Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial. Let $\theta$ be a root of $f(X)$. Let $A = \mathbb{Z}[\theta]$, $K$ be its field of fractions. Let $p$ be a prime number. Suppose $p$ does not divide the discriminant of $f(X)$.

Let $f(X) \equiv g_1(X)...g_e(X)$ (mod $p$), where $g_1(X), ..., g_e(X)$ are monic irreducible mod $p$. Since $f(X)$ mod $p$ has no multiple root, they are distict. By this, $P = (p, g_1(\theta))$ is a prime ideal of $A$.

Let $\Psi(\theta) = g_2(\theta)...g_e(\theta)$. Let $\alpha \in A$.

If $\Psi(θ)^n \alpha \equiv 0$ (mod $p^nA$), but not $\Psi(θ)^{n+1} \alpha \equiv 0$ (mod $p^{n+1}A$), we define ord$_P(\alpha$) = $n$. If there's no such $n$, we define ord$_P(\alpha$) = $\infty$.

My question: Is the following proposition correct? If yes, how would you prove this?

Proposition The following assertions hold.

(1) ord$_P$ can be extended to a unique discrete valuation of $K$ and its valuation ring is $A_P$.

(2) ord$_P(p$) = 1.

Motivation Let $m$ be the degree of $f(X)$. Let $\beta \in A$. $\beta$ can be written uniquely as $\beta = b_0 + b_1\theta + \dots + b_{m-1}\theta_{m-1}$, where $b_i \in \mathbb{Z}$. Hence $\beta \equiv 0$ (mod $p^nA$) if and only if $b_i \equiv 0$ (mod $p^n$) for $i = 0, \dots, m - 1$. Hence, when $\alpha \in A$ is given, it's rather easy to determine ord$_P(\alpha$). It's easy to see that ord$_P(\alpha) > 0$ if and only if $p$ divides the norm of $\alpha$. Therefore, if the norm of $\alpha$ is relatively prime to the discriminant of $f(X)$, we can compute the prime decomposition of $\alpha$.

Related question

share|improve this question

1 Answer 1

We first prove that ord$_P$ is a discrete valuation satisfying ord$_P(p) = 1$. We write ord instead of ord$_P$ for simplicity.

Let $\varphi: \mathbb{Z} \rightarrow \mathbb{Z}/p\mathbb{Z}$ be the canonical homomorphism. Let $g(X) \in \mathbb{Z}[X]$. We denote by $\bar g(X)$ the reduction of $g(X)$ (mod $p$).

Lemma 1 Let $\Omega$ be the algbraic closure of $\mathbb{Z}/p\mathbb{Z}$. Let $i$ be an integer such that $1 \leq i \leq e$. Let $\omega_i$ be a root of $\bar g_i(X)$ in $\Omega$. Then there exists a homomorphism $\psi_i:A \rightarrow \Omega$ extending $\varphi$ such that $\psi_i(\theta) = \omega_i$. Moreover Ker($\psi_i$) = $P_i$, where $P_i = (p, g_i(\theta))$.

Proof: Let $h(X) \in \mathbb{Z}[X]$. Suppose $h(\theta) = 0$. Since $f(X)$ is irreducible, there exists $r(X) \in \mathbb{Q}[X]$ such that $h(X) = f(X)r(X)$. Since $f(X)$ is monic, $r(X) \in \mathbb{Z}[X]$. Hence $\bar h(X) = \bar f(X)\bar r(X)$. Hence $\bar h(\omega_i) = \bar f(\omega_i)\bar r(\omega_i) = 0$. Therefore there exists a homomorphism $\psi_i:A \rightarrow \Omega$ extending $\varphi$ such that $\psi_i(\theta) = \omega_i$.

Suppose $h(\theta) \in$ Ker($\psi_i$), where $h(X) \in \mathbb{Z}[X]$. There exists $r(X) \in \mathbb{Z}[X]$ such that $\bar h(X) = \bar g_i(X) \bar r(X)$. Hence there exists $U(X) \in \mathbb{Z}[X]$ such that $h(X) = g_i(X)r(X) + pU(x)$. Hence $h(\theta) = g_i(\theta)r(\theta) + pU(\theta)$. Therefore $h(\theta) \in (p, g_i(\theta))$. Hence Ker($\psi_i) \subset (p, g_i(\theta))$.

Conversely suppose $h(\theta) \in (p, g_i(\theta))$, where $h(X) \in \mathbb{Z}[X]$. There exist $U(X), r(X) \in \mathbb{Z}[X]$ such that $h(\theta) = pU(\theta) + g_i(\theta)r(\theta)$. Hence $\psi_i(h(\theta)) = p\bar U(\omega_i) + \bar g_i(\omega_i)\bar r(\omega_i) = 0$. Hence $h(\theta) \in$ Ker($\psi_i$). Hence $(p, g_i(\theta)) \subset$ Ker($\psi_i$). QED

Let $\omega_i$ be a root of $\bar g_i(X)$ in $\Omega$ for $i = 1,\dots,e$. By Lemma 1, there exists a homomorphism $\psi_i:A \rightarrow \Omega$ extending $\varphi$ such that $\psi_i(\theta) = \omega_i$ for $i = 1,\dots,e$.

Lemma 2 Let $h(X) \in \mathbb{Z}[X]$. Suppose $\psi_i(h(\theta)) = 0$ for $i = 1,\dots,e$. Then $h(\theta) \equiv 0$ (mod $pA$).

Proof: Since $\bar h(\omega_i)$ for $i = 1,\dots,e$, $\bar h(X)$ is divisible by $\bar g_i(X)$. Hence $\bar h(X)$ is divisible by $\bar f(X)$. Hence $h(X) = f(X)H(X) + pR(X)$ for some $H(X), R(X) \in \mathbb{Z}[X]$. Therefore $h(\theta) = pR(\theta)$. QED

Lemma 3 Let $\alpha \in A$. $\Psi(θ) \alpha \equiv 0$ (mod $pA$) if and only if $\alpha \in P$.

Proof: Let $\psi_1:A \rightarrow \Omega$ be the homomorphism of Lemma 1. Suppose $\Psi(θ) \alpha \equiv 0$ (mod $pA$). $\psi_1(\Psi(θ) \alpha) = \bar \Psi(\omega_1) \psi(\alpha) = 0$. Since $\bar \Psi(\omega_1) = \bar g_2(\omega_1)...\bar g_e(\omega_1) \neq 0$, $\psi_1(\alpha) = 0$. Hence $\alpha \in$ Ker($\psi_1$). By Lemma 1, $\alpha \in P$.

Conversely suppose $\alpha \in P$. Since $\psi_1(\alpha) = 0$, $\psi_1(\Psi(\theta)\alpha) = 0$. If $i \neq 1$, $\psi_i(\Psi(\theta)) = \bar\Psi(\omega_i) = 0$. Hence $\psi_i(\Psi(\theta)\alpha) = 0$. Hence $\psi_i(\Psi(\theta)\alpha) = 0$ for all $i$, $i = 1,\dots,e$. By Lemma 2, $\Psi(\theta)\alpha \equiv 0$ (mod $pA$). QED

Lemma 4 Let $\alpha \in A$. Let $k \geq 0$ be an integer. Suppose $\Psi(\theta)^k\alpha \equiv 0$ (mod $p^kA$). Let $\beta = (\Psi(\theta)^k\alpha)/p^k \in A$. Then ord($\alpha$) = $k$ if and only if $\beta$ is not divisible by $P$.

Proof: $\Psi(\theta)\beta \equiv 0$ (mod $pA$) if and only if $\Psi(\theta)^{k+1}\alpha \equiv 0$ (mod $p^{k+1}A$). Hence the assertion follows immediately from Lemma 3. QED

Lemma 5 Let $k, l \geq 0$ be integers. Let $\alpha, \beta \in A$. Suppose ord($\alpha$) = $k$ and ord($\beta$) = $l$. Then ord($\alpha\beta$) = $k + l$.

Proof: Since $\Psi(\theta)^k\alpha \equiv 0$ (mod $p^k$), there exists $\lambda \in A$ such that $\Psi(\theta)^k\alpha = p^k\lambda$. Similarly there exists $\mu \in A$ such that $\Psi(\theta)^l\beta = p^l\mu$. Then $\Psi(\theta)^{k+l}\alpha\beta = p^{k+l}\lambda\mu$. Since $\lambda\mu$ is not divisible by $P$ by Lemma 4, ord($\alpha\beta$) = $k + l$. QED

Lemma 6 Let $\alpha \neq 0$ be an element of $A$. There exists an integer $k \geq 0$ such that ord($\alpha$) = $k$.

Proof: Suppose $\Psi(\theta)^k\alpha \equiv 0$ (mod $p^kA$) for every integer $k \geq 0$. There exists $\beta_k \in A$ such that $\Psi(\theta)^k\alpha = p^k\beta_k$ for every $k$. Since $\Psi(\theta)^{k+1}\alpha = \Psi(\theta)p^k\beta_k = p^{k+1}\beta_{k+1}$, $\Psi(\theta)\beta_k = p\beta_{k+1}$. Hence $\beta_k = \pi\beta_{k+1}$, where $\pi = p/\Psi(\theta)$. Since $\pi \in PA_P$, $\beta_kA_P \subset \beta_{k+1}A_P$ for every integer $k \geq 0$. Since $A_P$ is Noetherian, there exists an integer r such that $\beta_rA_P = \beta_{r+1}A_P$. Hence there exists $u \in A_P$ such that $\beta_{r+1} = u\beta_r$. Since $\beta_r = \pi\beta_{r+1}$, $\beta_r = u\pi\beta_r$. Hence $(1 - u\pi)\beta_r = 0$. Since $\pi \in PA_P$, $1 - u\pi$ is invertible in $A_P$. Hence $\beta_r = 0$. Since $\Psi(\theta)^r\alpha = p^r\beta_r$, $\alpha = 0$. This is a contradiction. QED

Lemma 7

ord($p$) $= 1$.

Proof: Clearly $\Psi(\theta)p \equiv 0$ (mod $p$). Suppose $\Psi(\theta)^2 p \equiv 0$ (mod $p^2$). Then $\Psi(\theta)^2 \equiv 0$ (mod $p$). Since $p \equiv 0$ (mod $P$), $\Psi(\theta)^2 \equiv 0$ (mod $P$). Hence $\Psi(\theta) \equiv 0$ (mod $P$). Since $\Psi(\theta) = g_2(\theta)\dots g_e(\theta)$, this is a contradiction. QED

Lemma 8 Let $\pi = p/\Psi(\theta)$. Every non-zero element $x$ of $A$ can be uniquely written as $x = \pi^k y$, where $k \geq 0$ is an integer and $y \in A$ and $y$ is not divisible by $P$.

Proof: Let ord($x$) = $k$. Then $\Psi(\theta)^k x \equiv 0$ (mod $p^kA$). Hence there exists $y \in A$ such that $\Psi(\theta)^k x = p^k y$. Hence $x = \pi^k y$. By Lemma 4, $y$ is not divisible by $P$. QED

Lemma 9 Let $x, y$ be non-zero elements of $A$. Then ord($x + y$) $\geq$ min{ord($x$), ord($y$)}.

Proof: We can assume $x + y \neq 0$. Let ord($x$) = $k$, ord($y$) = $l$. We can assume that $k \leq l$. By Lemma 8, we can write $x = \pi^ku$, where $u$ is not divisible by $P$. Similarly $y = \pi^lv$. Then $x + y = \pi^ku + \pi^lv = \pi^k(u + \pi^{l-k}v)$. Hence, by Lemma 5, ord($x + y$) = ord($\pi^k(u + \pi^{l-k}v)$) = $k +$ ord($u + \pi^{l-k}v$) $\geq k$. QED

Lemma 10 Let $\mathbb{Z}_{\infty} = \mathbb{Z} \cup$ {$\infty$}. There exists a unique map ord:$K \rightarrow \mathbb{Z}_{\infty}$ extending ord with the following properties.

(1) ord($K^*$) = $\mathbb{Z}$.

(2) ord($xy$) = ord($x$) + ord($y$) for $x, y \in K^*$.

(3) ord($x + y$) $\geq$ min{ord($x$), ord($y$)}

Proof: Let $x \in K^*$. If $x = a/b$ with $a, b \in A$, we define ord($x$) = ord($a$) - ord($b$). If $x = c/d$ with $c, d \in A$, $x = a/b = c/d$. Since $ad = bc$, ord($a$) + ord($d$) = ord($b$) + ord($c$). Hence ord($a$) - ord($b$) = ord($c$) - ord($d$). Therefore ord($x$) is well defined. Then (1), (2), (3) are clear.

The uniqueness is also clear. QED

Lemma 11 Let $\pi = p/\Psi(\theta)$. Let $U$ = {$s/t$; $s, t \in A - P$}. Let $x$ be a non-zero element of $K$. Then $x$ can be uniquely written as $x = \pi^k u$, where $k$ is an integer and $u \in U$.

Proof: Let ord($x$) = $k$. Let $u = x/\pi^k$. Then ord($u$) = ord($x$) - $k = 0$. Let $u = a/b$ with $a, b \in A$. Let $a = \pi^i s$ by Lemma 8, where $s$ is not divisible by $P$. Similarly let $b = \pi^j t$, where $t$ is not divisible by $P$. Then $u = \pi^{i-j}s/t$. Since ord($u$) = 0, $i = j$. Hence $u = s/t$.

The uniqueness is clear. QED

Proposition The following assertions hold.

(1) ord can be extended to a unique discrete valuation of $K$ and its valuation ring is $A_P$.

(2) ord($p$) = $1$.

Proof: By Lemma 10, ord can be extended to a unique discrete valuation of $K$. By Lemma 7, ord($p$) = $1$.

It remains to prove that $A_P$ is its valuation ring. Let $x \in K^*$. Suppose ord($x$) $\geq 0$. By Lemma 11, $x = \pi^k s/t$, where $s, t \in A - P$. Since $\pi \in PA_P$, $x \in A_P$.

Conversely suppose $x \in A_P$. $x$ can be written as $x = a/s$, where $a \in A$, $s \in A - P$. Then ord($x$) = ord($a$) - ord($s$) = ord($a$) $\geq 0$.

Therefore $A_P$ = {$x \in K$; ord($x$) $\geq 0$} as claimed. QED

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.