# Math Help - Quotent Ring

1. ## Quotent Ring

I need to prove that $F[x]/\langle p(x)\rangle$ contains both roots of $p(x)$ where p is irreducible and has degree 2

Here is my proof but I dont think I am going at it the right way

Suppose p(x) is irreducible and has degree 2. We want to prove that $F[x]/\langle p(x)\rangle$ contains both roots of p(x). Since p(x) has degree 2 it is of the form $ax^2 + bx + c$. Thus the roots of $ax^2 + bx + c$ are $\frac{-b + \sqrt{b^2 -4ac}}{2a}$ and $\frac{-b + \sqrt{b^2 -4ac}}{2a}$. Since though p(x) is a generator $a,b,2 \in F$[x]. All we need to do now is prove that $-b + \sqrt{b^2 -4ac}$ and $-b - \sqrt{b^2 -4ac}$ are in $F[x]/\langle p(x)\rangle$. This is obtained by multiplying by 2a and subtracting $\frac{-b}{2a}$ But since F is a field all we need to prove is that $-b + \sqrt{b^2 -4ac}$ is in F because F is also abelian group.

If i am on the right path I cannot think of a way to prove the rest any help would be great

2. ## Re: Quotent Ring

Originally Posted by compNewBe
I need to prove that $F[x]/\langle p(x)\rangle$ contains both roots of $p(x)$ where p is irreducible and has degree 2

Here is my proof but I dont think I am going at it the right way

Suppose p(x) is irreducible and has degree 2. We want to prove that $F[x]/\langle p(x)\rangle$ contains both roots of p(x). Since p(x) has degree 2 it is of the form $ax^2 + bx + c$. Thus the roots of $ax^2 + bx + c$ are $\frac{-b + \sqrt{b^2 -4ac}}{2a}$ and $\frac{-b + \sqrt{b^2 -4ac}}{2a}$. Since though p(x) is a generator $a,b,2 \in F$[x]. All we need to do now is prove that $-b + \sqrt{b^2 -4ac}$ and $-b - \sqrt{b^2 -4ac}$ are in $F[x]/\langle p(x)\rangle$. This is obtained by multiplying by 2a and subtracting $\frac{-b}{2a}$ But since F is a field all we need to prove is that $-b + \sqrt{b^2 -4ac}$ is in F because F is also abelian group.

If i am on the right path I cannot think of a way to prove the rest any help would be great
Note that the quadratic formula does not hold if your field is of characteristic 2, so you need some other argument for that case.

I will make a series of assertions. It is up to you to prove them or provide reasoning for them.

Let $\alpha = \frac{-b+\sqrt{b^2-4ac}}{2a}$ be a root of $p(x)$ and consider the substitution homomorphism $\varphi: F[x] \to F[\alpha]$. Since the ideal generated by $p(x)$ is maximal, $p(x)$ generates the kernel of this map. Thus $F[x]/\langle p(x)\rangle$ is isomorphic to the image of $\varphi$, which is $F[\alpha]$ itself. Moreover, $F[x]/\langle p(x)\rangle$ is a field, and so $F(\alpha)=F[\alpha]$. That is, $F[x]/\langle p(x)\rangle \cong F\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)$. This field has $F$-basis $(1,\alpha)$ and so we can recover the other root by writing

$\frac{-b-\sqrt{b^2-4ac}}{2a}=-\frac{2b}{2a}-\frac{-b+\sqrt{b^2-4ac}}{2a}$

3. ## Re: Quotent Ring

I must be stupid I dont get what you mean

4. ## Re: Quotent Ring

What do you specifically not understand?

Do you get that $F[x]/\langle p(x)\rangle \cong F(\alpha)$ where $\alpha$ is a root of the irreducible polynomial $p(x)$?