Treating splitting fields as the same dangerous?

Question

Let $f(x)$ be a polynomial over a field $F$. I know the definition of a splitting field and I know that if $E$ and $E'$ are two splitting fields for $f$ over $F$, then $E$ and $E'$ are isomorphic as fields.

I was reading these notes (book?) by J.S Milne. Here he has a warning 2.6 on page 30-that says

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Warning 2.16 Let $f\in F[X]$. If $E$ and $E'$ are both splitting fields of , then we know that there exists an $F$-isomorphism $E\to E'$, but there will in general be no preferred such isomorphism. Error and confusion can result if the fields are simply identified. Also, it makes no sense to speak of “the field $F[\alpha]$ generated by a root of $f''$ unless $f$ is irreducible (the fields generated by the roots of two different factors are unrelated). Even when $f$ is irreducible, it makes no sense to speak of “the field $F[\alpha, \beta]$ generated by two roots $\alpha, \beta$ of $f''$ (the extensions of $F[\alpha]$ generated by the roots of two different factors of $f$ in $F[\alpha][X]$ may be very different).

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I don't understand the comment about "error and confusion can result if the fields are simply identified"

Can someone explain what this is getting at? (I assume we can still talk about the splitting field?)

Answer 1

Let's consider the polynomial $f(x) = x^2 + 1$ over $\mathbb{Q}$. Then one way to construct a splitting field for $f$ is to take the subfield of $\mathbb{C}$ generated by its roots, which of course is $\mathbb{Q}[i]$ where $i$ is the usual complex number. Another way to construct a splitting field for $f$ is to again take the subfield generated by the roots but in the $p$-adic numbers $\mathbb{Q}_p$ for any prime $p \equiv 1 \bmod 4$; these are the primes for which there exists a $p$-adic number $i_p \in \mathbb{Q}_p$ satisfying $i_p^2 = -1$.

Question: Are these splitting fields "the same"? If they are, then does $i_p$ equal $i$ or $-i$? If $p \neq q$ are two distinct primes then does $i_p$ equal $i_q$ or $-i_q$?

Take a second to actually think about this!

The answer is that there is no canonical way whatsoever to identify $i_p$ with either $i$ or $-i$, and the same with identifying $i_p$ with either $i_q$ or $-i_q$. To be really explicit, take $p = 5$; there are two square roots of $-1$, one congruent to $2 \bmod 5$ and one congruent to $3 \bmod 5$. Which one is $i$ and which one is $-i$? When $p = 13$ similarly there are two square roots of $-1$, one congruent to $5 \bmod 13$ and one congruent to $8 \bmod 13$. Which one corresponds to which $5$-adic square root of $-1$?

The situation is exactly as Milne says: there simply are two isomorphisms between any pair of these splitting fields and no canonical way to choose between them. So if you actually want to use such an isomorphism you need to specify which one. Otherwise you can end up, for example, implicitly applying a bunch of isomorphisms back and forth over the course of an argument in such a way that you start in one field and end up in that field again, but via a nontrivial automorphism without realizing it.