Essential preliminary for algebraic number theory (1)

Zhan Ziqian, Stroll About in Spring (游春图).

In this essay, $R$$R$ is always an $\mathrm{U}\mathrm{F}\mathrm{D}$$\rm UFD$. $F$$F$ is the fraction field of $R$$R$​.

It is a preparation for algebraic number theory, reveal the connections between $R[T]$$R[T]$ and $F[T]$$F[T]$​​.

Let us consider a integral domain $S$$S$. We could define a preorder on $S-\{0\}$$S-\{0\}$ as $a|b⟺b=sa⟺(a)\supseteq (b)$$a|b\iff b=sa\iff (a)\supseteq (b)$​.

Act the posetalization functor on $S-\{0\}$$S-\{0\}$ we get a poset $\mathcal{P}=S-\{0\}/S^{\ast }$$\mathcal{P}=S-\{0\}/S^*$.

Definition. For $s\ne 0\in S$$s\ne 0\in S$, if $d|s⟹(d)=(s)$$d|s\implies (d)=(s)$, then we say $s$$s$ is irreducible.

If every element in $\mathcal{P}$$\mathcal P$ can be uniquely written as product of irreducible elements $p_1^{s_1}...p_n^{s_n}$$p_1^{s_1}...p_n^{s_n}$, then we say $S$$S$ is UFD.

Some basic observation.

$†.$$\dagger.$ Prime is irreducible. Since if $p=ab$$p=ab$, suppose $p|a$$p|a$, then $p|a\wedge a|p$$p|a\wedge a|p$. Hence $(p)=(a)$$(p)=(a)$.

$†.$$\dagger.$ Let ${\mathcal{P}}^{\prime }$$\mathcal P'$ be the poset of principle ideal in $R$$R$, then $(\mathcal{P},|)\cong ({\mathcal{P}}^{\prime },\supseteq )$$(\mathcal P,|)\cong (\mathcal P',\supseteq)$.

$r\in \mathcal{P}$$r\in\mathcal P$ is irreducible iff $(r)$$(r)$ is a maximal element in $({\mathcal{P}}^{\prime },\subseteq )$$(\mathcal P',\subseteq)$​​.

$†.$$\dagger.$ $({\mathcal{P}}^{\prime },\subseteq )$$(\mathcal P',\subseteq)$ satisfies ACC. That is, $(a_1)\subseteq (a_2)...$$(a_1)\subseteq (a_2)...$ must become stable. Since $a_1$$a_1$ only have finite many factors.

Conversely, if $(P^{\prime },\subseteq )$$(P',\subseteq)$ satiefies ACC, then every $a\in \mathcal{P}$$a\in \mathcal P$ can be factor to product of irreducible elements.

Since $(a_1)\subseteq (a_2)⟺a_2|a_1⟺a_1=r_1{a}_2...$$(a_1)\subseteq (a_2)\iff a_2|a_1\iff a_1=r_1 a_2...$

$†.$$\dagger.$ If all the irreducible elements are prime, for $a\in \mathcal{P}$$a\in \mathcal P$ then if $a=p_1^{e_1}...p^{e_s}$$a=p_1^{e_1}...p^{e_s}$, then it is the unique way.

Since if $a=p_1^{e_1}...p_n^{e_s}=q_1^{i_1}...q_m^{i_m}$$a=p_1^{e_1}...p_n^{e_s}=q_1^{i_1}...q_m^{i_m}$, then $p_1|q_j$$p_1|q_j$ for $1\le j\le m$$1\le j\le m$. But both $p_1$$p_1$ and $q_j$$q_j$ are irreducible, $p_1=q_j$$p_1=q_j$...

Collorary. For an integral domain $R$$R$, $R$$R$ is UFD iff $({\mathcal{P}}^{\prime },\subseteq )$$(\mathcal P',\subseteq)$ satisfies ACC and all the irreducible elements are prime.

Definition.

Definition. A content of a polynomial $f=a_n{T}^n+...+a_0\in R[T]$$f=a_nT^n+...+a_0\in R[T]$ is $c(f)\cong gcd(a_n,...,a_0)$$c(f)\cong \gcd(a_n,...,a_0)$​. The notation $\cong$$\cong$ here means equal but up to unit. For example, $gcd(3,4)\cong 1\cong -1$$\gcd(3,4)\cong 1\cong -1$. If $c(f)$$c(f)$ is a unit, then we say that $f$$f$ is primitive.

Lemma. Product of two primitive polynomial is primitive as well. Further, $c(fg)\cong c(f)c(g).$$c(fg)\cong c(f)c(g).$

Proof.

Let $f=\sum _{i=0}^n{a}_i{T}^i,g=\sum _{j=0}^m{b}_j{T}^j.$$f=\sum_{i=0}^n a_i T^i,g=\sum_{j=0}^m b_jT^j.$ Suppose $fg$$fg$ is not primitive, then there exists a $p,p|c(fg).$$p,p|c(fg).$ Since $f,g$$f,g$ is primitive, $p$$p$ does not divide $c(f)$$c(f)$ and $c(g)$$c(g)$. Hence $p$$p$ does not dived all the $a_i$$a_i$ and $b_j$$b_j$. Let $m,k$$m,k$ be the smallest number that $p$$p$ does not dived $a_m$$a_m$ and $b_k$$b_k$. Then let us consider the coefficient of $T^{m+k}$$T^{m+k}$, i.e. $\sum _{i+j=m+k}{a}_i{b}_j=...+a_{m-1}{b}_{k+1}+a_m{b}_k+a_{m+1}{b}_{k-1}$$\sum_{i+j=m+k} a_ib_j=...+a_{m-1}b_{k+1}+a_mb_k+a_{m+1}b_{k-1}$.

Notice that $p|a_{m-i}{b}_{k+j}$$p|a_{m-i}b_{k+j}$ since $p|b_{k+j}$$p|b_{k+j}$. Similarly, $p|a_{m+i}{b}_{k-j}$$p|a_{m+i}b_{k-j}$. By assumption, $p|c(fg)$$p|c(fg)$, hence $p|\sum _{i+j=m+k}{a}_i{b}_j$$p|\sum_{i+j=m+k} a_ib_j$, hence $p|a_m{b}_k$$p|a_{m}b_{k}$. But $p$$p$ is prime, hence $p|a_m$$p|a_m$ or $p|b_k$$p|b_k$. That is a contradiction. Hence $fg$$fg$ is also primitive.

For $c(fg)\cong c(f)c(g)$$c(fg)\cong c(f)c(g)$, let $f=c(f)f_0,g=c(g)g_0$$f=c(f)f_0,g=c(g)g_0$. Then $f_0,g_0$$f_0,g_0$ will be primitive and $fg=c(f)c(g)f_0{g}_0.$$fg=c(f)c(g)f_0 g_0.$ Since $c(f_0{g}_0)$$c(f_0g_0)$ is unit and $gcd(b{a}_n,...,b{a}_1)=bgcd(a_n,...,a_1)$$\gcd(ba_n,...,ba_1)=b\gcd(a_n,...,a_1)$, $c(fg)\cong c(f)c(g)$$c(fg)\cong c(f)c(g)$. $◻$$\square$

Corollary. If $p\in R$$p\in R$ is prime, then $p\in R[T]$$p\in R[T]$​ is prime as well. Hence $p$$p$ is primitivr in $R[T_1,...,T_n]$$R[T_1,...,T_n]$ by induction.

Proof. Since $R$$R$ is integral domain, $R^{\times }=R[T{]}^{\times }$$R^\times=R[T]^\times$, so $p$$p$ is not a unit. Let $f\in R[T]$$f\in R[T]$, $p|f⟺p|c(f)$$p|f\iff p| c(f)$. $◻$$\square$

Lemma. Let $f,g\in R[T]$$f,g\in R[T]$ and $g$$g$ is primitive. Then if $g|f\in F[T]$$g|f\in F[T]$, then $g|f\in R[T]$$g|f\in R[T]$.

Proof. Let $f=hg$$f=hg$ where $h\in F[T]$$h\in F[T]$. Let $a$$a$ be the lcm of the denominator of the coefficients of $h$$h$, then $ah\in R[T]$$ah\in R[T]$.

Hence $ah=c(ah)h_0,af=c(ah)h_{0}g.$$ah=c(ah)h_0,af=c(ah)h_0g.$ Since both $h_0$$h_0$ and $g$$g$ are primitive, $h_{0}g$$h_0g$ is primitive. Take the content both side, we get that $ac(f)\cong c(ah)⟹a|c(ah)=gcd(\frac{a{s}_1^{\prime }}{s_1},...,\frac{a{s}_n^{\prime }}{s_n})$$ac(f)\cong c(ah)\implies a| c(ah)=\gcd(\frac{as_1'}{s_1},...,\frac{as_n'}{s_n})$. Hence $h\in R[T]$$h\in R[T]$. $◻$$\square$​

Corollary. Let $g\in R[T]$$g\in R[T]$ be primitive. If $g$$g$ is prime in $F[T]$$F[T]$, then $g$$g$ is prime in $R[T]$$R[T]$. We will use this to prove that If $R$$R$ is UFD, then $R[T]$$R[T]$ is UFD as well.

Lemma. Gauss Lemma.

Let $f\in R[T]$$f\in R[T]$ be primitive. $f$$f$ is irreducible in $R[T]$$R[T]$, if and only if $f$$f$ is irreducible in $F[T]$$F[T]$.

Proof. Suppose $f$$f$ is reducible in $F[T]$$F[T]$, then $f=gh$$f=gh$. Let $g=g^{\prime }/d_1,h=h^{\prime }/d_2\in R[T]$$g=g'/d_1,h=h'/d_2\in R[T]$, where $d_1,d_2\in R-\{0\}.$$d_1,d_2\in R-\{0\}.$

We have $f=g^{\prime }{h}^{\prime }/d_1{d}_2$$f=g'h'/d_1d_2$, hence $d_1{d}_{2}f=g^{\prime }{h}^{\prime }$$d_1d_2f=g'h'$. Act content both sides we get $d_1{d}_{2}c(f)\cong c(g^{\prime })c(h^{\prime })$$d_1d_2c(f)\cong c(g')c(h')$.

Oberve that $f$$f$ is primitive, $c(f)=1$$c(f)=1$, hence $d_1{d}_2\cong c(g^{\prime })c(h^{\prime })$$d_1d_2\cong c(g')c(h')$. $f=(g^{\prime }/c(g^{\prime }))(h^{\prime }/c(h^{\prime }))$$f=(g'/c(g'))(h'/c(h'))$, where $g^{\prime }/c(g^{\prime }),h^{\prime }/c(h^{\prime })\in R[T]$$g'/c(g'),h'/c(h')\in R[T]$.

That is a contradiction to $f$$f$ is irreducible in $R[T]$$R[T]$. $\Leftarrow$$\Leftarrow$is obviously. $◻$$\square$

Proposition. $R$$R$ is UFD implies $R[T]$$R[T]$ is UFD.

Proof. Let $f\in R[T]$$f\in R[T]$, then $f=c(f)f_0$$f=c(f)f_0$. Since $R$$R$ is UFD, $c(f)$$c(f)$ is product of irreducible elements in $R$$R$. Hence product of irreducible elements in $R[T]$$R[T]$. So we only need to deal with primitive polynomial. Let $f_0=gh$$f_0=gh$. Here $gh\in R[T]$$gh\in R[T]$ are primitive as well. By induction on the degree, we can factor $f_0$$f_0$ to primitive irreducible.

Now we need to show uniqueness of factorization. i.e. All the irreducible elements are prime.

Let $f$$f$ be irreducible,in $R[T]$$R[T]$ hence $f$$f$ is irreducible in $F[T]$$F[T]$. Hence $f$$f$ is prime in $F[T]$$F[T]$. By the previous corollary, i.e.

Corollary. Let $g\in R[T]$$g\in R[T]$ be primitive. If $g$$g$ is prime in $F[T]$$F[T]$, then $g$$g$ is prime in $R[T]$$R[T]$. We will use this to prove that If $R$$R$ is UFD, then $R[T]$$R[T]$ is UFD as well.

$f$$f$ is prime. Hence $R[T]$$R[T]$ is UFD as well. $◻$$\square$

Eisenstein's criterion.

Let $R$$R$ be a commutative ring, $f(T)=a_0+a_{1}T+...+a_n{T}^n\in R[T]$$f(T)=a_0+a_1T+...+a_n T^n\in R[T]$. Let $\mathfrak{p}$$\mathfrak p$ be a prime ideal.

If $a_n\notin \mathfrak{p},a_0,...,a_{n-1}\in \mathfrak{p},a_0\notin {\mathfrak{p}}^2$$a_n\not\in\mathfrak p,a_0,...,a_{n-1}\in\mathfrak p,a_0\notin\mathfrak p^2$. Then $f$$f$ is irreducible.

Proof. Consider $R[T]/\mathfrak{p}$$R[T]/\frak p$, wchich is an integral domian. Suppose $f(T)=h(T)g(T)$$f(T)=h(T)g(T)$, and $\mathrm{deg}h=\alpha ,\mathrm{deg}g=\beta .$$\deg h=\alpha,\deg g=\beta.$

Then $f(T)\equiv h_{\alpha }{T}^{\alpha }\cdot g_{\beta }{T}^{\beta }\mathrm{mod}\mathfrak{p}$$f(T)\equiv h_{\alpha}T^{\alpha}\cdot g_{\beta}T^{\beta}\mod\mathfrak p$. Hence $h_0{g}_0=a_0\in \mathfrak{p}$$h_0 g_0=a_0\in\mathfrak{p}$. That is a contradiction. $◻$$\square$

Example. For $R[T]$$R[T]$, define $f(p(T))=p(T+r),r\in R$$f(p(T))=p(T+r),r\in R$. Easy to see this is a ring isomorphism.

Consider $p(T)=T^{p-1}+T^{p-2}+...+T+1=\frac{T^p-1}{T-1}.$$p(T)=T^{p-1}+T^{p-2}+...+T+1=\frac{T^p-1}{T-1}.$ Then

$$\begin{array}{}\text{(1)}& f(p(T))=\frac{(T+1{)}^p-1}{T}=T^{p-1}+(\genfrac{}{}{0ex}{}{p}{1})T^{p-2}+...+(\genfrac{}{}{0ex}{}{p}{p-2})T+(\genfrac{}{}{0ex}{}{p}{p-1})\end{array}$$

By Eisenstein's criterion, $f(p(T))$$f(p(T))$ is irreducible. Since $f$$f$ is isomorphism, $p(T)$$p(T)$ is irreducible.