# Multiplication tables of rings

1. Oct 16, 2008

### phyguy321

1. The problem statement, all variables and given/known data
construct a multiplication table for the ring Z$$_{3}$$[$$\alpha$$], $$\alpha$$$$^{2}$$ + 1(bar) = 0(bar)

2. Relevant equations

3. The attempt at a solution
I'm actually confused on how to find the elements of the ring. My book and notes have thrown me off a bit and I can't find them. Hint: there are 9 elements

2. Oct 16, 2008

### jacobrhcp

I think for advanced courses like yours it might help me and others if you remind us of what things like (bar) mean

and do you mean by Z^3 actually Z_3 or ZxZxZ or something completely different outside my knowledge?

3. Oct 16, 2008

### phyguy321

yea sorry the text editor on here sucks....it is supposed to be Z_3 (which is what i put) and (bar) refers to being part of that equivalence class so 1 bar is ...-2,1,4,7... and 0 bar is ...-6,-3,0,3,6... for the integer set Z_3 so the congruent mod 3.

4. Oct 17, 2008

### HallsofIvy

Staff Emeritus
Then, strictly speaking, each member of Z3[/sup] is an equivalence class, each class containing exactly one non-negative integer less than 3. That is, there is an equivalence class containing 0, and equivalence class containing 1, and an equivalence class containing 2. That is, I presume, what you mean by $\bar{0}$, $\bar{1}$, and $\bar{2}$. Since that does NOT have 9 members. Am I to assume that you mean the set of all numbers of the form $a\alpha+ b$ where a and b are in Z3 and $\alpha$ satisfies the equation above? That has nine members: 0, 1, 2, $\alpha$, $\alpha+ 1$, $\alpha+ 2$, $2\alpha$, $2\alpha+ 1$, and $2\alpha+ 2$.

Set up your table so it has those both along the top and verticaclly on the left. Multiply each of the 81 pairs and reduce to one of those 9 forms by using the equation $\alpha$ satisfies. For example, $\left(\alpha+ 1\right)\left(2\alpha+ 1\right)= 2\alpha^2+ 3\alpha+ 1$. Since $\alpha^2+ 1= 0$, $\alpha^2= -1$. Of course, 3 is equivalent to 0 mod 3 so this reduces to $-2+ 1= -1$ which is 2 mod 3:$\left(\alpha+ 1\right)\left(2\alpha+ 1\right)= 2$.

No, the LaTex editor here does not "suck" but it is a bad idea to try to combine both LaTex and non-LaTex in the same formula: use LaTex for the entire formula.

5. Oct 17, 2008

### phyguy321

awesome thanks