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Show that $f(x)=x^2+2x-1 \in \mathbb{Z}_3[x]$ is irreducible over $\mathbb{Z}_3$. Using this fact construct a finite field $\mathbb{F}_9$ of $9$ elements. If $\alpha$ is a root of $f(x)$, then find the elements of this field. Also find the discrete logarithms of $(\alpha-1)$ to the base $\alpha$ and of $\alpha$ to the base $(\alpha-1)$.

My approach:

Since $f(x)$ has order $2$, it is reducible in $\mathbb{Z}_3$ iff it has a root. Since none of $0$ or $1$ or $2$ satisfy $f(x)=0$, so it is irreducible in $\mathbb{Z}_3$. Now a field of 9 elements is given by $\frac{\mathbb{Z}_3[x]}{<x^2+2x-1>}$. I'm not sure if this is the field that the question is asking me to construct. Also, I need help with the later part of the question, is the question asking me to find the elements of this field in terms of $\alpha$? If yes, how to do that ?

Thanks in advance!

ARROW
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    The field contains $0,1,\alpha,\alpha^2,\alpha^3,\dotsc$. But you can simplify these: $\alpha^2+2\alpha-1=0$, so $\alpha^2=\alpha+1$; similarly, every power of $\alpha$ can be expressed as $r\alpha+s$ for some $r,s$ in ${\bf Z}_3$. – Gerry Myerson Jul 20 '23 at 03:40
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    The same field with a different polynomial is looked at for example here. The difference being that if $\alpha$ is a root of $f(x)$, then $-\alpha$ is a root of $f(-x)=x^2+x+2$. – Jyrki Lahtonen Jul 20 '23 at 03:50
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    And about calculation of the discrete logarithms. In such a small field you can follow Gerry's suggestion and simply calculate the powers $\alpha^k$, write them in the form $r\alpha+s$, and check when you get $\alpha-1$. The usual change of bases rules on logarithms apply as long as you remember that here the logarithm takes values in the residue class ring $\Bbb{Z}_8$. – Jyrki Lahtonen Jul 20 '23 at 03:59
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    Actually, in this old answer of mine I seem to carry out calculations with exactly the same $\alpha$. Observe that $x^2+2x-1=x^2-x-1$. – Jyrki Lahtonen Jul 20 '23 at 04:00

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Taking a shortcut. In this field we have $$ \alpha(\alpha-1)=\alpha^2-\alpha=1. $$ So $\alpha$ and $\alpha-1$ are each others inverses. Hence the two discrete logarithms are both ____ (you fill in the blank, observing the ring the discrete logarithm takes its values in).

Jyrki Lahtonen
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