Extended field for $u^2+1$

Question

I'm reading about extended field.

And I wonder what actually is field $F$ : $F_{q^2}= F (u)$, with $u^2 + 1=0$ means?

Answer 1

You are starting with the field $F_q$. Then $F_q(u)\cong \frac {F_q[X]}{(x^2+1)}\cong F_{q^2}$. Thus $u$ is being adjoined, where $u^2+1=0$.

Another way of describing it is as the two-dimensional vector space over $F_q$, with basis $\{1,u\}$.

If you start with $\Bbb R$, on the other hand, you get $\Bbb R(i)\cong\frac{\Bbb R[X]}{(x^2+1)}\cong \Bbb C$.

Answer 2

Think about the complex numbers built from the reals by adding $i$. Then the field is the set of sums $a+bi$ with the familiar rules. The rule for multiplication relies on the fact that $i$ is a root of the irreducible polynomial $x^2 + 1$.

The same construction works over any finite field in which $-1$ does not have a square root. For example, the $9$ element field is (formally) the set of sums $a+bi$ with $a$ and $b$ in $F_3$ (arithmetic mod $3$) and $i^2 = -1 \equiv 2 \pmod{3}$.

Answer 3

Consider the field $F_p$ and the polynomial $X^2+1$.

If $X^2+1=0$ has roots in $F_p$ (for example, when $p=5$, the polynomial has $u=2,u=3$ as roots), nothing happens when you "add” u.

If $X^2+1=0$ doesn't have roots in $F_p$, then $X^2+1$ is irreducible in $F_p[X]$ and hence $$\frac{F_p[X]}{<X^2+1>}$$ is a field with $p^2$ elements. Note that $u = X+<X^2+1>$ becomes a root of this polynomial in this extension, and $\{1,u\}$ becomes a basis of the extension over $F_p$.