# characterization of prime ideals

We start with a general ring $R$.

###### Theorem 1.

Let $R$ be a ring and $P\subsetneq R$ a two-sided ideal  . Then the following statements are equivalent:

1. 1.

Given (left, right or two-sided) ideals $I,J$ of $P$ such that the product of ideals $IJ\subseteq P$, then $I\subseteq P$ or $J\subseteq P$.

2. 2.

If $x,y\in R$ such that $xRy\subseteq P$, then $x\in P$ or $y\in P$.

###### Proof.
• 1$\Rightarrow$2”:

Let $x,y\in R$ such that $xRy\subseteq P$. Let $(x)$ and $(y)$ be the (left, right or two-sided) ideals generated by $x$ and $y$, respectively. Then each element of the product of ideals $(x)R(y)$ can be expanded to a finite sum of products    each of which contains or is a factor of the form $\pm xry$ for a suitable $r\in R$. Since $P$ is an ideal and $xRy\subseteq P$, it follows that $(x)R(y)\subseteq P$. Assuming statement 1, we have $(x)\subseteq P$, $R\subseteq P$ or $(y)\subseteq P$. But $P\subsetneq R$, so we have $(x)\subseteq P$ or $(y)\subseteq P$ and hence $x\in P$ or $y\in P$.

• 2$\Rightarrow$1”:

Let $I,J$ be (left, right or two-sided) ideals, such that the product of ideals $IJ\subseteq P$. Now $RJ\subseteq J$ or $IR\subseteq I$ (depending on what type of ideal we consider), so $IRJ\subseteq IJ\subseteq P$. If $I\subseteq P$, nothing remains to be shown. Otherwise, let $i\in I\setminus P$, then $iRj\subseteq P$ for all $j\in J$. Since $i\notin P$ we have by statement 2 that $j\in P$ for all $j\in J$, hence $J\subseteq P$.

###### Theorem 2.

Let $R$ a commutative ring and $P\subsetneq R$ an ideal. Then the following statements are equivalent:

1. 1.

Given ideals $I,J$ of $P$ such that the product of ideals $IJ\subseteq P$, then $I\subseteq P$ or $J\subseteq P$.

2. 2.
3. 3.

The set $R\setminus P$ is a subsemigroup of the multiplicative semigroup of $R$.

4. 4.

Given $x,y\in R$ such that $xy\in P$, then $x\in P$ or $y\in P$.

5. 5.

The ideal $P$ is maximal in the set of such ideals of $R$ which do not intersect a subsemigroup $S$ of the multiplicative semigroup of $R$.

###### Proof.

If $R$ has an identity element  $1$, statements 2 and 3 of the preceding theorem become stronger:

###### Theorem 3.

Let $R$ be a commutative ring with identity element $1$. Then an ideal $P$ of $R$ is a prime ideal   if and only if $R/P$ is an integral domain  . Furthermore, $P$ is prime if and only if $R\setminus P$ is a monoid with identity element $1$ with respect to the multiplication in $R$.

###### Proof.

Let $P$ be prime, then $1\notin P$ since otherwise $P$ would be equal to $R$. Now by theorem 2 $R/P$ is a cancellation ring. The canonical projection $\pi\colon R\to R/P$ is a homomorphism, so $\pi(1)$ is the identity element of $R/P$. This in turn implies that the semigroup $R\setminus P$ is a monoid with identity element $1$. ∎

Title characterization of prime ideals CharacterizationOfPrimeIdeals 2013-03-22 15:22:01 2013-03-22 15:22:01 GrafZahl (9234) GrafZahl (9234) 9 GrafZahl (9234) Result msc 13C05 msc 16D25 characterisation of prime ideals Localization  QuotientRingModuloPrimeIdeal