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My question comes from the top solution of A ring is a field iff the only ideals are $(0)$ and $(1)$. Here, at the end the solver states that $\langle a \rangle$ is automatically an ideal of $A$. Why is that?

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Jeze Ken
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    This is one of many things in higher mathematics that is completely straightforward provided you have fully grasped the definitions involved. If you know what an ideal is and what $\langle a \rangle$ means, you have some routine checking to do. Which part, if any, is troubling you? – Pete L. Clark Sep 04 '15 at 18:34
  • An ideal $I$ of $A$ is an additive subgroup of $A$ such that $rI \subseteq I$ for all $r \in A$. I get why $\langle a \rangle$ is an additive subgroup of $A$ but I don't understand how the solver knows it also satisfies $r\langle a \rangle \subseteq \langle a \rangle$. – Jeze Ken Sep 04 '15 at 21:10

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I just learned that $(a)$ represents the ideal generated by $a$ and not a subgroup. In other words, $$(a):=\{ra:r\in A\}.$$ After noting this, the solution makes sense.

Jeze Ken
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  • +1 for doing the research to answer your own question. Next time do the research first, save this site for more interesting things. – Ethan Bolker Sep 04 '15 at 21:28
  • I wanted to ask this question on the original solution, but I didn't have enough points to be able to leave comments. – Jeze Ken Sep 04 '15 at 21:53