I have this definition for the splitting field of a polynomial: let $f$ be a polynomial with coefficients in the field $F$. A field $E$ containing $F$ is called a splitting field for $f$ if it satisfies:

i) $f$ splits in $E[X]$, i.e $f(X)=a\displaystyle\prod_i (X-\alpha_i)$, with $a,\alpha_i\in E$

ii) $E$ is generated over $F$ by the roots of $f$, i.e. $E=F[\alpha_1,\ldots,\alpha_n]$

Further in my textbook i can find a proof that any two splitting fields for $f$ are $F$-isomorphic.

I can't understand the meaning of this result. By definition, condition ii) seems to state that a splitting field is minimal among all fields containing $F$ where $f$ completely splits, hence i thought that a splitting field was unique. Why then we need a theorem stating that all splitting fields are isomorphic?

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    $\begingroup$ The polynomial $x^2-2$ has two splitting fields over the rationals: $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}[x]/(x^2-2)$. They aren't the same. But they are isomorphic! $\endgroup$ – Jyrki Lahtonen Aug 29 '13 at 13:24
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    $\begingroup$ In other words: where did those zeros $\alpha_i$ come from? They don't exist in a vacuum, and we may have to construct them. The essence of the result is that we don't have to worry about the method of construction. The end results of two different constructions will be isomorphic. $\endgroup$ – Jyrki Lahtonen Aug 29 '13 at 13:27
  • $\begingroup$ @JyrkiLahtonen I think i've understood your observation: the field $\mathbb{Q}[X]/(X^2-2)$ is also of the form $\mathbb{Q}[\alpha]$, if i take $\alpha$ to be the coset $X+(X^2-2)$ and $\alpha$ is a root of $X^2-2$ in this field. This root is expressed in a "different" form, but the extension is isomorphic $\endgroup$ – bateman Aug 29 '13 at 13:51
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    $\begingroup$ Correct. Early on in one's studies it is tempting to think that all the roots of all the polynomials are complex numbers - FTA and all that. But this is not the case, when we have no control over the field of coefficients. We might have positive characteristic anyway, or be in a field that's too big to fit inside the complex numbers. $\endgroup$ – Jyrki Lahtonen Aug 29 '13 at 13:55
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    $\begingroup$ @niloderoock There are certainly extensions of complex numbers, say $\Bbb{C}(X)$ with $X$ an indeterminate. But that is actually isomorphic to a subfield of $\Bbb{C}$. To make that impossible pick a set of algebraically independent variables $S$ such that $S$ is larger than $\Bbb{C}$. For example $S$ indexed by the power set of $\Bbb{C}$. Then the field $\Bbb{Q}(S)$ has cardinality exceeding that of $\Bbb{C}$. Use $\Bbb{F}_p$ in place of $\Bbb{Q}$ if you want another characteristic. Those fields have algebraic closures as well. $\endgroup$ – Jyrki Lahtonen May 1 '17 at 11:39

Summary of the comments: The splitting field is uniquely determined by the polynomial, if you consider it inside a fixed extension field $E$ of $F$ where the polynomial happens to split. For example, we could use an algebraic closure of $F$ as $E$.

The point of this isomorphism result is to relieve us of the obligation to work inside any previously known field. It allows us to construct the zeros and the extension field any which way we see fit. Secure in the knowledge that the end result will be unique (up to isomorphism).

  • $\begingroup$ That captures it nicely. The first statement is separate from the result on splitting fields, in that there can be different but isomorphic extensions of F within the given E. It is the simpler fact that a polynomial uniquely determines the set of its roots and their multiplicities. For a splitting extension of the polynomial this implies it is a uniquely determined subfield of E. $\endgroup$ – zyx Aug 29 '13 at 14:42

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