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

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!

• Welcome to Maths SX! What is $R$? Feb 8 '19 at 12:55
• Bernard, it is supposed to be the group of real numbers. Sorry, I didn't know how to make that symbol in particular and used upper case R. Feb 9 '19 at 12:44

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 $$z = \pm (\sqrt{\alpha_0} + (\sqrt{\alpha_0} \pm \sqrt{\alpha_0 - \alpha_1})x$$.
• Thank you so much :) Feb 9 '19 at 12:43