Show that $A=$ $ \{$$(3x, y) | x, y\in Z$$\}$ is a maximal ideal of $\mathbb{Z}\oplus \mathbb{Z}$. What happens if $3x$ is replaced by $4x$? Generalize.
How can I do this problem?
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$\begingroup$ Are you familiar with the theorem that says an ideal is maximal if and only if the quotient of the original space and the ideal is a field? $\endgroup$– ClaytonApr 6, 2013 at 15:50
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2$\begingroup$ Or, you can also proceed from first principle definitions! Please think about this, instead of our helping you to even get started. For instance, write down the definition and state clearly what it is that you want to prove. $\endgroup$– knsamApr 6, 2013 at 15:52
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2 Answers
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Hint: Show $(\Bbb Z\bigoplus \Bbb Z)/(3\Bbb Z\bigoplus \Bbb Z)\cong \Bbb Z/3\Bbb Z$, which is a field, hence $3\Bbb Z\bigoplus \Bbb Z$ is a maximal ideal. In this way, we see any ideal of the form $p\Bbb Z\bigoplus Z$ is a maximal ideal since $\Bbb Z/p\Bbb Z$ is a field where $p$ is a prime.
With $3$ replaced by $4$, we know $\Bbb Z/4\Bbb Z$ isn't an integral domain. In particular, it isn't a field, therefore the ideal isn't maximal. Here, we're only using the property that $4$ is composite, not that it is in the form $p^2$.
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Hint $\rm\ \ r \not\in A = \{(3x,y)\}\,\Rightarrow\: r = (3x,y) \pm(1,1)\: \Rightarrow\: (1,1)\in (r)+A\:\Rightarrow\: A\,$ maximal.
Alternatively, use $\rm\ A\ max\!\iff\! R/A\:$ is a field, and, for insight, check how it relates to above.
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$\begingroup$ And, of course, it is clear that $\rm\:A\neq \Bbb Z\oplus \Bbb Z\:$ since $\rm\:(\color{#C00}1,1)\in A\!\iff\! \color{#C00}1/3\in \Bbb Z.\ \ $ $\endgroup$ Apr 6, 2013 at 17:31