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Given the ring $(\Bbb Z_{6}, +, *)$ (where addition and multiplication are mod 6), how do I find specific values of $a \neq0$ and $c \neq 0$ with $ab=c$, where there are two or more values of b?

I think I would start with the zero divisors of 6 (2 and 3) for $a$ and $c$, but how to I determine what b would be? Or is this even the right approach to this problem?

2 Answers2

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Suppose you have your $a,c \neq 0$ and $b_1,b_2$ with $ab_1=ab_2=c$. Then $a(b_1-b_2)=0$. In other words, you need to choose $a$ to be a zero-divisor and the difference in any choice of $b$'s to be a corresponding zero-divisor.

So let $a=2$. Pick a $b_1$ so $ab_1 \neq 0$, say $b_1=1$. So $c=2$. Then we can let $b_2=b_1+3=4$ since $2 \cdot 3 =0$. So we have $2 \cdot 1=2 \cdot 4=2$.

ET93
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It is easy to completely characterize the solutions of such (affine) linear equations.

Lemma $\ $ If $\ a\color{#0a0}{b_0}=c\ $ then $\ ab=c\iff b = \color{#0a0}{b_0}+\color{#c00}z\,$ for some root $\,\color{#c00}z\,$ of $\,a\color{#c00}z = 0$

Proof $\ \ (\Rightarrow)\ \ \ ab=c = ab_0\ $ so $\ 0 = a(b-b_0) =: a\color{#c00}z$

$(\Leftarrow)\ \ \ ab = a(\color{#0a0}{b_0}+\color{#c00}z) = a\color{#0a0}{b_0} + a\color{#c00}z = c + 0 = c$

Remark $ $ This result may be familiar from the study of linear differential or difference equations (recurrences), which shows the general solution of such inhomogeneous equations has the form of a fixed $\rm\color{#0a0}{particular}$ solution $\,\color{#0a0}{b_9}$ added to the general solution $\,\color{#c00}b\,$ of the associated homogeneous equation, as in the above solution $\,\color{#0a0}{b_0} +\color{#c00} z\,\,$ where $\,x =\color{#0a0}{ b_0}\,$ is a particular solution of $\,ax = c\,$ and where $\,x = \color{#c00}z\,$ denotes the general solution of the associated $\,\rm\color{darkorange}{homogeneous}\,$ equation: $\ ax = \color{darkorange}0.\,$

These results are all special cases of the above Lemma on the structure of the solution space of such inhomogeneous (affine) linear equations. The algebraic essence of the matter will be clarified upon studying linear algebra and module theory (= linear algebra with coefficient algebra a ring vs. field).

Bill Dubuque
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