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I have two problems quite similar. The first:

In $\mathbb{Z}_8$ find the identity of the following commutative operation: $$\overline{a}\cdot\overline{c}=\overline{a}+\overline{c}+2\overline{a}\overline{c}$$

I say: $$\overline{a}\cdot\overline{i}=\overline{a}+\overline{i}+2\overline{a}\overline{i} = \overline{a}$$ Since $\overline{a}$ is always cancelable in $\mathbb{Z}_9$ I can write: $$\overline{i}+2\overline{a}\overline{i} = \overline{0}$$ $$\overline{i}(\overline{1}+2\overline{a}) = \overline{0}$$ so $i=0$ whatever $\overline{a}$.

Second question:

In $\mathbb{Z}_9\times\mathbb{Z}_9$ find the identity of the following commutative operation: $$(\overline{a}, \overline{b})\cdot (\overline{c}, \overline{d})= (\overline{a}+ \overline{c}, \overline{8}\overline{b}\overline{d})$$

So starting from: $$(\overline{a}, \overline{b})\cdot (\overline{e_1}, \overline{e_2})= (\overline{a}+ \overline{e_1}, \overline{8}\overline{b}\overline{e_2})=(\overline{a}, \overline{b})$$ that is: $$\overline{a}+\overline{e_1}=\overline{a}\qquad (1)$$ $$\overline{8}\overline{b}\overline{e_2}=\overline{b}\qquad (2)$$ In (1) there's always cancellable element for $\overline{a}$ since $\overline{-a}$ is always present in $\mathbb{Z}_9$. In (2) I should multiply both member for $\overline{8^{-1}}$ and $\overline{b^{-1}}$ to know exactly $\overline{e_2}$. This happens only if both number are invertible. $\overline{8}$ is easily to demonstrate that it's invertible, cause $gcd(8,9)=1$. But what about $b$.

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4 Answers 4

up vote 1 down vote accepted

You have a slight misunderstanding about proof that's causing you problems here. Look at your first proof. Did you notice you did not prove "0 is the identity of this operation"? Instead, you proved the following statement:

  • If $a$ is an element $i$ is the identity, then $i = 0$ or $2a+1 = 0$

I'll grant it's easy to check that $2a+1 = 0$ can't ever be true, so I'll assume you thought that through and just didn't think it important enough to write, and so you proved

  • If $a$ is an element $i$ is the identity, then $i = 0$

First, note that you have not yet proven that $0$ is the identity. You've merely proven that if $a$ is an element, and $i$ is the identity, then $i = 0$. To prove that $0$ is the identity means you should actually compute $0 \cdot a$ (for a general $a$) and show that the result is $a$.

This is a common misconception that people develop early on in their algebra classes. They learn how to work backwards -- how to take what they're trying to solve, and simplify it to derive a condition on the variables. But they forget that once they think they have the answer, they need to turn around and work forwards to show that it really is the answer.

Now, in this particular case, you have two things going for you, that let you use the work you already did to produce a different proof:

  • You can assume that there is an identity, because the problem pretty much tells you that
  • You can plug in any value you like for $a$

and once you've done that, the hypotheses of the theorem you proved are satisfied, and you can conclude $0$ is the identity.

Now, apply either method to the second problem....

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So since you want $$(\overline{a},\overline{b})=(\overline{a}+\overline{e_1},\overline{8}\overline{b}\overline{e_2})$$

You must have $\overline{a}=\overline{a}+\overline{e_1}$, meaning $\overline{e_1}=\overline{0}$, and from the second entry we see that $\overline{b}=\overline{b}(\overline{8}\overline{e_2})$; since this holds for any $\overline{b}$ we must have $\overline{8}\overline{e_2}=1$, and as you said $\overline{8}$ has a multiplicative inverse, so you have that $\overline{e_2}=\overline{8}^{-1}(1)=\overline{8}^{-1}=\overline{8}$ since $8\cdot 8=64=1$ modulo $9$. So your answer is $\overline{e_1}=\overline{0}$ and $\overline{e_2}=\overline{8}$.

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Add hats everywhere. –  Daniel Montealegre Mar 9 '12 at 20:00

In the first one, you should justify the "so $\overline{i}=0$". The actual argument is that $2\overline{a}+1$ is necessarily invertible in $\mathbb{Z}_8$ (under the usual multiplication), and so the fact that $8$ divides $\overline{i}(1+2\overline{a})$ implies that $8$ divides $\overline{i}$, hence $\overline{i}=\overline{0}$.

You should also verify explicitly that $\overline{0}$ works as an identity.

In the second, you can multiply by $\overline{8}^{-1} = \overline{-1}$; but you cannot multiply by $\overline{b}^{-1}$ unless you happen to know ahead of time that $\overline{b}$ is invertibel in $\mathbb{Z}_9$. (What if $b=3$?)

Instead, pick a specific $\overline{b}$... say, $b=1$. Then you must have $8\overline{e}_2 = \overline{1}$. So that means that $\overline{e}_2$ is actually....

And then show that this conclusion actually works as an identity.

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Hint $\ $ Identity elements $\rm\:e\:$ are idempotent $\rm\:e^2 = e\:$. Therefore

$\rm(1)\ \ mod\ 8\!:\ \ e = e\cdot e = 2e+2e^2\ \Rightarrow\ e\:(1+2e) = 0\ \Rightarrow\ e = \ldots\:$ by $\rm n^2 \equiv 1\:$ for $\rm\:n\:$ odd

$\rm(2)\ \ mod\ (9,9)\!:\ \ (a,b) = (a,b)^2 = (2a,-bb)\ \Rightarrow\ (-a, b\:(b+1)) = (0,0)\ \Rightarrow\ \ldots$

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