+x+1$ is irreducible. But $\mathbb{Z}_4$ is not a splitting field. It is also missing with respect to which base field, I suppose it is $\mathbb{Z}_4$. The splitting field of $x^2+x+1$ over $\mathbb{Z}_2$ is $\mathbb{Z}_2(\alpha)$ for a root $\alpha$ in some closure of $\mathbb{Z}_2$.