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.

I am reading the proof in Marcus' number fields of the following fact:

Let $p$ be a prime in $\Bbb{Z}$ and suppose $p$ is ramified in a number ring $R$ (of some number field $K$). Then $p | \textrm{disc} (R)$.

The proof goes as follows. Suppose $\sigma_1,\ldots,\sigma_n$ are the number of embeddings of $K$ into $\Bbb{C}$ and extend all $\sigma_i's$ to the Galois closure $L$ of $K$. Now if $p$ is ramified in $R$ then we can write $pR = IP$ where $I$ is divisible by all primes lying over $p$ and $P$ some prime of $R$ lying over $p$. Choose some $\alpha \in I - pR$ (we can do this since $I$ properly contains $pR$). We see that $\alpha$ is in every prime lying over $p$ in $R$. Now let $S = \mathcal{O}_L$ and $Q$ a prime of $S$ that lies over $p \in \Bbb{Z}$.

My question is: Why should $\sigma(\alpha) \in Q$ for every $\sigma \in \textrm{Gal}(L/K)?$

In Marcus he claims to show this as follows: "To see this, notice that $\sigma^{-1}(\alpha)$ is a prime of $\sigma^{-1}(S) = S$ lying over $p$, and hence contains $\alpha$. In particular we have $\sigma_i(\alpha) \in Q$ for all $i$."

I am confused because what does he mean by "$\sigma^{-1}(\alpha)$ being a prime lying over $p$"? $\sigma^{-1}(\alpha)$ is just an element of $L$ no?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

It seems to be a typo in Marcus. It should read

... $\sigma^{-1}(Q)$ is a prime of $\sigma^{-1}(S)=S$ lying over $p$ ...

share|improve this answer

Your Answer


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