Exercise 11.2.11

Proof.

(a)

Let $$T^{\prime }:\{\begin{array}{ccc}\hfill R^{\prime }\hfill & \hfill \text{→}\hfill & \hfill R^{\prime }\hfill \\ \hfill y=(g_1+⟨f_1⟩,\dots ,g_k+⟨f_k⟩)\hfill & \hfill \mapsto \hfill & \hfill y^p=(g_1^p+⟨f_1⟩,\dots ,g_k^p+⟨f_k⟩)\hfill \end{array}$$

We show first that

$$T^{\prime }=\phi \circ T\circ {\phi }^{-1},$$

which means that the following diagram is commutative:

Let $y=(g_1+⟨f_1⟩,\dots ,g_k+⟨f_k⟩)$ be any element of $R^{\prime }$. By Exercise 10, $\phi$ is bijective, so there exists a unique $x=g+⟨f⟩\in R$ such that $\phi (x)=y$. So there exists $g\in {𝔽}_p[x]$ such that $(g+⟨f_1⟩,\dots ,g+⟨f_k⟩)=(g_1+⟨f_1⟩,\dots ,g_k+⟨f_k⟩)$. We have so proved the Chinese Remainder Theorem: there exists $g\in {𝔽}_p[x]$ such that

$$g\equiv g_1(\mathit{mod}{f}_1),\dots ,g\equiv g_k(mod{f}_k).$$

This implies

$$g^p\equiv g_1^p(\mathit{mod}{f}_1),\dots ,g^p\equiv g_k^p(mod{f}_k),$$

thus

$$\begin{array}{llll}\hfill (\phi \circ T\circ {\phi }^{-1})(y)& =(\phi \circ T)(g+⟨f⟩)& \hfill & \\ \hfill & =\phi (g^p+⟨f⟩)& \hfill & \\ \hfill & =(g^p+⟨f_1⟩,\dots ,g^p+⟨f_k⟩)& \hfill & \\ \hfill & =(g_1^p+⟨f_1⟩,\dots ,g_k^p+⟨f_k⟩)& \hfill & \\ \hfill & =T^{\prime }(y).& \hfill & \end{array}$$

Therefore $T^{\prime }=\phi \circ T\circ {\phi }^{-1}.$

Now we prove that, for all $\bar{g}=g+⟨f⟩\in R$,

$$T(\bar{g})=\bar{g}⟺T^{\prime }(\phi (\bar{g}))=\phi (\bar{g}).$$

$\text{∙}$

If $T(\bar{g})=\bar{g}$, then $$T^{\prime }(\phi (\bar{g}))=(\phi \circ T\circ {\phi }^{-1})(\phi (\bar{g}))=\phi (T(\bar{g}))=\phi (\bar{g}).$$

$\text{∙}$

If $T^{\prime }(\phi (\bar{g}))=\phi (\bar{g})$, then $(\phi \circ T\circ {\phi }^{-1})(\phi (\bar{g}))=\phi (\bar{g})$, so $(\phi \circ T)(\bar{g})=\phi (T(\bar{g}))=\phi (\bar{g})$. Since $\phi$ is bijective, $T(\bar{g})=\bar{g}$, and the equivalence is proved.

Thus, if $\bar{g}\in \ker <!--\; FUNCTION\; APPLICATION\; -->(T-1_R)$, then $T(\bar{g})=\bar{g}$, $T^{\prime }(\phi (\bar{g}))=\phi (\bar{g})$, $\phi (\bar{g})\in \ker <!--\; FUNCTION\; APPLICATION\; -->(T^{\prime }-1_{R^{\prime }})$. Conversely, if $y\in \ker <!--\; FUNCTION\; APPLICATION\; -->(T^{\prime }-1_{R^{\prime }})$, then $y=\phi (g),g\in R$, so $T^{\prime }(\phi (\bar{g}))=\phi (\bar{g})$, therefore $T(\bar{g})=\bar{g}$, so $g\in \ker <!--\; FUNCTION\; APPLICATION\; -->(T-1_R)$. We have proved

$$\phi (\ker <!--\; FUNCTION\; APPLICATION\; -->(T-1_R))=\ker <!--\; FUNCTION\; APPLICATION\; -->(T^{\prime }-1_{R^{\prime }}),$$

so $\phi$ induces an isomorphism between the kernel of $T-1_R$ and the kernel of $T^{\prime }-1_{R^{\prime }}$.

(b)

Let $y=(g_1+⟨f_1⟩,\dots ,g_k+⟨f_k⟩)$. Then $$y\in \ker <!--\; FUNCTION\; APPLICATION\; -->(T^{\prime }-1_{R^{\prime }})⟺f_1\mid g_1^p-g,\dots <!--\; FUNCTION\; APPLICATION\; -->,f_k\mid g_k^p-g_k.$$

Let $T_i:R_i\to R_i$ defined by $T_i(g_i+⟨f_i⟩)=g_i^p+⟨f_i⟩$, where $1\le i\le k$. Then

$$\ker <!--\; FUNCTION\; APPLICATION\; -->(T^{\prime }-1_{R^{\prime }})=\ker <!--\; FUNCTION\; APPLICATION\; -->(T_1-1_{R_1})\times \cdots \times \ker <!--\; FUNCTION\; APPLICATION\; -->(T_k-1_{R_k}).$$

Moreover

$$g_1\in \ker <!--\; FUNCTION\; APPLICATION\; -->(T_1-1_{R_1})⟺f_1\mid g_1^p-g_1.$$

The set of $\alpha \in R_1={𝔽}_p[x]∕⟨f_1⟩$ such that ${\alpha }^p-\alpha =0$ is a subfield isomorphic to ${𝔽}_p$, the prime field $\{[0]+⟨f_1⟩,\cdots ,[p-1]+⟨f_1⟩\}$, whose dimension is 1 as subspace of $R_1$:

$${\dim <!--\; FUNCTION\; APPLICATION\; -->}_{{𝔽}_p}\ker <!--\; FUNCTION\; APPLICATION\; -->(T_1-1_{R_1})=1.$$

Similarly ${\dim <!--\; FUNCTION\; APPLICATION\; -->}_{{𝔽}_p}\ker <!--\; FUNCTION\; APPLICATION\; -->(T_i-1_{R_i})=1$ for $i=1,\dots ,k$. Therefore

$${\dim <!--\; FUNCTION\; APPLICATION\; -->}_{{𝔽}_p}\ker <!--\; FUNCTION\; APPLICATION\; -->(T^{\prime }-1_{R^{\prime }})=k.$$

(c)

By part (a), $\phi (\ker <!--\; FUNCTION\; APPLICATION\; -->(T-1_R))=\ker <!--\; FUNCTION\; APPLICATION\; -->(T^{\prime }-1_{R^{\prime }})$, where $\phi$ is a vector space isomorphism, thus $$\dim <!--\; FUNCTION\; APPLICATION\; -->(\ker <!--\; FUNCTION\; APPLICATION\; -->(T-1_R))=\dim <!--\; FUNCTION\; APPLICATION\; -->\ker <!--\; FUNCTION\; APPLICATION\; -->(T^{\prime }-1_{R^{\prime }})=k.$$

Therefore, by the Rank Theorem, $T-1_R$ has rank $n-k$. In particular,

$f$ is irreducible over ${𝔽}_p$ $⟺$ $k=1$ $⟺$ $T-1_R$ has rank $n-1$.

This is another proof of Theorem 11.2.9.