1
$\begingroup$

Let $E/F$ be a finite field extension and let $K/E$ be an extension such that $K/F$ is the splitting field of some polynomial $f(x)$ over $F$. If $K$ is a minimal extension with respect to this property, then Joseph Rotman calls this the $\textit{split closure}$ of $E/F$. In the case that $f(x)$ is separable, it may also be called the $\textit{normal closure}$.

The issue is that I am uneasy about the existence and uniqueness of the split closure. It is an exercise in the book to show that there exists a field extension $K/E$ which is a splitting field over $F$, and I was able to show this. I assume the existence of a minimal such extension just follows from the fact that $K$ is a finite extension. However, I am very unsure whether a split extension of $E/F$ is unique up to isomorphism. Does anyone know?

Edit: Let me explain further the issue. It is true that if you fix a polynomial $f(x)\in F[x]$, then all of its splitting fields are isomorphic. But the definition of split closure doesn't specify any single polynomial. So what if I had two field extensions $K/E$ and $K'/E$ which are the splitting fields of different polynomials $f(x),g(x)\in F[x]$ respectively, and such that $K$ and $K'$ do not contain any field extensions of $E$ which are splitting fields over $F$ except for themselves? Then can we infer that $K\cong K'$? Could it happen that, for example, that $[K:E]=2$ but $[K':E]=3$? These are splitting fields of potentially different polynomials, so I don't see why they should be unique up to isomorphism.

$\endgroup$
11
  • $\begingroup$ What properties and characterizations do you know of splitting fields? Do you allow splitting fields of (posdibly infinite) sets of polynomials, or just single polynomials, or just irreducible polynomials? $\endgroup$ Commented Mar 28, 2023 at 4:27
  • 1
    $\begingroup$ If $K$ and $K'$ are both splitting fields and subfields of some extension $M$ of $E$, show that $K\cap K'$ is also a splitting field over $E$. $\endgroup$ Commented Mar 28, 2023 at 4:29
  • $\begingroup$ @Arturo Magidin The definition is that $K$ is the splitting field of a single (possibly reducible) polynomial $f(x)\in F[x]$. Thank you for your hint! That gives me some direction. $\endgroup$ Commented Mar 28, 2023 at 4:31
  • 1
    $\begingroup$ Do you know the following property of splitting fields? A finite extension $K/F$ is a splitting field over $F$ if and only if for every irreducible polynomial $f(x)\in F[x]$, either $f$ splits over $K$ or $f$ is irreducible over $K$. This proves the hint handedly. $\endgroup$ Commented Mar 28, 2023 at 4:43
  • $\begingroup$ Oooo no, I was not aware of that property! But I can see how the hint easily follows from it. $\endgroup$ Commented Mar 28, 2023 at 4:55

1 Answer 1

Reset to default
1
$\begingroup$

It is unique "up to isomorphism" but one has to be careful about how this is described.

The typical situation is that you have $f(x)$ in $F[x]$ and you want to compare two splitting fields of $f(x)$ over $F$, say $K$ and $K'$. Being a splitting field of $f(x)$ over $F$ means $K$ and $K'$ are both field extensions of $F$ generated by a full set of roots of $f(x)$, and the main theorem about splitting fields is that there is a field isomorphism $K \to K'$ fixing the elements of $F$ and often there is more than one such isomorphism. This result is in essentially every abstract algebra book discussing Galois theory. Look up splitting fields in the index of such a book and you'll find a version of that theorem.

Your setup, with three fields $K/E/F$ and an $f(x)$ in $F[x]$, is not quite the same because that field $E$ may have nothing to do with $f(x)$: a splitting field of $f(x)$ over $F$ need not contain $E$ at all.

Example. Let $F = \mathbf Q$, $f(x) = x^4 - 2$, and $E = \mathbf Q(\sqrt{5})$. A splitting field $K$ of $f(x)$ over $F$ is $\mathbf Q(\sqrt[4]{2},i)$ and this does not contain $E$ or any field even isomorphic to $E$. Thus your premise "let $K/E$ be an extension such that $K/F$ is the splitting field of some polynomial $f(x)$ over $F$" makes no sense here. Maybe you meant to let $K$ be a splitting field of $f(X)$ over $E$, rather than over $F$. Think carefully about how you want $E$ and $f(x)$ to be related to each other.

$\endgroup$
2
  • $\begingroup$ I understand what you're saying, but the definition is correct as stated. It is equivalent to say that $K$ is a splitting field of $f(x)$ over $E$ because it is encoded in the definition that $K$ must contain $E$. You're certainly right that if I pick a polynomial out of a hat, there is no reason for its splitting field to contain $E$. But that doesn't matter. All that matters is that there do exist some polynomials in $F[x]$ whose splitting field contains $E$. $\endgroup$ Commented Mar 28, 2023 at 3:43
  • $\begingroup$ Ah, I had misread the start of your question. The polynomial $f(x)$ does not come first, but last, only as part of a description of $K$ as an extension of $E$ (and $F$). $\endgroup$ Commented Mar 28, 2023 at 8:24

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.