# Thread: Quotient Ring

1. ## Quotient Ring

Just to make sure I get it right.
$\displaystyle \mathbb F_5[x]/(x^2+2)$ is a field.
We want to express $\displaystyle (x^2 + x) \cdot (x^3 + 1) - (x + 1)$ as a polynomial of degree less than 2.
$\displaystyle (x^2+2-2+x)\cdot(x^3+2 -2x+1)-(x + 1)$
$\displaystyle (x-2) \cdot (1-2x) - (x + 1)$
$\displaystyle 4x - 2x^2-3$
$\displaystyle 4x - 2x^2-4+1$
$\displaystyle 4x +1$
Thank you!
Also, I have to find the roots of $\displaystyle t^2+3$
$\displaystyle t=\pm\sqrt{-3}=\pm\sqrt{2}$ but 2 is not a perfect square in $\displaystyle \mathbb Z/5\mathbb Z$
$\displaystyle t^2+3$ has no root.
Is that right????

2. Correct.

Also note that in your field $\displaystyle t^2+3$ reduces to just $\displaystyle 1$.

3. But $\displaystyle t^2 +3$ reduce to 1 only if t=x. Therefore there might be roots even if x is not a root. For example, in this field, $\displaystyle t^2 +3$ has two roots that are x and -x.
Yes it's a cheap example but it's late and it still gives the idea. x is not a variable anymore in this field. 'Polynomials' becomes roots.
Anyway, good night.

4. Originally Posted by vincisonfire
Thank you!
Also, I have to find the roots of $\displaystyle t^2+3$
$\displaystyle t=\pm\sqrt{-3}=\pm\sqrt{2}$ but 2 is not a perfect square in $\displaystyle \mathbb Z/5\mathbb Z$
$\displaystyle t^2+3$ has no root.
Is that right????
The field $\displaystyle \mathbb{F}_5$ can be identified as a subfield of $\displaystyle \mathbb{F}_5/(x^2+2)$.
If $\displaystyle \alpha = x + (x^2+2)$ then $\displaystyle \alpha^2 + 2 = 0 \implies \alpha^2 = 3$

The polynomial $\displaystyle t^2 + 3 \in \mathbb{F}_5[t]$ has a root in $\displaystyle \mathbb{F}_5/(x^2+2)$. Consider $\displaystyle \pm 2\alpha$. Then $\displaystyle ( \pm 2 \alpha)^2 = 4 \alpha^2 = 4\cdot 3 = 2$. Therefore, $\displaystyle (\pm 2 \alpha)^2 + 3 = 0$. Which means $\displaystyle 2\alpha,-2\alpha$ are its roots.

5. Another (similar) way is to see that the discriminant $\displaystyle b^2-4ac=3$ is a square in the field since $\displaystyle x^2+2=0$ --> $\displaystyle x^2=3$

Then I used the usual formula for quadratic functions and got $\displaystyle t = (\pm\sqrt{3})/2 = \pm x * 3$ (since 1/2 = 3 in Z/5Z), which is the same answer as TPH since $\displaystyle \pm 3 = \pm 2$ in this field