I need to fully understand why if I have epimorphism $f: G_{\Bbb Q}\to \Bbb Z/2\Bbb Z\;$ , with $\;\Bbb G_{\Bbb Q}=\;$ full Galois group of rationals $\;\Bbb Q\;$ , and when last group corresponds with extension $\;\Bbb Q(i)/\Bbb Q\;$, then it cannot factor through (sub)group $\;\Bbb Z/4\Bbb Z\;$ …?
My idea is that if exists $\;g:\Bbb Z/4\Bbb Z\to\Bbb Z/2\Bbb Z\;$ such that $\;f=g\pi\;$ , with $\;\pi:G_{\Bbb Q}\to\Bbb Z/4\Bbb Z\;$ , then $\;\Bbb Q\;$ must have cyclic extension of order $\;4\;$ (corresponding to $\;\Bbb Z/4\Bbb Z\;$) which also contains extension $\;\Bbb Q(i)/\Bbb Q\;$, and this is impossible as then it must contain $\;i\;$ . This very last I completely understand, but the factoring fact corresponding to cyclic extension containing $\;i\;$ still isn't clear to me very much, though I think (feel is more exact) that this is pretty close to truth as I "feel" like $\;\pi\;$ should be canonical projection...
Any idea can be very helpful.
The fact that your epimorphism $f: G_{\mathbf Q} \to Gal(\mathbf Q(i)/\mathbf Q)$ cannot factor through $\mathbf Z/4$ is equivalent to saying that $\mathbf Q(i)$ cannot be embedded in a cyclic Galois extension of $\mathbf Q$ of degree $4$. For simplicity, write $K=\mathbf Q(i)$ and consider a chain of successive quadratic extensions $\mathbf Q < K < L$. Since we are in characteristic $\neq 2$, $L$ has the form $L=K(\sqrt a)$, where $a\in K^*$ and $a\notin {K^*}^2$. A priori $L/\mathbf Q$ is not always normal. A necessary and sufficient condition for normality is that $\gamma (a)/a \in {K^*}^2$, where $\gamma$ is the complex conjugation of $K$. (Hint: The splitting field of $X^2 - a$ over $\mathbf Q$ is obtained by adjoining to $\mathbf Q$ the roots of $X^2 - a$ and $X^2 - \gamma(a)$. It coincides with $L$ iff $\gamma (a)/a \in {K^*}^2$).
If $L/\mathbf Q$ is normal of degree $4$, its Galois group is either $(\mathbf Z/2)\times (\mathbf Z/2)$ or $\mathbf Z/4$. To distinguish between the two cases, note that the normality condition can be written as $\gamma (a)/a =x^2$, with $x\in K^*$, so $N(x)=\pm 1$, where $N$ is the norm of $K/\mathbf Q$. The condition $N(x)=1$ means that $x$ has the form $x=\gamma(b)/b$ (Hilbert's thm.90 or direct computation), hence $\gamma(ab^2)=ab^2$, or equivalently $ab^2=c\in\mathbf Q^*$, i.e. $L$ has the form $L=\mathbf Q (i, \sqrt c)$, with Galois group $(\mathbf Z/2)\times (\mathbf Z/2)$. The remaining condition $N(x)=-1$ means that $-1$ is the sum of 2 squares of rationals : impossible.