# characterization of prime ideals

This entry gives a number of equivalent http://planetmath.org/node/5865characterizations of prime ideals in rings of different generality.

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 $J\subseteq P$.

There are some additional properties if our ring is commutative.

###### 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.

$R/P$ is a cancellation ring.

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.
• 1$\Rightarrow$2”:

Let $\bar{x},\bar{y}\in R/P$ be arbitrary nonzero elements. Let $x$ and $y$ be representatives of $\bar{x}$ and $\bar{y}$, respectively, then $x\notin P$ and $y\notin P$. Since $R$ is commutative, each element of the product of ideals $(x)(y)$ can be written as a product involving the factor $xy$. Since $P$ is an ideal, we would have $(x)(y)\subseteq P$ if $xy\in P$ which by statement 1 would imply $(x)\subseteq P$ or $(y)\subseteq P$ in contradiction with $x\notin P$ and $y\notin P$. Hence, $xy\notin P$ and thus $\bar{x}\bar{y}\neq 0$.

• 2$\Rightarrow$3”:

Let $x,y\in R\setminus P$. Let $\pi\colon R\to R/P$ be the canonical projection. Then $\pi(x)$ and $\pi(y)$ are nonzero elements of $R/P$. Since $\pi$ is a homomorphism and due to statement 2, $\pi(x)\pi(y)=\pi(xy)\neq 0$. Therefore $xy\notin P$, that is $R\setminus P$ is closed under multiplication. The associative property is inherited from $R$.

• 3$\Rightarrow$4”:

Let $x,y\in R$ such that $xy\in P$. If both $x,y$ were not elements of $P$, then by statement 3 $xy$ would not be an element of $P$. Therefore at least one of $x,y$ is an element of $P$.

• 4$\Rightarrow$1”:

Let $I,J$ be ideals of $R$ such that $IJ\subseteq P$. If $I\subseteq P$, nothing remains to be shown. Otherwise, let $i\in I\setminus P$. Then for all $j\in J$ the product $ij\in IJ$, hence $ij\in P$. It follows by statement 4 that $j\in P$, and therefore $J\subseteq P$.

• 4$\Rightarrow$5”:

The condition 4 that the set  $S=R\setminus P$  is a multiplicative semigroup.  Now $P$ is trivially the greatest ideal which does not intersect $S$.

• 5$\Rightarrow$4”:

We presume that $P$ is maximal of the ideals of $R$ which do not intersect a semigroup $S$ and that  $xy\in P$.  Assume the contrary of the assertion, i.e. that  $x\notin P$  and  $y\notin P$.  Therefore, $P$ is a proper subset of both  $(P,\,x)$  and  $(P,\,y)$.  Thus the maximality of $P$ implies that

 $(P,\,x)\cap S\neq\{\},\quad(P,\,y)\cap S\neq\{\}.$

So we can choose the elements $s_{1}$ and $s_{2}$ of $S$ such that

 $s_{1}=p_{1}+r_{1}x+n_{1}x,\quad s_{2}=p_{2}+r_{2}y+n_{2}y,$

where  $p_{1},\,p_{2}\in P$,   $r_{1},\,r_{2}\in R$  and  $n_{1},\,n_{2}\in\mathbb{Z}$.  Then we see that the product

 $s_{1}s_{2}=(p_{1}+r_{2}y+n_{2}y)p_{1}+(r_{1}x+n_{1}x)p_{2}+(r_{1}r_{2}+n_{2}r_% {1}+n_{1}r_{2})xy+(n_{1}n_{2})xy$

would belong to the ideal $P$.  But this is impossible because $s_{1}s_{2}$ is an element of the multiplicative semigroup $S$ and $P$ does not intersect $S$.  Thus we can conclude that either $x$ or $y$ belongs to the ideal $P$.

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