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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?

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    It is. But then the question becomes; why is the quotient of a ring by a maximal ideal a field? – 2017-01-01
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    @Servaes This we had in lecture, so it is legitimate to use this fact. – 2017-01-01
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    Two (near) duplicates: [1](http://math.stackexchange.com/questions/101157/a-ring-is-a-field-iff-the-only-ideals-are-0-and-1), [2](http://math.stackexchange.com/questions/1795054/prove-that-a-commutative-ring-without-proper-ideals-is-a-field) – 2017-01-01

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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.

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    Nice alternate proof! Thanks. +1 – 2017-01-01