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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.

Sirajuddin Asjad
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1 Answers1

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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 $x=-2i$.

user729424
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    This might be a silly question, but how do we know $2i\neq -2i$? – Isaac Browne Dec 14 '19 at 21:25
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    If $2i=-2i$ then $4i=0$. Since we are in a field, this would imply that either $4=0$ or $i=0$, neither of which is true in $\mathbb{F}_7[i]$. – user729424 Dec 14 '19 at 21:28
  • So the answer I originally got is correct? The final answer is x = 2i or x = -2i? If so then why do I need to think about finite fields in this equation? I got my answers from factorization and did not think about $\mathbb{F}_7$ at all. That is why I assumed that I did something wrong. – Sirajuddin Asjad Dec 14 '19 at 21:38
  • Well, I guess you have to use that $7$ does not divide $4$, but other than that you don't :D – Isaac Browne Dec 14 '19 at 21:51
  • I don't know where the question came from, but I think you had the right idea. Let $F=\mathbb{F}_7$ and let $K=\mathbb{F}_7[i]$. You can solve the equation the way you normally would in $K$ because $K$ has an element $i$ with $i^2=-1$. Using this element, we can write $x^2+4$ as $x^2-4i^2$ and then use the difference of squares identity, which works in any commutative ring. In $F$, there is no element $i$ with $i^2=-1$, so we cannot do this. And the equation $x^2+4=0$ has no solution in $F$. – user729424 Dec 15 '19 at 19:32