## 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 is “prime” if the only numbers that divide it are and itself. So, we might make the guess that a prime ideal is one so that the only ideals containing it are 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 is prime if and only if whenever then either or . Let’s turn this into ideals. We’re defining a property of an ideal in terms of two other ideals and . In the case of integers, these are the principal ideals and since all ideals in are principal. The product of two integers generates the ideal — the product of the two ideals, so we’ll also consider the product ideal . Now we can state our property: an ideal is prime if whenever then either or . We also insist that is not the whole ring, just as we insist that 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 of a commutative ring by a prime ideal , and elements and in this quotient ring. If their product then so . Now we can show that , so either or since is prime. In particular or , so or . That is, if the product of two elements in is zero, then one or the other must be — is an integral domain!

What happens if we use a maximal ideal in this construction? Given any element in , we have an element . If we try to make an ideal containing all of and also , then we get the whole ring . In particular we get for some . Then in , so is an inverse of — 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 that’s not a field. Then the ideal is prime, since is an integral domain, but it’s not maximal since 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: . Then the second copy is an ideal in the direct sum, and is an integral domain but not a field. Thus is a prime ideal, but not a maximal one.

[...] contains the unit . A good place to get such multiplicatively closed sets is as complements of prime ideals — given two elements and in but not in the prime ideal , their product must also be [...]

Pingback by Fractions « The Unapologetic Mathematician | May 19, 2007 |

[...] 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 [...]

Pingback by Set — the card game « The Unapologetic Mathematician | July 17, 2008 |

[...] 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 [...]

Pingback by The Complex Numbers « The Unapologetic Mathematician | August 7, 2008 |

[...] 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 [...]

Pingback by Local Rings « The Unapologetic Mathematician | March 28, 2011 |

[…] 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 […]

Pingback by Locally ringed spaces insights « tomkojar | July 20, 2014 |