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I just started my first upper level undergrad course, and as we were being taught vector spaces over fields we quickly went over fields. What confused me was when the book (Advanced Linear Algebra, Cooperstein) said that the the set {0, 1, 2} was a field. However, to my understanding, that set doesn't satisfy the axiom, "For every element a in F, there is an element b such that a+b=0", among others. Can someone help clarify where my understanding is off.

Also Cooperstein states "for every prime power p^n, there exists a field with p^n elements" and then doesn't expand on it. If someone could give an example or proof i would be very grateful.

Thanks for reading.

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    $\begingroup$ This is a little confusing indeed, but what is meant is that there is a (unique) field with three elements. It may be constructed as the quotient ring $\Bbb Z/3\Bbb Z$, and as such its elements are the three cosets $0+\Bbb Z, 1+\Bbb Z$ and $2+\Bbb Z$. For the definition of quotient rings, look up en.wikipedia.org/wiki/Quotient_ring $\endgroup$
    Olivier Bégassat
    –  Olivier Bégassat
    2014-01-25 00:34:01 +00:00
    Commented Jan 25, 2014 at 0:34
  • $\begingroup$ check [ math.stackexchange.com/tags/finite-fields/info ] for specific concepts $\endgroup$
    janmarqz
    –  janmarqz
    2014-01-25 00:34:57 +00:00
    Commented Jan 25, 2014 at 0:34
  • $\begingroup$ Thank you for your answer, unfortunately I'm not yet familiar with quotient rings as I don't take abstract algebra until next semester and the wiki entry was over my head. I got my answer though. $\endgroup$
    user123429
    –  user123429
    2014-01-25 00:59:33 +00:00
    Commented Jan 25, 2014 at 0:59

2 Answers 2

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$\{0,1,2\}$ is a field if you do all arithmetic modulo 3, that is, adding/subtracting an appropriate multiple of $3$ after each operation to make the result one of $0$, $1$ or $2$.

In aritmetic modulo 3, the negative of $1$ is $2$, because $1+2=3$ and subtracting $3$ to get that into the range $\{0,1,2\}$ makes $0$.

This construction will give you the fields with $p$ elements. Getting to the fields with $p^n$ elements for $n\ge 2$ requires more algebraic prerequisites than will probably fit comfortably into a MSE answer.

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  • $\begingroup$ Ahh ok, doing the arithmetic in modulo 3 makes sense. Thanks. $\endgroup$
    user123429
    –  user123429
    2014-01-25 01:00:23 +00:00
    Commented Jan 25, 2014 at 1:00
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There's a nice series of lectures from MIT that I watched recently. Try this out:

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-451-principles-of-digital-communication-ii-spring-2005/lecture-notes/lecture-9-introduction-to-finite-fields/

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