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Let's say I have the polynomial $f(x)=x^4 +1 \in \mathbb{Z}_{3}[x]$. I know that this polynomial is reducible in $\mathbb{Z}_{3}[x]$ since it can be decomposed as follows: $$ f(x) = (x^2 +x +2)(x^2 +2x +2) $$ where both factors are irreducible since neither of them have roots in $\mathbb{Z}_{3}$.

My goal is to find two nonisomorphic extension fields in which $f(x)$ has roots. Here's what I did:

The first polynomial has roots $\frac{-1 \pm \sqrt{7}}{2}i$ and the second has roots $-1 \pm i$. I created these field extensions:

$\mathbb{Z}_{3}[\frac{-1 + \sqrt{7}}{2}i]$ and $\mathbb{Z}_{3}[-1+i]$

Now if I did everything right, $f(x)$ has roots over these fields. My worry is showing that these are not isomorphic. I would have to show that there does not exist an isomorphism between the two, but this seems less straight forward than showing that there does exist one.

PS, would it be any different if I had $\mathbb{Z}_{3}[i]$ instead of $\mathbb{Z}_{3}[\frac{-1 + \sqrt{7}}{2}i]$?

egreg
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Ninosław Brzostowiecki
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1 Answers1

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Your polynomials have coefficients in $\mathbb{Z}_3$, not in the complex numbers.

However, if you consider the splitting field of $x^2+x+2$ (by adding a root $\alpha$ of it) and the splitting field of $x^2+2x+2$ (by adding a root $\beta$), you get isomorphic fields, because both are fields with $3^2=9$ elements.

The field with $3^4=81$ elements contains the nine element field as a subfield and so $x^4+1$ has roots in it.

egreg
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  • Perhaps the intention is to take the field with 9 and the field with 81 elements? This will trivially give two non-isomorphic fields "in which the polynomial has roots" ... – DonAntonio Apr 03 '16 at 23:14
  • @Joanpemo Any field with $3^{2n}$ elements ($n>1$) will do; the fact I tried to explain is that just adding a root of either factor is not sufficient. – egreg Apr 03 '16 at 23:16
  • Why isn't it enough to take $\mathbb{Z}_{3}$ and adjoin both $\alpha$ and $\beta$? This would give me 27 elements containing the 9 from adjoining say only $\alpha$. – Ninosław Brzostowiecki Apr 03 '16 at 23:38
  • @NinosławCiszewski No: $\mathbb{Z}_3[\alpha]$ contains the two roots of $x^2+2x+2$; they are $2\alpha$ and $1+\alpha$, check it. – egreg Apr 03 '16 at 23:42
  • @egreg I think I understand. For simplicity, let's say that both my factors yielded roots that were linear combinations of 1 and i, like: 2 + i, 2 + 2i and i. Then it would suffice to construct $\mathbb{Z}_{3}$[i] as this would contain all the roots. Then to create an nonisomorphic field that also contains these, could I just adjoin some other value like "e" or $\pi$? – Ninosław Brzostowiecki Apr 04 '16 at 01:23
  • @NinosławCiszewski Yes, you could also add a transcendental element, but $e$ or $\pi$ wouldn't make sense in this context. – egreg Apr 04 '16 at 07:28
  • @egreg, yes, and I see why, there is no way to construct a transcendental number from the ones we have from our field. But we can construct something like $\sqrt{2}$, so that would work. – Ninosław Brzostowiecki Apr 04 '16 at 23:43