For each $a$ in $A=\mathbb R[x]/(x^2+x)$ find how many solutions $z^2 = a$ has.

Question

Exercise 15 of chapter two of "Algebra lineal y geometria," Castellet & Llerena.

To be fair, I don't think I even understand the question. What is $z$ supposed to be? A scalar? A polynomial?

I noticed that $p(x)=x^2+x=x*(x+1)$, so if $a|x$ or $a|(x+1)$ then $a$ is a divisor of zero in $A$. But I am not sure that's helpful at all.

Any help would be appreciated, thanks!

Answer 1

Here $z$ is supposed to be another element of $A$. The question is: given an element $a \in A$, how many elements $z \in A$ are there such that $z^2 = a$?

You have a basis of $A$ given by $(1,x)$. It's clearly free and any higher-degree monomial can be written as a multiple of $x$: you have $x^2 = -x$, $x^3 = x$ and so on. So suppose that $a = \alpha_0 + \alpha_1 x$ and that $z = \xi_0 + \xi_1 x$ for some scalars $\alpha_i, \xi_i$. Compute $$z^2 = (\xi_0 + \xi_1 x)^2 = \xi_0^2 + 2 \xi_0 \xi_1 x + \xi_1^2 x^2 = \xi_0^2 + (2\xi_0 \xi_1 - \xi_1^2)x.$$

Therefore the equation $z^2 = a$ is equivalent to the system of equations $\{\xi_0^2 = \alpha_0, 2\xi_0\xi_1 - \xi_1^2 = \alpha_1\}$. You now have several cases:

• If $\alpha_0 < 0$ then there are no solutions.
• If $\alpha_0 = 0$, then $\xi_0 = 0$. You need to solve $-\xi_1^2 = \alpha_1$:
  • If $\alpha_1 > 0$ then there are no solutions.
  • If $\alpha_1 = 0$ there is one solution: $z = 0$.
  • If $\alpha_1 < 0$ there are two solutions: $z = \pm \sqrt{-\alpha_1} x$.
• If $\alpha_0 > 0$, you have two solutions for $\xi_0 = \pm \sqrt{\alpha_0}$. Plugging this back in the second equation gives $\xi_1^2 \mp 2 \sqrt{\alpha_0} \xi_1 + \alpha_1 = 0$. The discriminant is $\Delta = 4 \alpha_0 - 4\alpha_1$.
  • If $\alpha_0 < \alpha_1$, there are no solutions.
  • If $\alpha_0 = \alpha_1$, you get one solution for each of the possibilities for $\xi_0$, so two solutions in total: $z = \pm (\sqrt{\alpha_0} + \sqrt{\alpha_0}x)$.
  • If $\alpha_0 > \alpha_1$, you get two solutions for each of the possibilities for $\xi_0$, so four solutions in total $z = \pm (\sqrt{\alpha_0} + (\sqrt{\alpha_0} \pm \sqrt{\alpha_0 - \alpha_1})x$.