Show that a commutative ring with unity having no proper ideals is a field

Question

Show that a commutative ring with unity having no proper ideals is a field.

No proper ideals means that $\{0\},R$ are the only ideals of $R$. But then $\{0\}$ is a maximal ideal and thus by applying the first isomorphism theorem we get that $$R/\{0\} \cong R$$ and since $\{0\}$ is maximal we also have that $R/\{0\}$ is a field. Thus $R$ is a field.

Is this reasoning correct?

It is. But then the question becomes; why is the quotient of a ring by a maximal ideal a field? — Servaes, Jan 01 '17 at 16:22
@Servaes This we had in lecture, so it is legitimate to use this fact. — TheGeekGreek, Jan 01 '17 at 16:23

score 10 · Accepted Answer · answered Jan 01 '17 at 16:23

10

Yes, that looks fine, although it can be done directly by inspecting the proof of the theorem you used.

Notice that if we pick $a\neq 0$ then the ideal generated by $a$, which is equal to $Ra$ is all of $R$, therefore $1$ is contained in $Ra$.

So there is an element $r\in R$ such that $ra=1$. so $a$ has an inverse.

answered Jan 01 '17 at 16:23

Asinomás

105,651

3

Nice alternate proof! Thanks. +1 – TheGeekGreek Jan 01 '17 at 16:27

Show that a commutative ring with unity having no proper ideals is a field

1 Answers1