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A ring is a field iff the only ideals are $(0)$ and $(1)$

Michael Artin's Algebra

in the introduction of maximal fields, there was a sentence stated that fields are characterized by the property of having exactly 2 ideals. what does mean? can someone elaborate on this ? I don't actually remember there are such proofs.

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marked as duplicate by Martin Sleziak, Chris Eagle, tomasz, Rudy the Reindeer, rschwieb Oct 6 '12 at 13:16

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Closed as a duplicate, but the question referred to is less correct than this one. Because in the zero ring $(0)$ and $(1)$ are (or rather "is") the only ideals, but it is not a field. –  Marc van Leeuwen Oct 6 '12 at 14:22

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up vote 2 down vote accepted

A commutative ring $R$ always has at least one ideal: $R$ and $\{0\}$ are examples (but they might not be two examples).

If $R$ has exactly one ideal, one apparently has $R=\{0\}$, and $R$ is not a field (because one of the field axioms is $0\neq1$, which is precisely listed to exclude this ring; all other field axioms are satisfied).

If $R$ has exactly two ideals, then firstly $R\neq\{0\}$, so these ideals are precisely $R$ and $\{0\}$. Then for any $a\in R\setminus\{0\}$, the ideal $aR$ it generates is not $\{0\}$ so it must be $R$. In particular $1\in aR$ and $a$ is invertible; this holds for any nonzero $a$ so $R$ is a field.

If $R$ has an ideal $I$ with $I\neq\{0\}$ and $I\neq R$, then $1\notin I$ (otherwise $a=1.a\in I$ for all $a\in R$ contradicting $I\neq R$), and choosing $a\in I\setminus\{0\}$ we have $1\notin aR\subseteq I$ so $a$ is non-invertible and $R$ is not a field.

If $R$ is not commutative, it depends on what "ideal" means. Taking it to mean "right ideal" the above remains valid, and one can conclude that the only rings with exacly two right ideals are the skew fields (division rings). But if one takes (more reasonably) "ideal" to mean "two-sided ideal", then the statement becomes false. There are non-commutative rings with exactly two two-sided ideals that are not division rings: a matrix ring over a field is an example. So one may assume that the context of your quotation is such that "characterized" is to be taken "among commutative rings".

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this sounds more detailed explaination. –  zinking Oct 7 '12 at 5:42

Let $F$ be a field, and let $I$ be an ideal other than the $0$-ideal. If $a\in I$ with $a\ne 0$, then $a^{-1}a$ is in $I$, so $1$ is in $I$. It follows that $x=(x)(1)$ is in $I$ for every $x$, and therefore $I=F$. Thus a field has precisely two ideals.

For a converse, let $A$ be a commutative ring with unit, with exactly two ideals. We show that $A$ is a field.

Let $w$ be a non-zero element of $A$. It is easy to verify that the set $I$ of all ring elements of the form $xw$, where $x$ ranges over $A$, is an ideal. Since $A$ has only two ideals, we have $I=A$. In particular, $1\in I$, meaning that $w$ is invertible.

Thus if $A$ has only two ideals, every non-zero element of $A$ is invertible, that is, $A$ is a field.

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The only ideals in a field ar $0$ and $F$ itself.

If on the other hand $A$ is a commutative ring with 1 and has no ideals apart from $0$ and $A$, then $0$ is a maximal ideal, hence $A/0$ is a field. But obviously $A\cong A/0$.

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