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How are multiplication tables for fields created? |
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| Apr7-11, 11:55 AM | #1 |
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How are multiplication tables for fields created?
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...ort/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. |
| Apr7-11, 12:12 PM | #2 |
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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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