**Motivation:** Take an algebraic number $\lambda$. In my research, I've stumbled upon the question in which cases the expression $\sum\_{\sigma \in S} \sigma(\lambda)$, where $S$ is a subset of the field embeddings of $K=\mathbb{Q}(\lambda)$, can be 'less irrational' then $\lambda$ itself.


**Question:** Take a finite Galois extension $L/\mathbb{Q}$ with Galois group $G$, let $H \leq G$ be a subgroup and let $K$ be the field fixed by $H$. Assume that $K$ has no proper subfields except for $\mathbb{Q}$.
Consider a subset $S$ of $G/H$ and consider the $\mathbb{Q}$-linear vector space homomorphism
$$
f\_S: K \to L, x \mapsto \sum\_{ \sigma \in S} \sigma(x).
$$
Is it true that this map is injective unless $S=G/H$?


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*Outdated Earlier version without the assumption that $K$ has no subfields (I leave this up only so Will's reply still makes sense)* If $SH$ is a subgroup of $G$ strictly bigger than $H$, this map will not be injective because $f\_S$ takes values in the subfield of $K$ fixed by $SH$. Intuitively, it seems likely that $SH$ being (the coset of) a subgroup is the only case in which this 'reduction of irrationality' can happen. Is this true, i.e. is $f\_S$ injective under the condition that $SH$ is not the coset of a subgroup?


EDIT: *As Will's example below shows below, one needs to exclude union of cosets, not only cosets. Indeed, if $U \subset G/H$ such that $UH$ is a subgroup, there is a linear relation which will carry over to any union of cosets of $UH$ as for $SH=UH \cup \sigma UH$,$f\_S=f\_U+\sigma \circ f\_U$. Apologies, I updated the question.*



Cancellation of irreducibility for Galois conjugates
No.


Let $a, b \in \mathbb Q(i)$. Let $\alpha\_1$ be a root of $x^3 + ax + b$. Let $L$ be the field generated by $i, \alpha\_1, \overline{\alpha}\_1$. Assume that $a,b$ are sufficiently general that $L$ has Galois group $S\_3 \wr \mathbb Z/2$, i.e. $(S\_3 \times S\_3 ) \rtimes \mathbb Z/2$.


Let $K$ be the subfield generated by $\alpha\_1 \overline{\alpha}\_1$. Then $K$ is stabilized by $S\_2 \wr \mathbb Z/2$, which is a maximal subgroup of index $9$, so $K$ has no proper subfields other than $\mathbb Q$.


Choose $\sigma\_1,\sigma\_2,\sigma\_3$ embeddings which send $\alpha$ to the the three roots $\alpha\_1,\alpha\_2,\alpha\_3$ of $f$ but preserve $\overline{\alpha}\_1$ (possible by our assumption on the Galois group.


Then $$\sigma\_1( \alpha\_1 \overline{\alpha}\_1) + \sigma\_2( \alpha\_1 \overline{\alpha}\_1)+ \sigma\_3( \alpha\_1 \overline{\alpha}\_1)=\alpha\_1 \overline{\alpha}\_1 + \alpha\_2 \overline{\alpha}\_1 + \alpha\_3 \overline{\alpha}\_1 = (\alpha\_1 + \alpha\_2 +\alpha\_3) \overline{\alpha}\_1 = 0 \overline{\alpha}\_1=0$$


so $x\mapsto \sigma\_1(x)+\sigma\_2(x) + \sigma\_3(x)$ is not injective.


