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Show $ \langle 5,7 \rangle $ and $ \langle 6,9 \rangle $ are principal ideals in $\Bbb Z$ .

I don't understand what the difference between a principal ideal and an ideal is?

I need some tips on this question.

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A principal ideal is one with a single generator. You may be familiar with the fact that $\mathbb{Z}$ is a principal ideal domain (because it is a Euclidean domain): every ideal in $\mathbb{Z}$ is principal.

However, invoking that theorem is probably against the spirit of the question. So you're going to need to find a single integer $n$ such that $\langle n \rangle = \langle 5,7\rangle$. Since $5 \in \langle n\rangle$, we must have $n \mid 5$; and since $7 \in \langle n \rangle$, we must have $n \mid 7$.

Now, once you've found your candidate $n$, show that it works: show that everything in $\langle 5,7\rangle$ is in $\langle n \rangle$, and that everything in $\langle n \rangle$ is in $\langle 5,7\rangle$.

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Hint: You need to show that there exists some number $a$ so that $$\langle5,7\rangle =\langle a\rangle.$$

Hint 2: Try $a=\gcd(5,7)$. To prove the equality of ideals use double inclusion.

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  • $\begingroup$ I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) $\endgroup$ – Patrick Stevens Dec 8 '16 at 20:54
  • $\begingroup$ @PatrickStevens Ty :) $\endgroup$ – N. S. Dec 8 '16 at 20:57
  • $\begingroup$ oh so you mean <1> right ? This question was easy after all ... $\endgroup$ – ggok Dec 8 '16 at 21:10
  • $\begingroup$ and for 6 9 we can use <3> or <1> ? $\endgroup$ – ggok Dec 8 '16 at 21:11
  • $\begingroup$ @ggok Is $\langle 6, 9 \rangle = \langle 1 \rangle$? Is it even the case that $\langle 6, 9 \rangle \supseteq \langle 1 \rangle$? $\endgroup$ – Patrick Stevens Dec 8 '16 at 21:35

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