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[parent] characterization of prime ideals (Result)

This entry gives a number of equivalent characterizations 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. 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\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$.

$ \qedsymbol$

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. 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 $ \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 means 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

    $\displaystyle (P,\,x)\cap S \neq\{\}, \quad (P,\,y)\cap S \neq\{\}.$
    So we can choose the elements $ s_1$ and $ s_2$ of $ S$ such that
    $\displaystyle s_1 = p_1+r_1x+n_1x, \quad s_2 = p_2+r_2y+n_2y,$
    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
    $\displaystyle s_1s_2 = (p_1+r_2y+n_2y)p_1+(r_1x+n_1x)p_2+(r_1r_2+n_2r_1+n_1r_2)xy+(n_1n_2)xy$
    would belong to the ideal $ P$. But this is impossible because $ s_1s_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$.
$ \qedsymbol$

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$. $ \qedsymbol$



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See Also: localization, quotient ring modulo prime ideal

Other names:  characterisation of prime ideals
Keywords:  prime, ideal, monoid, semigroup

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Cross-references: monoid, prime, integral domain, identity element, proper subset, associative, multiplication, closed under, canonical projection, contradiction, imply, intersect, semigroup, multiplicative, subsemigroup, cancellation ring, quotient ring, commutative ring, commutative, properties, factor, contains, products, sum, finite, expanded, ideal generated bies, product of ideals, ideals, right, two-sided ideal, rings, prime ideals, equivalent, number
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This is version 6 of characterization of prime ideals, born on 2005-06-26, modified 2005-07-01.
Object id is 7192, canonical name is CharacterisationOfPrimeIdeals.
Accessed 1970 times total.

Classification:
AMS MSC: 16D25 (Associative rings and algebras :: Modules, bimodules and ideals :: Ideals)
 13C05 (Commutative rings and algebras :: Theory of modules and ideals :: Structure, classification theorems)
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