A comment in Size of Galois Group based on roots and minimal polynomial says:
"By a change of variables a polynomial $f(X)$ in $\mathbb{Q}[X]$ can be made monic in $\mathbb{Z}[X]$ without changing its splitting field over $\mathbb{Q}$."
I guess the standard way to convert a polynomial $f(X)\in \mathbb{Q}[X]$ into a monic polynomial over $\mathbb{Z}$ is: First make it monic by multiplying it by a rational number and then replace it by $D^{ \, \text{deg}(f)} f(\frac{X}{D})$, where $D$ is a common denominator of the coefficients.
I don't see why the splitting field of the thus obtained polynomial is the same as that of the original polynomial? For example, the splitting of $5X^2 - 2$ is $E=\mathbb{Q}(\sqrt{\frac{2}{5}})$, the corresponding monic polynomial over $\mathbb{Z}$ is $X^2 - 10$, whose splitting field is $F=\mathbb{Q}(\sqrt{10})$. Are $E$ and $F$ isomorphic, and why so?
To state everything clearly:
We have $E = \Bbb Q[X]/(5X^2 - 2)$ and $F = \Bbb Q[Y]/(Y^2 - 10)$.
We define a homomorphism $f$ from $\Bbb Q[X]$ to $F$ by sending $X$ to the class of $\frac 1 5 Y$.
It is clear that the image of $5X^2 - 2$ under $f$ is zero, i.e. the ideal $(5X^2 - 2)$ lies in the kernel of $f$.
Therefore $f$ induces a quotient homomorphism $\phi: E \rightarrow F$, sending the class of $X$ to the class of $\frac 1 5 Y$.
In the same way, we may define a homomorphism $\psi: F \rightarrow E$, sending the class of $Y$ to the class of $5X$.
We then verify that $\phi \circ \psi$ is the identity, and $\psi \circ \phi$ is the identity.
This shows that $E$ and $F$ are isomorphic to each other.