The Unapologetic Mathematician

Mathematics for the interested outsider

Generators of ideals

Let’s say we’ve got a ring R and an element x\in R. What is the smallest left ideal that contains x? Well, we have to have all multiples rx for r\in R so it’s closed under left multiplication. If R has a unit, this is all we need. Otherwise, we have to make sure we include all the elements nx=x+...+x with n summands (and their negatives) to make sure it’s an abelian subgroup. Thus the subset \{rx+nx|r\in R,n\in\mathbb{Z}\} is a left ideal in R. If R has a unit, we just need the subset \{rx|R\in R\}. We call this the principal left ideal generated by x, and write Rx. We can do something similar for right ideals (xR), and for two-sided ideals we get the subset (x)=\{r_1xr_2+nx|r_1,r_2\in R,n\in\mathbb{Z}\}.

As for any submodules we can form the sum. If we have elements x_1,...,x_n\in R they generate the left ideal Rx_1+...+Rx_n, or a similar right ideal. For two-sided ideals we write (x_1,...,x_n). The term “principal”, however, is reserved for ideals generated by a single element.

Let’s look at these constructions in the ring of integers. Since it’s commutative, every ideal is two-sided. An integer n then generates the principal ideal (n)=\{kn\} of all multiples of n. In fact, every ideal in \mathbb{Z} is principal.

If I\subseteq\mathbb{Z} is an ideal, consider the subset of all its (strictly) positive elements. Since this is a subset of the natural numbers it has a least element n. I say that every element of I is a multiple of n. If not, then there is some m\geq n that n doesn’t divide. If we can apply Euclid’s algorithm to m and n, at the first step we get m=qn+r with r\leq n. The greatest common divisor d of m and n will thus be less than n, and Euclid’s algorithm gives us a linear combination d=xm+yn for integers m and n. Thus d\leq n must be in the ideal as well, contradicting the minimality of n.

So every ideal of \mathbb{Z} is principal. When this happens for a ring, we call it a “principal ideal ring”, or a “principal ideal domain” if the ring is also an integral domain.

So how do ideals of integers behave under addition and multiplication? The ideal (m)+(n) is the ideal (m,n). This it consists of all the linear combinations xm+yn. In particular, the smallest positive such linear combination is the greatest common divisor of m and n, as given by Euclid’s algorithm. The product of the ideals (m)(n) is the set of all products of multiples of m and n: k_1mk_2n=k_1k_2mn=kmn. Thus (m)(n)=(mn).

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May 13, 2007 - Posted by | Ring theory

4 Comments »

  1. [...] as well. But this is telling us that the kernel of the evaluation homomorphism for contains the principal ideal [...]

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

  2. [...] Well, if we have noninvertible elements and with invertible, then these elements generate principal ideals and . If we add these two ideals, we must get the whole ring, for the sum contains , and so must [...]

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

  3. [...] modulo . This is pretty straightforward to understand for integers, but it works as stated over any principal ideal domain — like — and, suitably generalized, over any commutative [...]

    Pingback by The Jordan-Chevalley Decomposition (proof) « The Unapologetic Mathematician | August 28, 2012 | Reply

  4. […] Well, if we have noninvertible elements and with invertible, then these elements generate principal ideals and . If we add these two ideals, we must get the whole ring, for the sum contains , and so must […]

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


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