• lol_nl
In summary, the use of field extensions in finding solutions of quadratic equations in Z_{n} is necessary when the solutions are not integers. This is done by creating a field extension F=\mathbb{Z}_2[c]/(c^2+c+1) and working with the elements of F as solutions. In this case, there is no need for an imaginary unit i, as it can be represented by the element 1 in the field F.

#### lol_nl

Working through A Book of Abstract Algebra, I encountered several exercises on roots of polynomials in $Z_{n}$ I was just wondering whether there exists something like a quadratic equation for polynomials of degree 2. If the solutions of the usual quadratic formula happen to be integers, can one simply take these modulo n to find solutions of the quadratic in $Z_{n}$? What if they are not integers? Clearly a field extension is needed to find solutions, but how does this precisely work?

As an example, consider the equation $x^{2} + x + 1 = 0$ in $Z_{2}$. The quadratic equation gives $x = -\frac{1}{2} \pm \frac{1}{2} i$. Suppose you name the positive root as c. Then $Z_{2}(c) = {0,1,c,1+c}$ is the field extension. Now can one work with c in the same way as with $x = -\frac{1}{2} \pm \frac{1}{2} i$? Clearly, you easily get contradictions like $1 = -1 = 2c-1 = 3i = i$ and so on.

Maybe it's a good first step to identify a good field extensions that allows you to have roots.

If we have the polynomial $X^2+X+1$ in $\mathbb{Z}_2$, then we can look at the polynomial ring $\mathbb{Z}_2[c]$.
Now, we let $F=\mathbb{Z}_2[c]/(c^2+c+1)$ the quotient of the polynomial ring with the ideal $(X^2+X+1)$. This is again a field, as is easily checked.

The elements of F are $\{0,1,c,c+1\}$ as you suspected.
Now, an important feature of this field is that $c^2=-c-1$. This allows you to multiply all numbers in the field.

Now, you can not work with c as you can work with $-\frac{1}{2}\pm \frac{1}{2}i$. One reason for that is already that $\frac{1}{2}$ doesn't make any sense in $\mathbb{Z}_2$

If you want to work with an analogous things as i then you will have to look for solutions to the equation $X^2+1=0$. Clearly, X=1 is the only solution to this equation. So in this case, there is no need to make an element i such that $i^2=-1$. We already have such an element. That is: i=1. So if you work in F or in $\mathbb{Z}_2$, then you can safely say i=1. Even better: you don't need to talk about i.