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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.

Arturo Magidin
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user938272
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  • 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 05 '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 05 '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 05 '11 at 17:41
  • Such quotient reciprocity arises from applying the Third Isomorphism Theorem in two different ways. – Bill Dubuque Oct 05 '20 at 22:40

2 Answers2

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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.

Arturo Magidin
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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).

Joel Cohen
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  • Sorry if this is a stupid question; I haven't quite got my head round quotients. What do you mean by "cancel"? – user938272 Jun 05 '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 05 '11 at 17:31