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There is a lemma from a lecture I attended where I scribbled down notes and try to make sense of the proof afterwards and there is a spot at which I am stuck. First I have to set up some notations and refresh memory on Frobenius element.

Let $l,p$ be distinct odd primes, $L = mathbbQ(zeta_l)$, $K$ be the unique quadractic subfield contained in $L$. (To see what $K$ is, we know that $Gal(L/mathbbQ) =(mathbbZ/lmathbbZ)^*$ has a unique subgroup $H$ of index 2, and by Galois theory, $K$ would then be $L^H$).

By definition, $Frob_p$ is defined as follows. We know $p$ is unramified in $L$. If $mathfrakqsubseteq O_L$ is any prime ideal containing $pcdot O_L$, then the decomposition group at $mathfrakq$ defined as $G_mathfrakq:=sigmain Gal(L/mathbbQ) : sigma(mathfrakq)=mathfrakq$ is isomorphic to $Gal((O_L/mathfrakq)/mathbbF_p)$. The latter has Frobenius element $xmapsto x^p$. The corresponding element in $G_mathfrakq$ is denoted $Frob_p$. Note that $Frob_p$ does not depend on the choice of $mathfrakq$ because the extension is Abelian.

Lemma: $Frob_pin Gal(L/mathbbQ)$ is in $H$ if and only if $p$ splits in $K$.

Proof: Let $mathfrakp_1 subseteq O_K$ be a prime factor of $pcdot O_K$. Then $Frob_pin H$ $Leftrightarrow$ $Frob_p$ acts trivially on $O_K$ $Leftrightarrow$ $xmapsto x^p$ is identity on $O_K/mathfrakp_1$ $Leftrightarrow$ $O_K/mathfrakp_1 = mathbbF_p$ $Leftrightarrow$ $p$ splits in $K$.

The line that I do not get is "$Frob_p$ acts trivially on $O_K$ $Leftrightarrow$ $xmapsto x^p$ is identity on $O_K/mathfrakp_1$". I think I have the forward implication, but could someone enlighten me on the backward implication?

I think you're missing the following useful observation: $Frob_p$ is characterized by the property $Frob_p x equiv x^p pmodmathfrakq$ for all $x$ in $O_L$. That $Frob_p$ has this property follows from your definition. To see that it is determined among all automorphisms of $L$ by this property, note that if for some automorphism $sigma$ of $L$ we have $sigma x equiv x^p pmodmathfrakq$, then letting $x$ be in $mathfrakq$, we see that $sigma$ fixes $mathfrakq$, hence is in the decomposition group of $mathfrakq$. Now as you point out, this group is isomorphic to the Galois group of $O_L/mathfrakq$ over $mathbbF_p$, so there is a unique such automorphism that acts as Frobenius.

Examine the above paragraph and note that nothing depends on the upper field being $L$ -- it could be any abelian extension of $mathbbQ$. In particular, there is a $Frob_p$ automorphism of $K/mathbbQ$. In fact, by the above characterization, this $Frob_p$ is the restriction to $K$ of the $Frob_p$ for $L/mathbbQ$.

$implies$: Choose the $mathfrakq$ to lie above $mathfrakp_1$. If $Frob_p$ acts trivially on $O_K$, then the $Frob_p$ for $K/mathbbQ$ is the identity automorphism. In particular, $x = Frob_p x equiv x^p pmodmathfrakp_1$ for all $x$ in $O_K$, so $x mapsto x^p$ is the identity on $O_K/mathfrakp_1$.

$impliedby$: If $x mapsto x^p$ is the identity on $O_K/mathfrakp_1$, then the identity automorphism of $K$ has the property that characterizes the $Frob_p$ for $K/mathbbQ$. Thus, this $Frob_p$ is the identity on $K$, so the $Frob_p$ for $L$ is trivial on $O_K$.