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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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. $\endgroup$
    – Anon
    Commented Mar 28, 2023 at 4:31
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    $\begingroup$ 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. $\endgroup$
    – Anon
    Commented Mar 28, 2023 at 4:55

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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.

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  • $\begingroup$ 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 $E$. $\endgroup$
    – Anon
    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 $F$). $\endgroup$
    – KCd
    Commented Mar 28, 2023 at 8:24

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