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I'm stuck on the following exercise from Herstein's "Topics in Algebra":

"show that ($n$ is prime) $\Leftrightarrow ([a][b]=[0]\Rightarrow [a]=[b]=[0]) $ in $J_n$".


for the rightward implication I have:

$[a][b]=[ab]=[0]\Rightarrow n|ab\Rightarrow n|a$ or $n|b$ (or both) $\Rightarrow [a]=[0]$ or $[b]=[0]$

and I'm stuck on the leftward implication.

Why do I get $[a]=[0]$ OR $[b]=[0]$ and not $[a]=[0]$ AND $[b]=[0]$ in the leftward implication? Where is it that I go wrong?

I'd also appreciate any comment/hint about how to prove the remaining left implication.

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    $\begingroup$ The problem statement is false: $2 \cdot 7 = 0$ in $J_7$ but $2$ is not $0$ in $J_7$. It should be one of the two being zero. $\endgroup$ – Student Mar 5 '17 at 13:14
  • $\begingroup$ @Student So that's why I got [a]=[0] or [b]=[0]; thank you for bringing up this counterexample. $\endgroup$ – lorenzo Mar 5 '17 at 13:20
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    $\begingroup$ It would be helpful to add the edition and section or page number from which this exercise is taken. $\endgroup$ – hardmath Mar 5 '17 at 13:33
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As I already commented, the problem statement should be:

$n$ is prime if and only if ($[a][b] = [0] \implies [a] = [0]$ or $[b] = [0]$).

For the leftward direction: Prove this by contraposition: this means that if we want to prove $P \implies Q$, then we do so by proving $\text{not } Q \implies \text{not }P$ (these are equivalent). Hence I will prove the following: If $n$ is not a prime, then we have that $[a][b] = [0]$ and $[a] \neq [0] \neq [b]$. (since the we have that $\text{not }(K \implies L)$ is equivalent with ($K$ and $\text{not } L$).

First of all, note that $n$ is prime if and only if its only divisors are $1, n$. Hence $n$ is not prime if it has at least one divisor (and then automatically also a second one) which is neither $1$ nor $n$!

Now for your proof:

Suppose $n$ is not prime, then there exists divisors $a,b$ of $n$ which are by definition different from 1 and different form $n$. Hence we can write $n = ab$. In particular, we have that since $ 1 < a <n$ and $1 < b <n$ that $[a] \neq [0]$ and $[b] \neq [0]$ in $J_n$. However, we have that $$[0] = [n] = [ab] = [a][b]$$ and this concludes the prove.

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  • $\begingroup$ by $c$ and $d$ do you mean $a$ and $b$? If so why should you be able to write $n=cd$? Shouldn't it be $ab=hn$ (where $h\in\mathbb{Z}$) since $[ab]=[0]$ (and thus $n|ab$) or am I getting something wrong about the meaning of congruence class? (see also my comment to the other answer) $\endgroup$ – lorenzo Mar 5 '17 at 13:59
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    $\begingroup$ @lorenzo yes: I denoted it by $c,d$ in order to prevent you of being confused, but this apparently backfired. I will edit my answer so that I can adress your other questions. $\endgroup$ – Student Mar 5 '17 at 14:02
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Let us prove the leftward implication. That is, we need to show that $$(i)\quad [a][b]=[0] \text{ implies either } [a]=[0] \text{ or } [b]=[0]\qquad\implies\qquad (ii)\quad n\text{ is prime}.$$

We prove this by contradiction. Assume that statement $(i)$ holds and suppose statement $(ii)$ is false. This means that $n$ is not prime, that is, $n$ is composite. Then we write $n=ab$, where $1<a<n$ and $1<b<n$. We know that $$n\equiv 0\pmod n.$$ Thus, $$ab\equiv 0\pmod n.$$ This implies that $[ab]=[0]$. Thus, $[a][b]=[0]$. By hypothesis, either $[a]=[0]$ or $[b]=[0]$. Hence, either $a\equiv 0\pmod n$ or $b\equiv 0\pmod n$. Hence, either $n|a$ or $n|b$ implying that either $n\leq a$ or $n\leq b$. We obtain a contradiction and this proves the result.

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    $\begingroup$ @lorenzo Look at my proof. I assumed that $n$ is composite. What is your definition of a composite? $\endgroup$ – Juniven Mar 5 '17 at 14:02

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