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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?2012-07-08
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    @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)2012-07-08
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    @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.2012-07-08
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    Thanks for your comments but I need to read a few sections and come back to your answer.2012-07-08
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    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.)2012-07-09
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    @POTUS: Look again at the comment I made above; the field extension is **connected** to irreducible polynomials. So, naturally, you will be looking for an irreducible polynomial (hence, the section onf irreducible polynomials). What is the connection? Quotients of rings by maximal ideals. Which comes *before* this section in Dummit and Foote. You are trying to find a polynomial of certain degree that is irreducible in $\mathbf{F}[x]$. What degree? Like I said, *figure out* what the connection is between the degree of $f$ and the size of $\mathbf{F}[x]/\langle f\rangle$.2012-07-09
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    Can you please comment on my last comment to Dylan?2012-07-09
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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.2012-07-08
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    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: $X,X+1,X^2+X+1,X^3+X^2+1,X^3+X+1,X^4+X^3+X^2+X+1,X^4+X^3+1,X^4+X+1.$ $K^{*}=\langle X \rangle$ ?2012-07-09
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    @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 $1$.2012-07-09
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    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, $K^{*}=\langle X \rangle$.2012-07-09
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    @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.2012-07-09
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    I agree with everything Prof Magidin has said. It might help to recall the statement: if $F$ is a field and $g \in F[X]$ is irreducible of degree $n$, then $F[X]/(g(X))$ has degree $n$ over $F$, and a basis is formed by the images of $1, X, \ldots, X^{n - 1}$. To show that it spans, you note that an element of the quotient has a representative $h(X)$. Perform polynomial division on this: $h(X) = b(X)g(X) + r(X)$ where $\deg r < n$. Then note that $g = 0$ in the quotient. A relation of linear dependence, on the other hand, would give you that $g$ divides a polynomial of lesser degree.2012-07-09
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    For the generator of the multiplicative group, it seems like this should be easier to find once you have the multiplication table written down.2012-07-09
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    @POTUS (and Dylan, Arturo) +1 to you all. This is either a good time or a bad time to disclose that as [a part of my answer](http://math.stackexchange.com/a/76136/11619) to another beginner in finite fields I explicitly describe $GF(8)$ as consecutive powers of a generatot. That answer is written with the aim of helping a programmer/engineer, so the presentation is not very mathematical. You have to scroll half way down to that answer to get to this stuff. The beginning is coding theory. It is not a good fit for the present question, but you can check your results against that calculation.2012-07-09