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 $P\mathrm{⊊}R$ a two-sided ideal. Then the following statements are equivalent:

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. 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 $P⊊R$, 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 $P\mathrm{⊊}R$ 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. 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 $\overline{x},\overline{y}\in R/P$ 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 $\overline{x}\overline{y}\ne 0$.

• “2$\Rightarrow$3”:

  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 $R$.

• “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

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

  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	2013-03-22 15:22:01
Last modified on	2013-03-22 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