Let $E_{1}$ and $E_{2}$ be two splitting fields of polynomial $p(x)\in \mathbb{F[x]}$ over $\mathbb{F}$.

My textbook has a long proof for proving that $E_{1}$ and $E_{2}$ are isomorphic.

But isn't this obvious? Aren't $E_{1}$ and $E_{2}$ equal, as both of them are the smallest field extensions of field $\mathbb{F}$ such that all the roots of $p(x)$ are contained within them?

EDIT: Obviously $E_{1}$ and $E_{2}$ can be constructed differently. But when they're fully constructed, they have to be the very same field! Hence, even if we considered different basis sets for them, $[E_{1}:F]=[E_{2}:F]$.

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    $\begingroup$ ...and your text book is...? I mean, it'd be interesting to know what exactly is the definition of "splitting field of polynomial" there. $\endgroup$ – DonAntonio May 23 '13 at 12:34
  • $\begingroup$ For the smallest field extension of $F$ such that all the roots... to make sense, you need some field extension that contains all the roots to begin with. If you consider the question inside, e.g. a fixed algebraically closed field containing $F$, it would be simple. In general you have to construct the extension from scratch, and there is no reason to think that two different methods of construction would lead to the same result. $\endgroup$ – Jyrki Lahtonen May 23 '13 at 12:36
  • $\begingroup$ Herstein. It says "a finite extension $E$ of $F$ is said to be a splitting field over $F$ for $f(x)$ if over $E[x]$, and not over any proper subfield, $f(x)$ can be factored as a product of linear factors" $\endgroup$ – fierydemon May 23 '13 at 12:37
  • $\begingroup$ IOW, a student's imagination is easily lead astray here by being exposed only to extensions, where we can assume that everything takes place inside the field of complex numbers. $\endgroup$ – Jyrki Lahtonen May 23 '13 at 12:37
  • $\begingroup$ @JyrkiLahtonen- from what I understand, there can only be one smallest field containing all the elements of $\mathbb{F}$, and then some extra elements, i.e. the roots of $p(x)$ not contained within $F$. Let these roots be ${a,b,c\dots n}$. So aren't $E_{1}=E_{2}=F[a,b,c\dots n]$? $\endgroup$ – fierydemon May 23 '13 at 12:41

Trying to shed some light to this by proffering several splitting fields of the polynomial $x^2+1$ over the rationals.

  1. One splitting field is the field $\mathbb{Q}[i]$, i.e. the usual subfield of complex numbers.
  2. Another splitting field is the quotient field $\mathbb{Q}[x]/\langle x^2+1\rangle$ of the ring of polynomials with rational coefficients by the maximal ideal generated by the polynomial $x^2+1$.
  3. Let us consider the skewfield of Hamiltonian quaternions $$ \mathbb{H}=\{a+bi+cj+dj\mid a,b,c,d\in\mathbb{R}\}, $$ where the usual relations are satisfied: $i^2=j^2=k^2=ijk=-1$. Let us pick any triple $(b,c,d)$ of real numbers such that $b^2+c^2+d^2=1$. Then the quaternion $$ u=bi+cj+dk $$ satisfies the equation $u^2=-1$. It is easy to see that the subset $$ \mathbb{Q}[u]=\{a+fu\in\mathbb{H}\mid a,f\in\mathbb{Q}\} $$ is then a splitting field of $x^2+1$ over the rationals.
  4. The set of matrices of the form $$ \left(\begin{array}{rr}a&b\\-b&a\end{array}\right), $$ with $a,b$ arbitrary rational numbers is yet another splitting field of $x^2+1$.

No two of these splitting fields are equal in an obvious sense. They are all isomorphic to each other, though.

Another example. You often think of the splitting field of $x^2-2$ over the rationals as $\mathbb{Q}[\sqrt2]$ - a subset of the real numbers. So then $\sqrt2$ may be a certain equivalence class of Cauch sequences. Why is that particular equivalence class of Cauchy sequence equal to the coset of $x$ in the quotient ring $\mathbb{Q}[x]/\langle x^2-2\rangle$?

IOW. This questions is philosophical in the sense that we really are talking about the identity of the roots. Not just up to isomorphism!

  • $\begingroup$ Shouldn't the splitting firld in (1.) be $\mathbb{Q[i,-i]}$? $\endgroup$ – fierydemon May 23 '13 at 13:22
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    $\begingroup$ Well, yeah, but when you adjoin $i$, you get $-i$ free of charge. $\endgroup$ – Jyrki Lahtonen May 23 '13 at 13:24

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