3
$\begingroup$

Could someone please explain to me why

$ \mathbb{F}_p [X] / \langle\bar{f_\alpha} (X)\rangle \,\ \cong \,\ \mathbb{Z}[X] / \langle p, f_\alpha (X) \rangle \,\ \cong \,\ \mathbb{Z} [\alpha] / \langle p\rangle $?

Where $p$ is prime in $\mathbb{Z}$, $ \bar{f_\alpha} $ is the polynomial obtained by taking the coefficients of $ f_\alpha $ modulo p (and $ f_\alpha $ is the minimal polynomial of $\alpha$ in $ \mathbb{Z}[X] $) I know that $ \mathbb{F_p} \cong \mathbb{Z}/p\mathbb{Z} $.

Thanks.

  • 0
    And I know the three isomorphism theorems, I just can't get it to work. I'm confused about $ \bar{f_\alpha} $. Is it the minimal polynomial of $ \alpha $ in $\mathbb{F}_p [X]$ ? I also know that $ \mathbb{Z}[X] / \cong \mathbb{Z} [\alpha] $.2011-06-05
  • 0
    Note that $f_{\alpha}$ and $\overline{f}_{\alpha}$ are essentially the same when you're in $\mathbb{F}_p[X]$. But rigourously speaking, $f_{\alpha}$ is not in $\mathbb{F}_p[X]$ (it coefficients are elements of $\mathbb{Z}$, not elements of $\mathbb{F}_p$), so we give the name $\overline{f}_{\alpha}$ to make the distinction clearer.2011-06-05
  • 0
    Also, I may add that $\overline{f}_{\alpha}$ is not necessarily irreducible in $\mathbb{F}_p[X]$ but this is not important here (your isomophism holds anyway). For an example, you make take $p = 2$, $\alpha = i$, and $f_{\alpha} = X^2 + 1$. Then $\overline{f}_{\alpha} = X^2 + 1 \equiv (X+1)^2 \mod 2$.2011-06-05

2 Answers 2