Exercise 11.2.10

Proof.

(a)

Let $g,h\in {𝔽}_p[x]$. If $g+⟨f⟩=h+⟨f⟩$, then $f\mid g-h$. Since $f_i\mid f,i=1,\dots ,k$, then $f_i\mid g-h$, so $g+⟨f_i⟩=h+⟨f_i⟩$. Therefore $\phi$ is well-defined.

Write $\bar{g}=g+⟨f⟩,g\in {𝔽}_p[x]$. For all $g,h\in {𝔽}_p[x]$,

$$\begin{array}{llll}\hfill \phi (\bar{g}+\bar{h})& =(g+h+⟨f_1⟩,\dots ,g+h+⟨f_k⟩)& \hfill & \\ \hfill & =(g+⟨f_1⟩+h+⟨f_1⟩,\dots ,g+⟨f_k⟩+h+⟨f_k)& \hfill & \\ \hfill & =(g+⟨f_1⟩,\dots ,g+⟨f_k⟩)+(h+⟨f_1⟩,\dots ,h+⟨f_k⟩)& \hfill & \\ \hfill & =\phi (\bar{g})+\phi (\bar{h}),& \hfill & \end{array}$$

and similarly

$$\phi (\bar{g}\bar{h})=\phi (\bar{g})\phi (\bar{h}).$$

Finally $\phi (1+⟨f⟩)=(1+⟨f_1⟩,\dots ,1+⟨f_k⟩)$ is the unit of $R_1\times \cdots \times R_k$ for the product. So $\phi$ is a ring homomorphism.

(b)

If $\bar{g}=g+⟨f⟩\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $(g+⟨f_1⟩,\dots ,g+⟨f_k⟩)=(0,\dots ,0)$, so $f_1\mid g,\dots ,f_k\mid g$. Since $f_1,\dots ,f_k$ are distinct monic irreducible polynomials, $f_1,\dots ,f_k$ are relatively prime, therefore $f=f_1\cdots f_k\mid g$. This implies that $\bar{g}=g+⟨f⟩=⟨f⟩=\bar{0}$.

$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{\bar{0}\}$, so $\phi$ is injective.

(c)

We know that $$\dim <!--\; FUNCTION\; APPLICATION\; -->R=\dim <!--\; FUNCTION\; APPLICATION\; -->({𝔽}_p[x]∕⟨f⟩)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f),$$

where $\dim <!--\; FUNCTION\; APPLICATION\; -->R={\dim <!--\; FUNCTION\; APPLICATION\; -->}_{{𝔽}_p}R$ is the dimension of $R$ over ${𝔽}_p$.

Similarly, since $f=f_1\cdots f_k$,

$$\begin{array}{llll}\hfill \dim <!--\; FUNCTION\; APPLICATION\; -->(R_1\times \cdots \times R_k)& =\dim <!--\; FUNCTION\; APPLICATION\; -->R_1+\cdots +\dim <!--\; FUNCTION\; APPLICATION\; -->R_k& \hfill & \\ \hfill & =\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f_1)+\cdots +\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f_k)& \hfill & \\ \hfill & =\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f).& \hfill & \end{array}$$

So

$$\dim <!--\; FUNCTION\; APPLICATION\; -->R=\dim <!--\; FUNCTION\; APPLICATION\; -->(R_1\times \cdots \times R_k).$$

(d)

Since $\phi$ is a ring homomorphism and $\phi (\lambda \bar{g})=\mathit{\lambda \phi }(\bar{g}),\lambda \in {𝔽}_p,\bar{g}\in R$, $\phi$ is linear ( $\phi$ is an algebra homomorphism).

$\phi :R\to R_1\times \cdots \times R_k$ is linear, injective, and $\dim <!--\; FUNCTION\; APPLICATION\; -->R=\dim <!--\; FUNCTION\; APPLICATION\; -->(R_1\times \cdots \times R_k)$, therefore $\phi$ is bijective. So $\phi$ is a ring isomorphism.