I’m turning my comment into an answer. First, the extension $H=\mathbb{Q}(\alpha,\sqrt{-23})$ contains a root of $f$ (it is an irreducible polynomial of degree $3$) and it contains the quantities $\beta+\gamma,\beta\gamma,(\beta-\gamma)(\alpha-\gamma)(\alpha-\beta)$ (where $\beta,\gamma$ are the other roots of $f$), so that it contains $\beta\pm \gamma$ and thus $\beta,\gamma$ and $H$ contains the splitting field of $f$.
Said splitting field contains $K=\mathbb{Q}(\alpha)$ and contains $\sqrt{-23}=\pm (\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)$ so it really is $H$. In particular $H/\mathbb{Q}$ is Galois.
Now, the extension $H/K$ is generated by the integral polynomial $X^2-X-6$ (the root is $\frac{1\pm \sqrt{-23}}{2}$) which is separable mod any prime $p\neq 23$, so $H/K$ unramified at any prime not above $23$. I use this polynomial because $X^2+23$ isn’t separable mod $2$.
Now, since $K/\mathbb{Q}$ is generated by a root of $f$, an irreducible integral polynomial whose discriminant is $-23$, it is unramified above any prime $p\neq 23$, and it must be ramified somewhere so it is ramified at $23$.
What is the ramification type of $K/\mathbb{Q}$ at $p=23$? We know that $\sum_{\mathfrak{p}}{e_{\mathfrak{p}/23}f_{\mathfrak{p}/23}}=3$, where $\mathfrak{p}$ runs over the prime ideals of $K$ above $23$, and these primes are in bijection with the prime factors of $X^3-X-1$ in $\mathbb{Q}_{23}$. Because $K/\mathbb{Q}$ is ramified at $23$, there are either two such primes (and the ramification indices are $1,2$ while the residual degrees are $1$) or one such prime (and the extension is totally ramified). Now, $X^3-X-1$ is $(X+13)^2(X+20)$ mod $23$ so by Hensel this polynomial isn’t irreducible.
Thus there are two primes $\mathfrak{p},\mathfrak{q}$ in $K/\mathbb{Q}$ above $23$ with ramification indices $2,1$ and residual degrees $1$.
Now, the ramification index of a prime of $H$ above $23$ does not depend on the choice of such a prime (as $H/\mathbb{Q}$ Galois). Let $\mathfrak{r}$ be a prime of $H$ above $\mathfrak{p}$. If $e_{\mathfrak{r}/\mathfrak{p}} >1$, then $e_{\mathfrak{r}/23}\geq 2e_{\mathfrak{p}/23}=4$, and $e_{\mathfrak{r}/23}|[H:\mathbb{Q}] \in \{3,6\}$ so $e_{\mathfrak{r}/23}=6=[H:\mathbb{Q}]$ and $H$ is totally ramified above $23$, and thus $K$ is totally ramified above $23$ too – a contradiction.
So $H/K$ is unramified at $\mathfrak{p}$, and thus $e_{H/\mathbb{Q},23}=e_{\mathfrak{p}/23}=2$. But this means that if $\mathfrak{r}$ is a prime ideal of $H$ above $\mathfrak{q}$, then $e_{\mathfrak{r}/\mathfrak{q}}=e_{\mathfrak{r}/23}=2$ so $H/K$ is ramified at $\mathfrak{q}$ (thanks to franz lemmermeyer for pointing it out).
Now, how to show that $H/L=\mathbb{Q}(\sqrt{-23})$ is unramified everywhere? If $p \neq 23$ is a prime, then $K/\mathbb{Q}$ is unramified at $p$, and $H/K$ is unramified at any prime of $K$ above $p$, so $H/\mathbb{Q}$ is unramified at $p$, and thus $H/L$ is unramified at any prime of $L$ above $p$.
Now, it remains only to show that $H/L$ is unramified at the prime $\mathfrak{l}=(\sqrt{-23})$. But $e_{H/L,\mathfrak{l}}=\frac{e_{H/\mathbb{Q},23}}{e_{\mathfrak{l}/23}}=\frac{2}{2}=1$ from what precedes. So $H/L$ is Galois and everywhere unramified.
Best Answer
Let $K = \mathbf Q(\sqrt{-5})$ and $L = K(i)$, so $L$ has the quadratic subfields $M = \mathbf Q(i)$, $K = \mathbf Q(\sqrt{-5})$, and $N = \mathbf Q(\sqrt{5})$. Since $M$ is unramified away from $2$ and $N$ is unramified away from $5$, $K/\mathbf Q$ is unramified away from $2$ and $5$. (I am not considering archimedean ramification.) Thus $K/\mathbf Q(i)$ is unramified away from $(1+i)$, $(1+2i)$, and $(1-2i)$. How do $(1+i)$, $(1+2i)$, and $(1-2i)$ decompose in $K$?
How $(1+i)\mathcal O_K$ factors: in $\mathcal O_M = \mathbf Z[i]$ we have $(2) = (1+i)^2$ and in $\mathcal O_N = \mathbf Z[(1+\sqrt{5})/2]$ the ideal $(2)$ is prime, so $2\mathcal O_K = \mathfrak p^2$ where ${\rm N}(\mathfrak p) = 4$. Thus $((1+i)\mathcal O_K)^2 = \mathfrak p^2$, so $(1+i)\mathcal O_K = \mathfrak p$: $K$ is unramified at $(1+i)$.
How $(1+2i)\mathcal O_K$ factors: in $\mathcal O_M = \mathbf Z[i]$ we have $(5) = (1+2i)(1-2i)$ and in $\mathcal O_N = \mathbf Z[(1+\sqrt{5})/2]$ we have $(5) = (\sqrt{5})^2$, so $5\mathcal O_K = \mathfrak q\mathfrak q'$ where ${\rm N}(\mathfrak q) = {\rm N}(\mathfrak q') = 5$. Thus $(1+2i)\mathcal O_K (1-2i)\mathcal O_K = \mathfrak q\mathfrak q'$, so the ideals $(1+2i)\mathcal O_K$ and $(1-2i)\mathcal O_K$ are both prime: $K$ is unramified at $(1+2i)$ and $(1-2i)$.