Sign up ×
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 $ \alpha_1, \ldots, \alpha_r $ be elements of a field $K \supseteq \mathbb Q$, which have the property that for $n_i \in \mathbb Z$, $i=1,\ldots,r$, it follows from $\alpha_1^{n_1}\cdots \alpha_r^{n_r}=1$ that $n_i=0 \quad \forall \ i=1,\ldots,r$.

How does one prove that from $\beta_1\log_p(\alpha_1) + \cdots + \beta_r\log_p(\alpha_r)=0\ $ with $ \beta_i \in \mathbb Q $ it follows: $ \beta_i = 0\ \forall\ i=1,\ldots,r $ ?

($\log_p$ denotes the $p$-adic logarithm)

share|cite|improve this question
You may assume that $\log_p(x)=0$ if and only if $x$ is a rational power of $p$ times a root of unity. – user10148 Apr 27 '11 at 21:00
Is the field $K$ contained in $\mathbb{C}_p$? – Alon Amit Apr 27 '11 at 22:49
Does it help to note that one may assume without loss of generality that the $\beta_i$ are integers (just by multiplying through by a common denominator)? – Gerry Myerson Apr 28 '11 at 1:15
@AlonAmit: $K|\mathbb Q$ is finite and $K \subseteq \mathbb C$. What I have already: $\beta_1 \log_p(\alpha_1)+...+\beta_r \log_p(\alpha_r)= \log_p(\alpha_1^{\beta_1})+...+\log_p(\alpha_r^{\beta_r})=0$. So $\log_p(\alpha_i^{\beta_i})=0\ \forall\ i$ and therefore $\alpha_i^{\beta_i}=p_i^{k_i}\zeta_i \ \forall\ i$. So it follows $M:=\alpha_1^{\beta_1}\cdot ...\cdot \alpha_r^{\beta_r}=q \zeta$ for a rational $q$ and a root of unity $\zeta$. Then $(M/q)^s=1$ for a $s$, but I can't conclude that all $\beta_i$ have to be zero. – user10148 Apr 28 '11 at 8:22

Your Answer


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

Browse other questions tagged or ask your own question.