# The Unapologetic Mathematician

## Prime Ideals

Now we know that we can talk about divisibility in terms of ideals, we remember a definition from back in elementary school: a number $p$ is “prime” if the only numbers that divide it are $1$ and $p$ itself. So, we might make the guess that a prime ideal $P$ is one so that the only ideals containing it are $P$ itself and the whole ring. Unfortunately, that’s not quite right.

There’s actually a different definition of a prime number, and it just so happens for numbers that the two definitions describe (almost) the same numbers. In more general rings, however, they’re different. What we’ve just described we’ll call a “maximal” ideal, since you can’t make it any bigger without getting the whole ring.

Here’s the other definition of a prime number: a number $p$ is prime if and only if whenever $p|ab$ then either $p|a$ or $p|b$. Let’s turn this into ideals. We’re defining a property of an ideal $P$ in terms of two other ideals $A$ and $B$. In the case of integers, these are the principal ideals $(a)$ and $(b)$ since all ideals in $\mathbb{Z}$ are principal. The product of two integers generates the ideal $(ab)=(a)(b)$ — the product of the two ideals, so we’ll also consider the product ideal $AB$. Now we can state our property: an ideal $P$ is prime if whenever $AB\subseteq P$ then either $A\subseteq P$ or $B\subseteq P$. We also insist that $P$ is not the whole ring, just as we insist that $1$ is not a prime number.

Prime ideals have a number of nice properties, especially when we’re just looking at commutative rings with units. For instance, let’s consider the quotient $R/P$ of a commutative ring $R$ by a prime ideal $P$, and elements $a+P$ and $b+P$ in this quotient ring. If their product $ab+P=0$ then $ab\in P$ so $(ab)\subseteq P$. Now we can show that $(a)(b)\subseteq(ab)\subseteq P$, so either $(a)\subseteq P$ or $(b)\subseteq P$ since $P$ is prime. In particular $a\in P$ or $b\in P$, so $a+P=0$ or $b+P=0$. That is, if the product of two elements in $R/P$ is zero, then one or the other must be — $R/P$ is an integral domain!

What happens if we use a maximal ideal $M$ in this construction? Given any element $a+M\neq0$ in $R/M$, we have an element $a\notin M$. If we try to make an ideal containing all of $M$ and also $a$, then we get the whole ring $R$. In particular we get $1=xa+ym$ for some $m\in M$. Then $(x+M)(a+M)=xa+M=(1-ym)+M=1+M$ in $R/M$, so $x+M$ is an inverse of $a+M$$R/M$ is a field!

Now we can be sure that there are rings with prime ideals that are not maximal, as indicated above. Take any integral domain $D$ that’s not a field. Then the ideal $\mathbf{0}$ is prime, since $D/\mathbf{0}\cong D$ is an integral domain, but it’s not maximal since $D$ isn’t a field. Of course I hear you cry out, “but maybe the only difference is ever the zero ideal!” Well, just take the direct sum of two copies of the ring: $D_1\oplus D_2$. Then the second copy is an ideal in the direct sum, and $(D_1\oplus D_2)/D_2\cong D_1$ is an integral domain but not a field. Thus $D_2$ is a prime ideal, but not a maximal one.

May 18, 2007 - Posted by | Ring theory

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2. […] left with either or , and this gives us all the integers in our expanded ideal. But we know that the quotient of a ring by a maximal ideal is a field! Thus adding and multiplying integers modulo gives us a field. For the three elements of […]

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3. […] quotient ring . But what we just discussed above further goes to show that is a maximal ideal, and the quotient of a ring by a maximal ideal is a field! Thus when we take the real numbers and adjoin a square root of to get a ring we might call , the […]

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4. […] the concept of a “local ring”. This is a commutative ring which contains a unique maximal ideal. Equivalently, it’s one in which the sum of any two noninvertible elements is again […]

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5. […] well define the concept of a “local ring”. This is a commutative ring which contains a unique maximal ideal. Equivalently, it’s one in which the sum of any two noninvertible elements is again […]

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