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Let $F$ be a field and let $f(x)\in F[x]$ be a polynomial. Recall the following two facts:

(1) algebraic closures are unique up to isomorphism

(2) splitting fields are unique up to isomorphism

Fix an algebraic closure $\overline F$ of $F$. Is it true that any two (necessarily isomorphic) splitting fields $E,E'\subseteq \overline F$ for $f$ are equal $as$ $sets$ in addition to being isomorphic?

My intuition tells me that this should be the case: fixing an algebraic closure allows us to fix roots of $f$ within this particular ambient field, and since any splitting field is the smallest field containing these roots, any two such fields must actually be equal as sets, in addition to being isomorphic. Does this sound about right? Is this totally trivial?

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2 Answers 2

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Yes, this is correct. The only splitting field for $f$ over $F$ in $\overline{F}$ is the subfield of $\overline{F}$ generated by $F$ and all the roots of $f$ in $F$.

(To be clear, for this to be true, by "splitting field for $f$ over $F$ in $\overline{F}$" we must mean a splitting field for $f$ over $F$ which is a subfield of $\overline{F}$ (so the field operations are the same) and whose copy of $F$ is the same as the copy of $F$ inside $\overline{F}$.)

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  • $\begingroup$ I don't think this is correct. For example, both $\;\Bbb Q(\sqrt2)\;$ and $\;\Bbb Q[x]/\langle\,x^2-2\,\rangle\;$ are splitting fields of $\;x^2-2\in\Bbb Q[x]\;$ over $\;\Bbb Q\;$...and they both are in the algebraic closure $\;\overline{\Bbb Q}\;$ of the rationals. Of course, they clearly are isomorphic...but not equal as sets as the elements in one are completely different than in the other one. And this is what I mentioned in may answer, which was downvoted without explanation. Am I missing something? $\endgroup$
    – DonAntonio
    Commented Feb 6, 2018 at 8:28
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    $\begingroup$ My only issue with this example is, it would seem, that $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}[x]/(x^2-2)$ are not actually in $the$ $same$ algebraic closure of $\mathbb{Q}$. From what I can tell, the first is a subset of the algebraic closure of $\mathbb{Q}$ which sits inside the complex numbers, and the other is a subset of the Artinian construction of the algebraic closure of $\mathbb{Q}$ (defined using Zorn's lemma and quotienting out by polynomials). $\endgroup$
    – Arbutus
    Commented Feb 6, 2018 at 16:28
  • $\begingroup$ @Jacob111 That may be an issue of language: in both cases the ground field, $\;\Bbb Q\;$ , is embedded in a rather canonical way: $\;q\mapsto q\cdot+0\cdot\sqrt2\;$ for $\;\Bbb Q\hookrightarrow\Bbb Q(\sqrt2)\;$ , and in the other $\;q\mapsto q+\langle x^2-2\rangle\;$ for $\;\Bbb Q\hookrightarrow\Bbb Q[x]/\langle x^2-2\rangle\;$ ...and thus both extensions are contained in the same algebraic closure of $\;\Bbb Q\;$ . Somehow this seems to be a pretty standard thing, as all the books I can remember always talk of "isomorphic splittingfields"...etc., never as "there's only one" ... $\endgroup$
    – DonAntonio
    Commented Feb 6, 2018 at 21:39
  • $\begingroup$ @DonAntonio: I have no idea what you think "the same algebraic closure" means. As far as I can tell you think the word "same" is completely meaningless in that phrase... $\endgroup$
    – Eric Wofsey
    Commented Feb 6, 2018 at 21:45
  • $\begingroup$ @EricWofsey Oh, that's simple to explain: it means what most mathematicians know it is: an algebraically closed field which extends the ground field and is algebraic over it. Isn't it the same for you? Perhaps I'm missing something else the OP wrote in his question, but not such a basic thing... $\endgroup$
    – DonAntonio
    Commented Feb 6, 2018 at 21:50
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I don't think they are equal as sets. Remember that in our way to get splitting field for $\;f(x)\in F[x]\;$ , we first get (assuming $\;f\;$ is irreducible, otherwise we take one of its irreducible factors) the quotient ring (field) $\;K:=F[x]/\langle f(x)\rangle\;$ .

Here, we already have no more $\;F\;$ but an isomorphic copy of $\;F\;$ within $\;K\;$ , and there could be several ways to obtain that copy...

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