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Let $x$ be a positive number in some real number field $K$ whose degree $[K:\mathbb{Q}]$ is odd, is it true that $N_{K\mid\mathbb{Q}}(x)$ is also a positive number?

Update By the norm of $x$, I mean the product of all $\sigma(x)$, where $\sigma$ runs over all embeddings of $K$ into $\mathbb{C}$.

Update As @KCd shows, the above statement is wrong. I'm actually not sure if the product of complex conjugates effects the sign of norm. So, please allow me to change the question as follows:

1, if there are only one real embedding $\mathbb{Q}(x)\to \mathbb{R}$, is it true that the norm $N_{\mathbb{Q}(x)\mid\mathbb{Q}}(x)$ have the same sign as $x$?

2, if all the real embeddings $\mathbb{Q}(x)\to \mathbb{R}$ preserves the sign of $x$, is it true that the norm $N_{\mathbb{Q}(x)\mid\mathbb{Q}}(x)$ have the same sign as $x$?

Update Sorry, I forget that if $\sigma$ is a complex embedding of $K$, then so is its complex comjugate $\bar{\sigma}$, and $\sigma(x)\overline{\sigma(x)}$ is just $|\sigma(x)|^2$, which is always positive.

Thanks @KCd's answer.

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    So you're thinking something like $\mathbb Z[\root 3 \of 2]$? – 2017-02-27
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    The answer to the question in the title is no; since $sqrt{2}$ has norm -2 in $\mathbb{Q}(\sqrt{2})$. But this doesn't address the case of an odd extension. – 2017-02-27
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    @RobertSoupe Yes. – 2017-02-27
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    If I've done my calculations correctly: $N(-1 + (\root 3 \of 2)^2) = 3$ but $N(1 - (\root 3 \of 2)^2) = -3$. – 2017-02-27
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    @RobertSoupe But $1-(\sqrt[3]{2})^2$ is negative. – 2017-02-27
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    Ah yes, that's an important detail. Then my empirical evidence suggests the answer is yes. – 2017-02-27

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Certainly not. Pick an irreducible cubic polynomial in $\mathbf Q[x]$ with leading coefficient $1$ and positive constant term $c$. The product of the three roots is $-c$, so that is the norm of any root in the field extension it generates.

For example, let $f(x) = x^3 - 3x + 1$. It is irreducible over $\mathbf Q$ (since it's irreducible mod $2$) with three real roots, which are approximately $1.53$, $.34$, and $-1.87$. The product of the three roots of $f(x)$ is $-f(0) = -1$. If $\alpha$ is any root of $f(x)$ and $K = \mathbf Q(\alpha)$ then $[K:\mathbf Q] = 3$ and ${\rm N}_{K/\mathbf Q}(\alpha) = -1$.

Another counterexample is $f(x) = x^3 - 9x + 3$. It has three real roots, which are approximately $2.81$, $.33$, and $-3.15$. The product of the roots of $f(x)$ is $-f(0) = -3$, so if $\alpha$ is any root of $f(x)$ and $K = \mathbf Q(\alpha)$ then $[K:\mathbf Q] = 3$ and ${\rm N}_{K/\mathbf Q}(\alpha) = -3$.

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    Thanks. What if I requires that all the real conjugates of $x$ is positive? – 2017-02-27
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    In that case, the norm must be positive. – 2017-02-27
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    Then of course the norm is positive. After all, the norm is the product of each $\mathbf Q$-conjugate of the number (with some common multiplicity $> 1$ if the number doesn't generate the field extension you're working in), and each non-real complex root multiplied by its complex conjugate is positive, so including your new condition will make the overall product positive. – 2017-02-27
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I haven't solved this yet, but one can reformulate the problem as follows. Let $x \in K$ be positive. We are looking at

$$N_{K/\mathbf{Q}}(x) = N_{\mathbf{Q}(x)/\mathbf{Q}} \circ N_{K/\mathbf{Q}(x)}(x) = N_{\mathbf{Q}(x)/\mathbf{Q}}(x)^{[K :\mathbf{Q}(x)]}$$

with $[K : \mathbf{Q}(x)]$ odd, so we are reduced to determining the sign of $N_{\mathbf{Q}(x)/\mathbf{Q}}(x)$ when $x$ is positive and $[\mathbf{Q}(x) : \mathbf{Q}]$ is odd. So we need to solve the following problem:

Assume $f(X) = a_0 + a_1X + \cdots + X^n$ ($n$ odd) is an irreducible polynomial over $\mathbf{Q}$ with at least one positive root. Does $a_0$ have to be negative?