Are all finite fields isomorphic to $\mathbb{F}_p$?

I've recently started taking some algebra courses and I was wondering whether or not every finite field is isomorphic to $\mathbb{F}_p$, where $p$ is prime.

Solution 1:

No. But every finite field is isomorphic to $$F_{p^n}$$ for a prime $p$ and a positive integer $n$. In particular, the finite field with $p^n$ elements is the splitting field of the polynomial

$$x^{p^n} - x$$

over $F_p$. There's a full Wikipedia article here about these.

Solution 2:

You're close, every finite field is isomorphic to $\mathbb{F}_{p^n}$, which is a quotient of $\mathbb{F}_p[x]$ by an irreducible polynomial of degree $n$. It's often difficult to see that things like this can exist without a concrete example. The easiest is $\mathbb{F}_4$:

$$\begin{array}{c|cccc} + & 0 & 1 & \alpha & 1+\alpha\\ \hline 0 & 0 & 1 & \alpha & 1+\alpha\\ 1 & 1 & 0 & 1+\alpha& \alpha \\ \alpha & \alpha & 1+\alpha& 0 & 1 \\ 1+\alpha& 1+\alpha& \alpha & 1 & 0 \\ \end{array}$$

$$\begin{array}{c|cccc} \times & \color{grey}{0} & 1 & \alpha & 1+\alpha\\ \hline \color{grey}{0} & \color{grey}{0} & \color{grey}{0} & \color{grey}{0} & \color{grey}{0} \\ 1 & \color{grey}{0} & 1 & \alpha & 1+\alpha\\ \alpha & \color{grey}{0} & \alpha & 1+\alpha& 1 \\ 1+\alpha& \color{grey}{0} & 1+\alpha& 1 & \alpha \\ \end{array}$$

You can see that $\mathbb{F}_4$ is a group under addition and $\mathbb{F}_4^\times=\mathbb{F}_4\setminus \{0\}$ is a group under multiplication.

In this case, I made $\mathbb{F}_4=\mathbb{F}_{2^2}$ out of the polynomial $x^2+x+1$, which is irreducible over $\mathbb{F}_2$. I get the multiplication table by taking all polynomials of degree $1$ or less (i.e. below the degree of the irreducible polynomial minus one), then multiplying them together and reducing them mod $x^2+x+1$. (For calculation, it doesn't matter whether you use $x$ or $\alpha$.)

Solution 3:

No. As the wikipage linked to by Jack explains there is a unique (up to isomorphism) finite field $\mathbb{F}_q$ with $q$ elements whenever $q=p^n$ is a power of a prime $p$ ($n$ a positive integer).

The simplest example is $\mathbb{F}_4$. It has elements $\{0,1,\alpha,1+\alpha\}$, where $\alpha$ is a mysterious constant that satisfies the equation $$\alpha^2=\alpha+1.\qquad(*)$$ For all $x\in\mathbb{F}_4$ we have $x+x=0$. This explains everything there is to know about the addition in $\mathbb{F}_4$, so for example $$ \alpha+(1+\alpha)=1+(\alpha+\alpha)=1+0=1. $$ That mysterious equation $(*)$ explains everything you need to know about multiplication in $\mathbb{F}_4$ (together with ring axioms). So for example $$ \alpha(1+\alpha)=\alpha+\alpha^2=\alpha+(1+\alpha)=1. $$ This means that $\alpha$ and $1+\alpha$ are (multiplicative) inverses to each other. From this it follows easily that $\mathbb{F}_4$ is, indeed, a field.

An exercise for you: Show that $x^3=1$ for all non-zero elements $x\in\mathbb{F}_4$.

Solution 4:

Not quite.

There is a finite field with $q$ elements, iff $q=p^k$ for some prime $p$. This field is unique up to isomorphy.

So finite fields with a prime number of elements are indeed isomorphic to $F_p$.

But there are also fields with a prime power number of elements, usually also denoted by $F_q$ where $q=p^k$, $k>1$. They can be constructed as following: $$F_q\simeq F_p[X]/(f(X))$$ where $f\in F_p[X]$ is an irreducible polynomial of degree $k$. This construction does not depend on the choice of $f$, as different polynomials (of the same degree) lead to isomorphic fields.