Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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] / <f_\alpha (X) > \cong \mathbb{Z} [\alpha] $. –  user938272 Jun 5 '11 at 17:03
    
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. –  Joel Cohen Jun 5 '11 at 17:15
    
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 $\overline{f}_{\alpha} = X^2 + 1 \equiv (X+1)^2 \mod 2$. –  Joel Cohen Jun 5 '11 at 17:41

2 Answers 2

The map $\mathbb{Z}\to\mathbb{F}_p$ given by reduction modulo $p$ induces a homomorphism $\mathbb{Z}[x]\to\mathbb{F}_p[x]$ (reducing coefficients modulo $p$); this is the universal property of the polynomial ring.

Now consider the composite map $$\mathbb{Z}[x] \to \mathbb{F}_p[x] \to \frac{\mathbb{F}_p[x]}{\langle \overline{f_{\alpha}(x)}\rangle}.$$ The map is onto, because the induced map $\mathbb{Z}[x]\to\mathbb{F}_p[x]$ is onto, and the canonical projection is onto. By the isomorphism theorems, you know that $\mathbb{F}_p[x]/\langle \overline{f_{\alpha}(x)}\rangle$ is isomorphic to the quotient of $\mathbb{Z}[x]$ modulo the kernel of the map.

It should be clear that the kernel contains both $p$ and $f_{\alpha}(x)$. So you just need to show that this is the entire kernel. If $g(x)$ lies in the kernel, then $g(x)\equiv f_{\alpha}(x)\pmod {p}$; this tells you that you can express $g(x)$ as a multiple of $f_{\alpha}(x)$ up to multiples of $p$ in the coefficients, which gives the equality you want.

For the other isomorphism, consider the map $\mathbb{Z}[x]\to\mathbb{Z}[\alpha]$ obtained by "evaluation at $\alpha$". The kernel of this map is precisely the ideal generated by the minimal polynomial of $\alpha$, so $$\mathbb{Z}[\alpha]\cong\frac{\mathbb{Z}[x]}{\langle f_{\alpha}(x)\rangle}.$$ Under this isomorphism, $p + \langle f_{\alpha}(x)\rangle$ maps to $p$ in $\mathbb{Z}[\alpha]$, so the isomorphism theorem tells you that the ideal corresponding to $(p+\langle f_{\alpha}(x)\rangle)$ in the quotient corresponds to the ideal of $\mathbb{Z}[x]$ generated by $p$ and $f_{\alpha}(x)$; this gives the second isomorphism.

share|improve this answer

In a nutshell, all of these rings are made of polynomials in one variable where you add the condition that $p = 0$ and $f_{\alpha}(X) = 0$ (making $X$ a root of $f_{\alpha}$). In the first one, you cancel $p$ then $f_{\alpha}(X)$ (which becomes $\overline{f_{\alpha}}$ once you cancel $p$), in the third you cancel $f_{\alpha}(X)$ then $p$, and in the second, you do it simultaneously. You can easily check that the order does not matter (and give explicit isomorphisms).

share|improve this answer
    
Sorry if this is a stupid question; I haven't quite got my head round quotients. What do you mean by "cancel"? –  user938272 Jun 5 '11 at 17:21
    
One can think of taking the quotient of a ring $A$ by an ideal $I$ like canceling the elements of $I$ in the ring (which is also like canceling the generator(s) of the ideal). With this point of view, it is clear why we have $\mathbb{Z}[X]/\langle f_{\alpha} \rangle \simeq \mathbb{Z}[\alpha]$, because taking a quotient by $\langle f_{\alpha} \rangle$ is exactly like decreeing that $f_{\alpha}(X) = 0$, which transforms $X$ into a root of $f_{\alpha}$. –  Joel Cohen Jun 5 '11 at 17:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.