The definition is provided in the upper figure. What if f(x) had repeated roots? Are the roots considered deduplicated such that there’ are less than n roots f(x) have?
As @reuns pointed out, I have got a wrong understanding on what a splitting field of $f(x)$ means (I do think the splitting field of $(x^2+1)^2$ is still $\mathbb{C}$... I think i can understand that $\mathbb{R}/(x^2+1)$ is not the same with $\mathbb{R}/(x^2+1)^2$, but I dont think that the splitting field of $(x^2+1)^2$ is $\mathbb{R}/(x^2+1)^2$).
So I resupply my constructively understanding of splitting field here: Given $f(x)$ and $k$, we know that in the extended field $k[x]/(f(x))$, $f(x)$ take $z_1 = x + (f(x))$ as a root. So we can repeatedly divide $f(x)$ by $x - z_1$ to get $g(x)$. Repeat this step we can get a extended field $K$. And I think it's the splitting field of $f(x)$.
I think my understanding to $k(z_1, z_2, ..., z_n)$ is similar with the concept of subgroup generated by elements in a group. let $K/k$ be a field that $f(x)$ has all n (maybe repeated) roots. Then the image of the valued map, from the formal n-valued frac field $k(y_1, y_2, ..., y_n)$ to $K$ $$ \phi: k(y_1, y_2, ..., y_n) \rightarrow K $$ , is a splitting field.
Is it wrong?
Updated again: The confusion maybe come from the proof below(theorem A-5.3):
the proof supposes the roots is distinct. I think a more stable proof should be argued in the set of distinct roots. And then using the result of Sm is a subgroup of Sn.
And in fact the conclusion can be claimed stronger: the galois group is of subgroup of Sm, where m is the number of distinct roots of f(x). Right?
Remember that an element of the Galois group is a field automorphism of $E$ which fixes $k$. For such an automorphism, it is irrelevant if the roots of $f$ are simple or multiple.
A polynomial has the same Galois group as the polynomial given by its irreducible factors to the first power, and for this reason people don't usually worry about multiple roots, since they are irrelevant.