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Could someone prove that in a commutative ring $R$ the following two definitions of prime ideal are equivalent:

1) An ideal $P$ is prime if $P\neq R$ and if for all ideals $A,B$ of $R$ with $AB\subseteq P$ we have that $A \subseteq P$ or $B \subseteq P$.

2) An ideal $P$ is prime if $P\neq R$ and if $ab \in P \implies a \in P$ or $b\in P$.

I have read (in Artin as well as online) that these are equivalent, but I cannot find a suitable proof.

Artin's proof of $1) \implies 2)$ is one line: Take $A=(a)$ and $B=(b)$. I do not see how this is a proof. If someone could explain the implication as if I were a complete novice to ring theory (which I am), that would be most helpful.

Note: I am not assuming that $R$ is unital, only that it is commutative.

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In (2), you would also need to include the condition $P\neq R$, wouldn't you? –  Arturo Magidin Oct 17 '11 at 1:44
    
Which Artin? The one I have (Algebra, Prentice-Hall 1991) explicitly defines a ring to be commutative and unital. –  Henning Makholm Oct 17 '11 at 1:56
    
I am using the same version of Artin. My current professor does not require a ring to be unital, but I still want to use this result. Sorry I phrased it poorly. I meant that the result is true even if the ring is not unital, but I cannot find a clear proof where the author does not assume the ring is unital. –  nullUser Oct 17 '11 at 2:12
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up vote 7 down vote accepted

(2)$\implies$(1) always holds: assume that $AB\subseteq P$, and $A$ is not contained in $P$. Then there exists $a\in A$ such that $a\notin P$. But for every $b\in B$, $ab\in AB\subseteq P$, hence by (2) we must have $b\in P$. Thus, $B\subseteq P$. Hence, (2) holds. Note that this holds without the need to assume commutativity of existence of a unity. Condition (2) is called "totally prime" or "completely prime", and is stronger than primality in noncommutative rings.

The problem with (1)$\implies$(2) is just the question of your definition of $(a)$. For commutative rings with unity, $(a)=Ra=\{ra\mid r\in R\}$. However, for commutative rings without unity, since $(a)$ is the smallest ideal that contains $a$, then it is equal to $$(a) = \{ra + na\mid r\in R,n\in\mathbb{Z}\},$$ where $na$ is defined inductively: $0a=0_R$, $(n+1)a = (na)+a$, and $(-n)a = -(na)$. In particular, $a\in (a)$ always holds.

(1)$\implies$(2). Assume that $ab\in P$. Then $(ab)\subseteq P$, hence $(a)(b)=(ab)\subseteq P$. Since $P$ satisfies (1), then either $(a)\subseteq P$ or $(b)\subseteq P$. If $(a)\subseteq P$, then $a\in (a)\subseteq P$ so $a\in P$. If $(b)\subseteq P$, then $b\in(b)\subseteq P$, so $b\in P$. This proves (2).

But perhaps you were thinking that $(a)=Ra$ and so were confused about why this argument would work, given that if $R$ has no unity there is no reason to expect that $a\in Ra$? (it might be there anyway even if $R$ has no unity; e.g., in the direct sum of infinitely many copies of $\mathbb{Z}$ there is no unity, but for every $a$ there is an $x$ such that $ax=a$). The argument as is doesn't work, but the implication still holds: just consider $A$ to be the smallest ideal of $R$ that contains $a$, $B$ to be the smallest ideal of $R$ that contains $b$, notice that $ab\in AB$, and then prove, as I do below, that $AB$ is in fact the smallest ideal that contains $ab$.

The condition that $(a)(b)=(ab)$ is established by noting that every element of $(a)(b)$ is a sum of elements of the form $xy$, with $x\in (a)$ and $y\in (b)$. Each of this is of the form $$(ra+na)(sb+mb) = (rs)ab + (na)sb + ra(mb) + (na)(mb) = (rs + ns + mr)ab + (nm)ab\in (ab).$$ So $(a)(b)\subseteq (ab)$. Since $ab\in (a)(b)$, then $(ab)\subseteq (a)(b)$, which gives the equality.


For not-necessarily-commutative rings, an ideal satisfying condition (2) is known, as I mentioned, as "completely prime" or "totally prime". Completely prime always implies prime, but the converse does not necessarily hold. For an example, take the ring of $n\times n$ matrices with coefficients in a field $F$, $n\gt 1$; the only ideals of this ring are $(\mathbf{0})$ and $R$, so in particular $(\mathbf{0})$ is prime. But it is not completely prime, since for example the matrix $E_{1n}$ ($1$ in the $(1,n)$ coordinate, zeros elsewhere) satisfies $E_{1n}^2 = \mathbf{0}\in (\mathbf{0})$, but $E_{1n}\notin(\mathbf{0})$.

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