How are multiplication tables for fields created?

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SUMMARY

This discussion focuses on the creation of multiplication tables for finite fields, specifically the 9-element field F[9], which is an extension of F[3]. The key concept involves using a monic quadratic polynomial, such as x² + 1, to facilitate calculations. The rationale for selecting a monic polynomial is that it simplifies the process by reducing the number of coefficients involved. The discussion also clarifies the evaluation of polynomials in finite fields, particularly how to compute f(2) in F[3].

PREREQUISITES
  • Understanding of finite fields, specifically F[3] and F[9]
  • Knowledge of polynomial functions and their properties
  • Familiarity with vector spaces in the context of abstract algebra
  • Basic modular arithmetic
NEXT STEPS
  • Study the properties of monic polynomials in field theory
  • Learn how to construct multiplication tables for finite fields
  • Explore the concept of vector spaces over finite fields
  • Investigate the implications of polynomial roots in finite fields
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Students of abstract algebra, mathematicians interested in finite fields, and educators seeking to teach multiplication in fields effectively.

phillyj
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The class I'm in is not modern algebra nor have I take those courses. The professor of the class [Learning how to read/write in math] decided to try to teach some abstract algebra. I am trying to understand how multiplication is done in a field. I am using this site to help me as it was most relevant to making tables like the professor wanted:

http://math.arizona.edu/~ura-reports/041/Patterson.Genevieve/Final/FinalReport/node8.html

I am stuck on 9 element fields. This is the second to last example at the end of the page. I understand that we can extend the F[3] field to get the table.

The site says "As a vector space, F[9] = F[3^2] = {a+bx, (a,b)[tex]\in[/tex] F[3]}. To find the multiplication table we need a monic-quadratic that has no zeros in F[3]
A monic-quadratic will have a coefficient of 1 on the highest degree term."

[1]Why do they use monic quadratic?

[2] When they try x^2+1, they get f(0)=1, f(1)=2 but why is f(2)=2? Are they not plugging into x?

I don't really have good study material as the professor wrote a short paper on this stuff but it is bare minimum.

Thanks for your help.
 
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F[3] is just the integers modulo 3. Given a number, its remainder when you divide by 3 is either 0, 1 or 2, and that's the corresponding number that you get in F[3]. So f(2)=5, but we have to take the remainder when you divide by 3, which gives f(2)=2.

The choice for a monic quadratic is just that it makes the calculations a little easier to look at (there are fewer coefficients flying around). x is going to be an element that satisfies f(x)=0. If we have [tex]ax^2+bx+c=0[/tex] the way we use this is by re-arranging:
[tex]x^2 = \frac{1}{a}(-bx-c)[/tex]

So given any arbitrary polynomial in x, we can write it as something of the form cx+d by repeatedly replacing x2s with the other side of that equality. We could have chosen a monic f(x) by just dividing both sides of [tex]ax^2+bx+c=0[/tex] and we would have ended up with the same solution for x2 at the end anyway. So we can basically just ignore the a if we assume our polynomial is monic to begin with
 

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