Galois group of $L/\mathbb{Q}$ is generated by inertia groups

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L is a Galois extension and we want Galois group G to be generated by the inertial groups for all primes

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Here is my proof (in process): The intermediate field $L_{I}$ for the inertia group over prime $p\in \mathbb{Q}$ is unramified $\Rightarrow p\nmid d_{\mathbb{L_{I}}}\Rightarrow [L_{I}:\mathbb{Q}]=1$. Also, every prime ideal in L lies over a rational prime

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If $H$ is the subgroup of $G$ generated by all the inertia groups $I(p)$, then $L^H \subset L^{I(p)}$ for each $p$, and so $L^H/\mathbb Q$ is unramified at every $p$. Since $\mathbb Q$ has no nontrivial unramified extensions, this means that $L^H = \mathbb Q$ and $H = G$.