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 $P\mathrm{\u228a}R$ a twosided ideal^{}. Then the following statements are equivalent:

1.
Given (left, right or twosided) ideals $I,J$ of $P$ such that the product of ideals $IJ\subseteq P$, then $I\subseteq P$ or $J\subseteq P$.

2.
If $x,y\in R$ such that $xRy\subseteq P$, then $x\in P$ or $y\in P$.
Proof.

•
Let $x,y\in R$ such that $xRy\subseteq P$. Let $(x)$ and $(y)$ be the (left, right or twosided) 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 $P\u228aR$, so we have $(x)\subseteq P$ or $(y)\subseteq P$ and hence $x\in P$ or $y\in P$.

•
Let $I,J$ be (left, right or twosided) 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 $P\mathrm{\u228a}R$ an ideal. Then the following statements are equivalent:

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.
The quotient ring^{} $R/P$ is a cancellation ring.

3.
The set $R\setminus P$ is a subsemigroup of the multiplicative semigroup of $R$.

4.
Given $x,y\in R$ such that $xy\in P$, then $x\in P$ or $y\in P$.
 5.
Proof.

•
Let $\overline{x},\overline{y}\in R/P$ be arbitrary nonzero elements. Let $x$ and $y$ be representatives of $\overline{x}$ and $\overline{y}$, 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 $\overline{x}\overline{y}\ne 0$.

•
Let $x,y\in R\setminus P$. Let $\pi :R\to R/P$ 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, $\pi (x)\pi (y)=\pi (xy)\ne 0$. Therefore $xy\notin P$, that is $R\setminus P$ is closed under multiplication. The associative property is inherited from $R$.

•
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 $P$.

•
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 $J\subseteq P$.

•
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 $S$.

•
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
$$(P,x)\cap S\ne \{\},(P,y)\cap S\ne \{\}.$$ So we can choose the elements ${s}_{1}$ and ${s}_{2}$ of $S$ such that
$${s}_{1}={p}_{1}+{r}_{1}x+{n}_{1}x,{s}_{2}={p}_{2}+{r}_{2}y+{n}_{2}y,$$ 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 $\mathrm{1}$. Then an ideal $P$ of $R$ is a prime ideal^{} if and only if $R\mathrm{/}P$ is an integral domain^{}. Furthermore, $P$ is prime if and only if $R\mathrm{\setminus}P$ is a monoid with identity element $\mathrm{1}$ 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 $\pi :R\to R/P$ 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 

Canonical name  CharacterizationOfPrimeIdeals 
Date of creation  20130322 15:22:01 
Last modified on  20130322 15:22:01 
Owner  GrafZahl (9234) 
Last modified by  GrafZahl (9234) 
Numerical id  9 
Author  GrafZahl (9234) 
Entry type  Result 
Classification  msc 13C05 
Classification  msc 16D25 
Synonym  characterisation of prime ideals 
Related topic  Localization^{} 
Related topic  QuotientRingModuloPrimeIdeal 