Solve $x^2+4=0$ in $\mathbb{F}_7[\sqrt{-1}]$

57 Views Asked by At

I'm trying to solve this equation and I'm not sure how to proceed. I'm having some difficulties to understand finite field equations with complex numbers.

The task is:

Solve $x^2+4=0$ in $\mathbb{F}_7[\sqrt{-1}]$

My progress so far:

$x^2 = -4$

$x=\sqrt{-1} \cdot \sqrt{4} = {^+_-}2i$

$(x-2i)(x+2i)$

I'm not sure how to continue from here. Do I need to calculate the modulo to proceed with the finite field calculations? Or is ${^+_-}2i$ the final answer?

Thanks a lot in advance.

1

There are 1 best solutions below

5
On BEST ANSWER

It might be better to write $i=\sqrt{-1}$, and to keep in mind that in this context, $i$ is not an element of $\mathbb{C}$. What we're doing is we're starting with the field $\mathbb{F}_7$, and we're adding a new element to $i$ to it. And this element $i$ has the property that $i^2=-1$. $\mathbb{F}_7[i]$ is the smallest field which contains both $\mathbb{F}_7$ and $i$.

To solve $x^2+4=0$ in $\mathbb{F}_7[i]$, note that:

$$0=x^2+4=x^2-4i^2=(x-2i)(x+2i).$$

Since $(x-2i)(x+2i)=0$, and we are in a field, it follows that either $x-2i=0$ or $x+2i=0$. So either $x=2i$ or $x=-2i$.