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Show that $f(x)=x^2+x+3 \in \mathbb{Z}_7$ is a primitive polynomial.

I have shown that it is irreducible but I am unsure how to show that it is primitive. Any help would be great.

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By which definition of primitive do you mean? Normally a primitive polynomial means its coefficients share no common nontrivial factor, but according to wikipedia there is another definition. Also please clarify on what $\mathbb{Z}_7$ is, is it the integers modulo $7$ or the $7$-adic integers? Both definitions lead to this question being meaningful strangely. –  dinoboy Jan 24 '13 at 21:28
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I think you do not mean primitive as defined by dinoboy –  Amr Jan 24 '13 at 21:30
    
I do not believe this is the definition I am familiar with. I believe I have to show that $\alpha$ is a primitive element of F but I am not sure how to go about it. And when I say Z_7 I am taking about integers modulo $7$. –  user59548 Jan 24 '13 at 21:35

1 Answer 1

Let $F = \mathbb{Z}_7[x] / \left< f(x) \right> $ represented as degree-$1$ polynomials in $\alpha$. $|F^*| = 48$.

Now all you have to do is prove that $o(\alpha)=48$. $48=2^4 \cdot 3$, so all you have to show is that $\alpha^{24} \ne 1$ and $\alpha^{16} \ne 1$. I would recommend repeated squaring:

$$\alpha^2 = 6\alpha + 4$$ $$\alpha^4 = 5\alpha + 6$$ $$\alpha^8 = 3$$

And it's easy from here.

(I used this calculator, which you might find useful for such computations, but it's not hard at all to square these by hand if you need to on a test.)

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Thank you for your answer. I am not 100% sure how you got F*=48. Am I correct in thinking that you did $7^2-1$ ? –  user59548 Jan 24 '13 at 22:12
    
Yes: $F^*$ consists of all of the elements of $F$ (of which there are $7^2$) except for $0$. –  Alfonso Fernandez Jan 24 '13 at 22:18
    
So next I would calculate $\alpha^{16}$? –  user59548 Jan 24 '13 at 22:26
    
Yes, and $\alpha^{16} = (\alpha^8)^2$ which is really easy to calculate, and $\alpha^{24} = (\alpha^8)^3$ which is also easy. –  Alfonso Fernandez Jan 25 '13 at 0:19
    
Thank you much appreciated. –  user59548 Jan 25 '13 at 8:17

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