Exercise 11.1.9

Proof.

(a)

If $q_1,q_2\in \mathbb{Z}[x]$ are such that $q_1(x)+\mathit{f\mathbb{Z}}[x]=q_2(x)+\mathit{f\mathbb{Z}}[x]$, then there exist $u_1,u_2\in \mathbb{Z}[x]$ such that $q_1(x)+f(x)u_1(x)=q_2(x)+f(x)u_2(x)$. Since $f(\alpha )=0$, then $q_1(\alpha )=q_2(\alpha )$, so the map $$\phi :\{\begin{array}{ccc}\hfill \mathbb{Z}[x]∕\mathit{f\mathbb{Z}}[x]\hfill & \hfill \to \hfill & \hfill \mathbb{Z}[\alpha ]\hfill \\ \hfill q(x)+\mathit{f\mathbb{Z}}[x]\hfill & \hfill \mapsto \hfill & \hfill q(\alpha )\hfill \end{array}$$

is well defined.

Let $\tilde{q}=q(x)+\mathit{f\mathbb{Z}}[x],\tilde{r}=r(x)+\mathit{f\mathbb{Z}}[x]$ be elements of $\mathbb{Z}[x]∕\mathit{f\mathbb{Z}}[x]$. Then

$$\phi (\tilde{q})+\phi (\tilde{r})=q(\alpha )+r(\alpha )=(q+r)(\alpha )=\phi (q(x)+r(x)+\mathit{f\mathbb{Z}}[x])=\phi (\tilde{q}+\tilde{r}).$$

Similarly, $\phi (\tilde{q})\phi (\tilde{r})=\phi (\tilde{q}\tilde{r})$, and $\phi (\tilde{1})=1$, so $\phi$ is a ring homomorphism.

If $\phi (\tilde{q})=0$, then $q(\alpha )=0$. As $f$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$, $f$ divides $q$ in $\mathbb{Q}[x]$, so $q=\mathit{uf}$, where $u\in \mathbb{Q}[x]$. But $f$ is a monic polynomial in $\mathbb{Z}[x]$, so the algorithm of the Euclidean division gives a quotient $u\in \mathbb{Z}[x]$, therefore $q\in \mathit{f\mathbb{Z}}[x]$, and $\tilde{q}=0$. $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{\tilde{0}\}$, so $\phi$ is injective.

Any element $z$ of $\mathbb{Z}[\alpha ]$ is of the form $z=q(\alpha ),q\in \mathbb{Z}[x]$, so $z=q(\alpha )=\phi (\tilde{q})$: $\phi$ is surjective.

$\phi$ is a ring isomorphism:

$$\mathbb{Z}[x]∕\mathit{f\mathbb{Z}}[x]\simeq \mathbb{Z}[\alpha ].$$

(b)

Let $R$ be the ring $\mathbb{Z}[x]$, and $I=\mathit{p\mathbb{Z}}[x]$, $J=\mathit{f\mathbb{Z}}[x]$ be the principal ideals of $\mathbb{Z}[x]$ generated by $p$ and $f$. By Exercise 8, $$\bar{I}=\{r+J|r\in I\}=\{\mathit{pq}(x)+\mathit{f\mathbb{Z}}[x]|q(x)\in \mathbb{Z}[x]\}$$

is an ideal of $\mathbb{Z}[x]∕\mathit{f\mathbb{Z}}[x]$, and

$$\mathbb{Z}[x]∕⟨p,f⟩=\mathbb{Z}[x]∕(\mathit{p\mathbb{Z}}[x]+\mathit{f\mathbb{Z}}[x])\simeq (\mathbb{Z}[x]∕\mathit{f\mathbb{Z}}[x])∕\bar{I}.$$

The isomorphism $\phi$ of part (a) sends $\mathbb{Z}[x]∕\mathit{f\mathbb{Z}}[x]$ on $\mathbb{Z}[\alpha ]$ and the ideal $\bar{I}$ on $\mathit{p\mathbb{Z}}[\alpha ]$, since $\phi (\mathit{pq}(x)+\mathit{f\mathbb{Z}}[x])=\mathit{pq}(\alpha )$ for all $q\in \mathbb{Z}[x]$.

Therefore $(\mathbb{Z}[x]∕\mathit{f\mathbb{Z}}[x])∕\bar{I}\simeq \mathbb{Z}[\alpha ]∕\mathit{p\mathbb{Z}}[\alpha ]$, so

$$\mathbb{Z}[x]∕⟨p,f⟩\simeq \mathbb{Z}[\alpha ]∕\mathit{p\mathbb{Z}}[\alpha ].$$

(c)

If we switch $I,J$, then by Exercise 8, $R∕(I+J)=R∕(J+I)\simeq (R∕I)∕\bar{J}$, where $$\bar{J}=\{s+I|s\in J\}=\{f(x)v(x)+\mathit{p\mathbb{Z}}[x]|v(x)\in \mathbb{Z}[x]\},$$

so

$$\mathbb{Z}[x]∕⟨p,f⟩\simeq (\mathbb{Z}[x]∕\mathit{p\mathbb{Z}}[x])∕\bar{J}.$$

Let

$$\psi :\{\begin{array}{ccc}\hfill \mathbb{Z}[x]∕\mathit{p\mathbb{Z}}[x]\hfill & \hfill \to \hfill & \hfill {𝔽}_p[x]\hfill \\ \hfill u+\mathit{p\mathbb{Z}}[x]\hfill & \hfill \mapsto \hfill & \hfill \bar{u},\hfill \end{array}$$

where $\bar{u}$ is the reduction of $u\in \mathbb{Z}[x]$ modulo the prime $p$.

As in part (a), $\psi$ is a well-defined ring isomorphism, and $\psi$ sends $\bar{J}$ on $⟨\bar{f}⟩$, since for all $v(x)\in \mathbb{Z}[x]$, $\psi (f(x)v(x)+\mathit{p\mathbb{Z}}[x])=\bar{f}\bar{v}$, and $\bar{v}$ takes all possible values in ${𝔽}_p[x]$ when $v\in \mathbb{Z}[x]$. Therefore $(\mathbb{Z}[x]∕\mathit{p\mathbb{Z}}[x])∕\bar{J}\simeq {𝔽}_p[x]∕⟨\bar{f}⟩$, so

$$\mathbb{Z}[x]∕⟨p,f⟩\simeq {𝔽}_p[x]∕⟨\bar{f}⟩.$$

Finally, if $\alpha$ is a root of the irreducible polynomial $f\in \mathbb{Z}[x]$,

and so $\mathbb{Z}[\alpha ]∕\mathit{p\mathbb{Z}}[\alpha ]$ is a finite field. □