# Quick question: finite extensions and norms

[Edit: "Suppose K is a field complete with respect to a discrete valuation, with valuation ring $\mathcal{O}$."]

I'm trying to solve the following problem: let $L/K$ be a finite extension of fields, and $x \in L$. Show if $N_{L/K}(x) \in \mathcal{O}$ then $N_{L/K}(1+x) \in \mathcal{O}$, where $N_{L/K}$ denotes the norm (i.e. the product of embeddings $\sigma_i:L \to \overline{K}$) and $\mathcal{O}$ denotes $\{x:|x|_K \leq 1\}$ under the absolute value $|\cdot|_K$ on $K$.

So, I did the obvious thing, and wrote out the product

$N_{L/K}(1+x)=\prod_{i=1}^n \sigma_i(1+x)=\prod_{i=1}^n (\sigma_i(1)+\sigma_i(x))=\prod_{i=1}^n (1+\sigma_i(x))$ $= \prod_{i=1}^n \sigma_i(x)+\sum_{j=1}^n \prod_{i\neq j} \sigma_i(x) + \sum_{j=1}^n \sum_{k=1}^n \prod_{i\neq j,k} \sigma_i(x)+\ldots +1$

where $\prod_{i=1}^n \sigma_i(x)=N_{L/K}(x)$. This unfortunately got me nowhere: you can pull a $N_{L/K}(x)$ out of every term to write the sum in a slightly different way, but I can't see any way to show that this should also be in $\mathcal{O}$. Obviously the norm is multiplicative but pulling out this factor with norm at most 1 didn't help anything. Could anyone help? I expect I'm probably missing something painfully obvious. Thanks! -BTS

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There has to be more assumptions. Consider $(K, | \cdot |_{K}) = (\mathbb{R}, | \cdot |)$ and $L = \mathbb{C}.$ Then $|N_{L/K}(1)| \leq 1$ but $|N_{L/K}(1 +1)| \not\leq 1.$ – jspecter Nov 8 '11 at 3:46
Are we talking about a $p$-adic norm here? and $\cal O$ is the ring of integers in $K$, which is the elements of $p$-adic norm at most 1? That would answer jspecter's objection, as the norm of 2 is either 1 or 1/2, depending on which $p$ determines the norm. – Gerry Myerson Nov 8 '11 at 6:08
Sorry, I should have said absolute value, not norm, though I'm not sure if that makes a difference or not. I have worked a lot with the p-adic absolute value previously, but the question doesn't specify that the norm is p-adic: it is exactly as I've written it. Is there any way to interpret it which would make the assertion correct? – BTS Nov 8 '11 at 8:43
Ok, i've spoken to my professor and he says he forgot to include at the start: "Suppose K is a field complete with respect to a discrete valuation, with valuation ring $\mathcal{O}$". So I guess the discreteness of the valuation prevents the absolute value from being that on $\mathbb{C}$: does that make the problem soluble? – BTS Nov 8 '11 at 11:45
The norm is not the product of images under field embeddings unless the extension $L/K$ is separable. So your norm formula is sometimes wrong. Are you secretly trying to show the formula $|x| = \sqrt[n]{|{\rm N}_{L/K}(x)|_K}$ is an absolute value on $L$, where $n=[L:K]$? I ask because that problem reduces to exactly what you are asking, which otherwise looks like a weird question to wonder about. – KCd Nov 8 '11 at 12:05