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The problem:

Show that $p(x)=x^2-x-1\in \mathbb Z/(3)[x]$ is irreducible over $\mathbb Z/(3)$. Show that there exists an extension K of $\mathbb Z/(3)$ with nine elements having all roots of $p(x)$.

What I did:

I almost solved the question, I proved that $p(x)$ is irreducible because there are no roots of $p(x)$ in $\mathbb Z/(3)$ I proved also the extension K is in the form: $K=\{a+bu; a,b \in \mathbb Z/(3)\}$ where $u$ is a root of $p(x)$ in an extension of $\mathbb Z/(3)$ with nine elements by Kronecker's Theorem. My problem is prove that any root of $p(x)$ is in $K$. Obviously $u \in K$ and the other root? Notice that p(x) has just two roots because $Z/(3)$ is an integral domain. Anyone can help me, please?


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You will be able to identify the sum of the roots from the polynomial ... ? – Mark Bennet Oct 23 '12 at 17:08
@MarkBennet no, I can't see what they are – user42912 Oct 23 '12 at 17:21
Note that $(x-a)(x-b)=x^2-(a+b)x+ab$ – Mark Bennet Oct 23 '12 at 17:23
@MarkBennet yes of course, since the equality holds, we have u+ u'=1, so u'= 1-u = 1+2u which is in the K with a=1 and b=2, correct? – user42912 Oct 23 '12 at 17:27

1 Answer 1

up vote 1 down vote accepted

Hint $\rm\ x^2-x-1\, =\, (x-u)(x-u')\:\Rightarrow\: u+u' =\, 1\, =\, -uu'$

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thank you, it helps a lot – user42912 Oct 23 '12 at 17:27

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