# The Unapologetic Mathematician

## 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 […]

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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 […]

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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 […]

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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 […]

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