# Multiply polynomials under field $Z_3$

As part of my journey in understanding finite fields , I have a little problem with multiplying polynomials . Given: $(x^2+x-1)(x^2-x-1)$ , a normal multiplication would be :

$(x^2+x-1)(x^2-x-1)$ = $1 - 3 x^2 + x^4$

But this is incorrect since the correct answer (by the solution of my homework) is : $x^4 +x^2 +1$

Can you please clarify how to compute the modulo here ?

Regards

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$(x^2+x-1)(x^2-x-1)=1 - 3 x^2 + x^4$ is right. In $\mathbb Z_3$, this is the same as $1+x^4$. Please check the question and the answer in the book. Perhaps it's in $\mathbb Z_4$ or $\mathbb F_4$, in which case your answer and the one given coincide. – lhf Mar 16 '12 at 12:20
Take for example the multiplication of : $(x^2 + 1) (x^4 + x^2 + 1)$ . A normal output is : $1 + 2 x^2 + 2 x^4 + x^6$ but the answer of the book is : $x^6 + x^4 + x^2 +1$ . So where is my mistake ? – ron Mar 16 '12 at 12:41
Here, too, you're right and the book, if you're quoting it correctly, is wrong. Which book is this? – joriki Mar 16 '12 at 12:53
From a book of my professor , but maybe I'm the one to blame . The actual question is : calculate the multiplication of all irreduceable polynomials of power 2 , above $Z_3$ . So , I've tried with the following polynomials : $x^2+1 ,x^2+x-1 ,x^2-x-1$ , but the multiplication result of my calculation and the book (of the professor) are different . So , which it is :) ? – ron Mar 16 '12 at 13:12
$(x^2+1)(x^2+x-1)(x^2-x-1)=x^6-2 x^4-2 x^2+1=x^6+ x^4+x^2+1$. – lhf Mar 16 '12 at 13:25