Is There a Quadratic Equation for Polynomials in Z_n?

[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.

 

[micromass]

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.