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This seems counter intuitive to me. For example, the roots of $p(x)=x^3-2$ over $\mathbb{Q}$ are $2^{(1/3)},\gamma2^{(1/3)}, \gamma^22^{(1/3)}$ where $\gamma$ is a primitive third root of unity. So, $\mathbb{Q}(\gamma^22^{(1/3)})$$\cong$$\mathbb{Q}(2^{(1/3)})$, okay, but something isn't quite clicking for me, and perhaps it's because i want to believe that two roots of $p(x)$ is in $\mathbb{Q}(\gamma^22^{(1/3)})$ when that's not true, I mean if it did have two roots then it would have to have all three and so that's obviously not true. So is it safe to say that in the splitting field generated by different roots of an irreducible polynomial, that irreducible polynomial will split in "isormorphic" ways in each respective splitting field? It must, it's just I've been discovering more and more places where i'm learing how "isomorphic" does not mean "equal" and so maybe that's why i'm so unsure. Sorry, i understand my phrasing is not perfect, but I hope that i'm getting my question across, and if not i hope someone just comes here and gives me general insight into the genereal area.

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    $\begingroup$ This is a good question that you truly need to strive to understand inside out. +1 for realizing the need to ask it! I'm not sure that I approached it from a direction useful to you (in my answer). Do comment! $\endgroup$ – Jyrki Lahtonen Oct 23 '18 at 10:29
  • $\begingroup$ thanks man, your answer helped a lot i appreciate it $\endgroup$ – Math is hard Oct 23 '18 at 10:30
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Let $L_1=\Bbb{Q}(\root3\of2)$ and $L_2=\Bbb{Q}(\gamma^2\root3\of2)$. We have the isomorphism of fields $\sigma:L_1\to L_2$ defined by the properties $\sigma(q)=q$ for all $q\in\Bbb{Q}$, and $\sigma(\root3\of2)=\gamma^2\root3\of2.$

My guess is that your troubles are related to some of the following:

  • The isomorphism $\sigma$ can be extended to an isomorphism between polynomial rings $\sigma:L_1[x]\to L_2[x]$ defined by applying $\sigma$ to the coefficients: $$\sigma(a_0+a_1x+a_2x^2+\cdots+a_nx^2)=\sigma(a_0)+\sigma(a_1)x+\sigma(a_2)x^2+\cdots+\sigma(a_n)x^n$$ for all $n\in\Bbb{N}$ and all $a_0,a_1,\ldots,a_n\in L_1$.
  • In the ring $L_1[x]$ we have the factorization $$f(x)=x^3-2=(x-\root3\of2)(x^2+\root3\of2x+\root3\of4).\qquad(*)$$ Call the factors $g(x)=x-\root3\of2$, $h(x)=x^2+\root3\of2x+(\root3\of2)^2$.
  • Applying the ring homomorphism $\sigma$ to the identity $f(x)=g(x)h(x)$ of the ring $L_1[x]$ we get the corresponding identity $$ \sigma(f(x))=\sigma(g(x))\sigma((h(x)) $$ in the ring $L_2[x]$.
  • Here $\sigma(f(x))=f(x)$ because all the coefficients of $f(x)$ are rational, and those are fixed by $\sigma$. But $$\sigma(g(x))=\sigma(1)x-\sigma(\root3\of2)=x-\gamma^2\root3\of2,$$ and similarly $$ \sigma(h(x))=x^2+\gamma^2\root3\of2x+(\gamma^2\root3\of2)^2=x^2+\gamma^2\root3\of2 x+\gamma\root3\of4. $$ Meaning that the factorization $(*)$ of $f(x)$ in $L_1[x]$ becomes a slightly different factorization of $f(x)$ this time in the ring $L_2[x]$.
  • The two resulting factorizations of $f(x)$, namely $g(x)h(x)$ and $\sigma(g(x))\sigma((h(x))$ are, in this sense, isomorphic but not equal. I approve of your choice of phrase here!

So the linear factors of $f(x)$ in $L_1[x]$ (resp. $L_2[x]$) are mapped to each other by the extension of $\sigma$. The same holds for the quadratic factors (and in a more general setting to all the factors - their degrees are preserved). But, because $\sigma$ may move the coefficients of those factors, their homomorphic images are not identical to the original versions unless all the coefficients of the factor are fixed by $\sigma$.

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  • $\begingroup$ and even though $\sigma$ may have more coefficients of those factors, it won't reduce the factor to a lower degree. I think that's the part that is surprising to me. $\endgroup$ – Math is hard Oct 23 '18 at 10:27
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    $\begingroup$ @MichaelVaughan One way to think about that is to observe that if $\sigma(h(x))$ should factor further in the ring $L_2[x]$, then an application of the extension of $\sigma^{-1}$ would show that $h(x)$ would then have a corresponding factorization in $L_1[x]$. But no such factorization exists, because $L_1$ is a subfield of the reals, and the zeros of $h(x)$ are both complex. $\endgroup$ – Jyrki Lahtonen Oct 23 '18 at 10:32
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    $\begingroup$ The polynomial $\sigma(h(x))$ does have the factor $g(x)=x-\root3\of2$ in $L[x]$ where $L$ is the splitting field. But $g(x)\notin L_2[x]$, so it cannot be a factor of $\sigma(h(x))$ in the ring $L_2[x]$. $\endgroup$ – Jyrki Lahtonen Oct 23 '18 at 10:36
  • $\begingroup$ Yeah, right, of course. In regards to the idea "having more coefficients to work with" (even though they factor the same), is there any insight in thinking about the fields generated by the roots of $x^3-2$ as lines passing through the origin and one of the roots on a circle of radius $2^{1/3}$ in the complex plane? $\endgroup$ – Math is hard Oct 23 '18 at 10:40

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