Question
Theorem 11.2 asserts that for any nonzero ideal $\mathfrak{a}\subset\mathcal{O}_K$ of the ring of integers of a number field $K/\mathbb{Q}$ there is a second ideal $\mathfrak{b}\subset\mathcal{O}_K$ such that $\mathfrak{a}\mathfrak{b}=(c)$ for some nonzero rational integer $c$. It's only the first step of the proof that I'm having trouble with, namely the assertion:
If $\rho_1,\dots,\rho_n$ are all field embeddings $K\to\mathbb{C}$, and if $\alpha_0,\alpha_1,\dots,\alpha_l\in\mathcal{O}_K$ are algebraic integers, then $$ F=\prod_{i=1}^n(\rho_i(\alpha_0)+\rho_i(\alpha_1)t+\cdots+\rho_i(\alpha_l)t^l) \in \mathbb{Z}[t]. $$
The reason given is "by symmetry, $F\in \mathbb{Z}[t]$". In the first section of Chapter 10, the author writes "Some knowledge of Galois Theory is useful [...] but not essential" and he says that in stead of Galois theory he will rely on the
Symmetric Function Theorem. The ring of symmetric polynomials $S\subset R[x_1,\dots,x_n]$ (for any ring $R$) is isomorphic to the polynomial ring $R[\sigma_1,\dots,\sigma_n]$ by mapping the variable $\sigma_k$ the the $k$-th symmetric polynomial: $$\begin{array}{cccc} R[\sigma_1,\dots,\sigma_n] & \longrightarrow & S\qquad\qquad\qquad\qquad\phantom{.} \\ \sigma_k & \longmapsto & \displaystyle\sigma_k(x_1,\dots,x_n)=\sum_{\substack{I\subset\{1,\dots,n\}\\\#I=k}}\prod_{i\in I}x_i.\end{array}$$
Question 1. How to prove $F\in\mathbb{Z}[t]$ "by symmetry" i.e. using only the symmetric function theorem and no Galois theory?
Solution attempt
It seems it would be simplest to prove something like the following:
Lemma 1. Let $f\in K[t]$ and write $f_i=\rho_i(f)\in \mathbb{C}[t]$ for $i=1,\dots,n$. Then $$\prod_{i=1}^nf_i\in\mathbb{Q}[t]$$
Induction on the degree of $f$ might work: if $f\in K$ is constant then this computes the norm of $f$, which is a rational number. And for the induction step: suppose the proposition is true for all degree $<l$ polynomials. If $f$ has degree $l$, then the proposition is trivial if $f(0)=0$ so that $f=tg(t)$ with $\deg(g)<l$. Otherwise, write $f=\alpha_0(1+tg(t))$ where $g\in K[t]$ has degree $<l$ and so $$ \begin{array}{rcl} \displaystyle \overbrace{\frac1{N(\alpha_0)}}^{\in\mathbb{Q}}\cdot \prod_{i=1}^nf_i & = & \displaystyle \prod_{i=1}^n(1+tg_i(t))\\ & = & \displaystyle 1 + \sum_{k=1}^nt^k\sigma_k(g_1,\dots,g_n) \end{array}$$
One might feel compelled to conjecture a slightly more general result:
Lemma 2. Let $f\in K[t]$ and write $f_i=\rho_i(f)\in \mathbb{C}[t]$ for $i=1,\dots,n$. Then for all $k=1,\dots,n$, $$\sigma_k(f_1,\dots,f_n)\in\mathbb{Q}[t]$$
Question 2. Can you prove Lemma 2?
I have a proof of the claim $F(t)\in\mathbb{Z}[t]$ but it's not in the "by symmetry"-spirit I was hoping for... At least it's very simple.
Proof. Express $F\in\mathbb{C}[t]$ in the Lagrange basis of the set $\{0,1,\dots,ln\}$: $$F=\sum_{k=0}^{ln}F(k)L_k$$ where $$L_k=\prod_{\substack{0\leq j\leq ln\\j\neq k}}\frac{X-j}{k-j}\in\mathbb{Q}[t]$$ We then notice that by $\mathbb{Q}$-linearity of the $\rho_j$: $$\forall q\in\mathbb{Q},\quad F(q)=N_{K/\mathbb{Q}}(f(q))\in\mathbb{Q}$$ where $f(t)=\alpha_0+\alpha_1t+\cdots+\alpha_l t^l$. It follows that $F\in\mathbb{Q}[t]$. Since the coefficients of $F$ are algebraic integers, $F\in\mathbb{Z}[T]$ and we are done.
The same proof works for Lemma 2 as well:
Proof. It's litterally the same: $$\underbrace{\sigma_k(f_1,\dots,f_n)}_{s_k(t)} = \sum_{i=0}^{ln}s_k(i)L_i$$ and by $\mathbb{Q}$-linearity of the $\rho_j$: $$s_k(i)=\sigma_k(f_1(i),\dots,f_n(i))\in\mathbb{Q}$$ as, up to sign, it's one of the coefficients of (a power of the) minimal polynomial of $f(i)\in K$.