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Give an explicit contruction of the finite field $K$ containing $8$ elements, as a quotient of an appropriate polynomial ring. Include the multiplication table of the group $K^{*}=K\setminus \{0\},$ and write $K^{*}=\langle \alpha \rangle$ for some $\alpha \in K.$

I have no idea how to approach this problem. Can anyone guide me in the right direction? Thanks.

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Start with a field $\mathbf{F}$ with $2$ elements. A field with $8$ elements must contain $\mathbf{F}$ and be an extension of degree $3$, by size considerations.

Do you know how to get an extension of degree $3$ of a given field? Once you have such a field, the rest of the problem will follow by simply staring at your field long enough.

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Field extensions was not taught to me when I took Abstract algebra II. Also, I checked the syllabus for the prelim and it does not have field extensions listed. Is this the only way to solve this problem? – Lyapunov Jul 8 '12 at 6:01
@POTUS: I'll wager you did, you just didn't call them that. In any case, if $\mathbf{F}$ is a field, then $\mathbf{F}[x]$ is a comutative ring with $1$; and if $I$ is a maximal ideal of a commutative ring with $1$, then $R/I$ is a field. The maximal ideals of $\mathbf{F}[x]$ are precisely the ideals $\langle f(x)\rangle$ where $f(x)$ is irreducible. Use the division algorithm to figure out the size of $\mathbf{F}[x]/\langle f\rangle$ when $\mathbf{F}$ is finite, in terms of the degree of $f$. (cont) – Arturo Magidin Jul 8 '12 at 6:03
@POTUS: Trying to construct a field of $8$ elements without knowing their relation with irreducible polynomials in $\mathbf{F}[x]$ (where $\mathbf{F}$ is a field of two elements) is likely to produce only frustration. PS If the syllabus includes fields, then it definitely include field extensions, though perhaps not explicitly. Any study of fields will, sooner rather than later, run into the idea of field extensions. – Arturo Magidin Jul 8 '12 at 6:04
Thanks for your comments but I need to read a few sections and come back to your answer. – Lyapunov Jul 8 '12 at 6:06
I did some studying and found out that this problem is from section 9.4 irreducibility criteria of Dummit and Foote's Abstract algebra. Field extensions comes much later. So, I think I am supposed to solve this without the knowledge of field extensions.(PS: I am in no way trying to question your authority on the subject.) – Lyapunov Jul 9 '12 at 0:42

Your textbook should explain you the passage from a cubic irreducible polynomial with coefficients in a given field $F$ (here $\mathbb{Z}/2\mathbb{Z}$) to an extension field $K$ of degree three (of the same field $F$). Then use the method described in your previous question for finding such a polynomial.

In your case $K^*$ will have seven elements. What do you know about groups of seven elements? In terms of having a generator?

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As a warm-up, how about we try to write down a field $k$ with $4$ elements? This is a degree $2$ extension of $\mathbf F_2 = \mathbf Z/2\mathbf Z$, so we need to find an irreducible polynomial of degree $2$ in $\mathbf F_2[X]$. We quickly find that $f(X) = X^2 + X + 1$ is the only one, so let $k = \mathbf F_2[X]/(f(X))$.

Using $\alpha$ to denote the image of $X$ in $k$, the set $\{1, \alpha\}$ is a basis for $k$ over $\mathbf F_2$. To perform multiplication, use the relation $\alpha^2 =\alpha + 1$ imposed by $f$. For example, $$(1 + \alpha)\alpha = \alpha + \alpha^2 = \alpha + \alpha + 1 = 1.$$

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Of course there's a bit of theory (mostly linear algebra) lying behind this. I don't want to give you a bunch of justifications that you already know, so I haven't included all of them; if any of it is unclear, then do leave a comment. – Dylan Moreland Jul 8 '12 at 6:00
So, to find a field $k$ with $8$ elements I need $k=\mathbf{F}_2[X]/(f(X))$ where $f(X)=X^4+X+1.$ The eight elements would be: $X,X+1,X^2+X+1,X^3+X^2+1,X^3+X+1,X^4+X^3+X^2+X+1,X^4+X^3+1,X^4+X+1.$ $K^{*}=\langle X \rangle$ ? – Lyapunov Jul 9 '12 at 1:12
@POTUS Heya. As Arturo and Jyrki mention, you want $f$ to be a cubic so that your field extension has degree $3$ over $\mathbf F_2$, since $8 = 2^3$. Your $8$ elements can then be written as $a_0 + a_1X + a_2X^2$ where each $a_i$ can be $0$ or $1$. – Dylan Moreland Jul 9 '12 at 1:15
Okay so I need $k=\mathbf{F}[X]/(f(X))$ where $f(X)=X^3+X+1$ or $f(X)=X^3+X^2+1.$ The eight elements are $0,\pm 1,X,X+1,X^2+X+1,X^3+X^2+1,X^3+X+1.$ Also, $K^{*}=\langle X \rangle$. – Lyapunov Jul 9 '12 at 2:02
@POTUS: Either polynomial is correct; but your elements are not. For example, if you take the second polynomial, then in the quotient you have $X^3=X^2+1$. But then $X^3+X^2+1 = X^2+1+X^2+1=0$, so you don't get a new element that way. And if you use the first polynomial instead, then $X^3+X+1=0$, so that's no good either. Use the remainders. As to the generator, remember that you are only looking at the powers, so you need to prove that $X$ generates, not just say it does. – Arturo Magidin Jul 9 '12 at 3:54

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