Why is $1<|\text{Nm}_{K|\Bbb{Q}}(\beta)|< |\text{Nm}_{K|\Bbb{Q}}(\alpha)|$

Question

Let $K$ be an algebraic number field. Let $\mathcal{O}_K$ be its ring of integers. Say $\alpha\in \mathcal{O}_K$ is not irreducible, then $\alpha=\beta\gamma$ where these are not units, Neukirchs text 'Algebraic Number Theory' page $16$ says that: $$1<|\text{Nm}_{K|\Bbb{Q}}(\beta)|< |\text{Nm}_{K|\Bbb{Q}}(\alpha)|$$

Why is this so? I know the following two ways of defining the norm 1) let each $x\in L$ correspond to an endomorphism of $K$-vector spaces: $$T_x:L\to L,\quad a\mapsto xa.$$$\text{Nm}_{K|\Bbb{Q}}(\alpha)=\text{Det}(T_\alpha)$.

2) $\text{Nm}_{K|\Bbb{Q}}(\alpha)=\prod_{\sigma\in\text{Aut}_{\Bbb Q}(K)}\sigma\alpha.$

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Not really sure how to begin. I know that the norm takes me into $\Bbb Q$, but I am not sure what is taken to $1$ or if anything can be taken to less than $1$, or why.

Answer 1

An element $\alpha$ has norm $\pm 1$ if and only if it is a unit in the ring of integers. Hence every non-unit $\gamma$ satisfies $|Nm(\gamma)|>1$. Hence we have $1<|Nm(\beta)|$ and $$ |Nm(\alpha)|=|Nm(\beta\gamma)|=|Nm(\beta)Nm(\gamma)|=|Nm(\beta)|\cdot |Nm(\gamma)|>|Nm(\beta)|. $$

Answer 2

Hint:: keep in mind that norm is multiplicative