Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

One way to prove that a field $K$ has no ideals except the entire field and the trivial ideal is to note the fact that every element $x$ has an inverse. By the definition of an ideal, if $x$ is in the ideal then $x^{-1}x$ is because $x^{-1} \in K$. But now we have that 1 is in the ideal, and so again by the definition of an ideal we have that every element is in the ideal. Therefore it is either the entire field or trivial.

However, this works for any would-be ideal that has a unit; hence my question. I don't see how this coheres particularly with the idea that ideals are generalizations of things like "multiple of $n$", or that we use them to form quotient rings.

Can someone please explain whether this has a deeper meaning or if it's not really important? I think it might have something to do with what is written in the "motivation" section in the Wikipedia article for ideals but I'm not really sure.

Edit: I do realize that not all subsets without a unit are ideals. Sorry for the confusion.

share|improve this question
    
Which idea are you referring to? –  Arturo Magidin Dec 28 '10 at 1:47
2  
I don't understand the problem. An ideal containing a unit is the full ring (equiv. it is not proper). Similarly, if $n$ is a unit, then every element of the ring is a multiple of $n$. –  George Lowther Dec 28 '10 at 1:48
    
there isn't really a "problem"; I was just wondering whether there is something more going on here. If you think this question is not well defined then I will delete it. –  Harry Stern Dec 28 '10 at 1:54
    
this is such an embarrassing question, looking back a year later. –  Harry Stern Jan 13 '12 at 4:31

3 Answers 3

up vote 6 down vote accepted

Yes, it's true that any ideal in a ring which contains a unit must be the whole ring. (Also: every left ideal that contains a left-invertible element must be the whole ring).

Now, there are two items here. One is the notion of generalizing "multiples of $n$", and the second is the notion of making quotients.

Let's start with the first. The notion that ideals generalize things like "all multiples of $n$" in $\mathbb{Z}$ goes back all the way to Dedekind, who in fact came up with the term "ideal" (it was a counterpart to Kummer's "ideal numbers"; at the risk of touting my own horn, check out the paper with David McKinnon, Gauss's Lemma for Number Fields, Amer. Math. Monthly 112 no. 5 (2005) pp. 385-416). Dedekind was looking at rings like $\mathbb{Z}[\sqrt{-5}]$, and he wanted something along the lines of unique factorization; Kummer had done this by introducing something called "ideal numbers", which were like the primes for integers but which did not actually occur in the ring in order to explain things like $2\times 3 = (1+\sqrt{-5})(1-\sqrt{-5})$, even though none of $2$, $3$, $1+\sqrt{-5}$ and $1-\sqrt{-5}$ can be written as a product of two numbers (from $\mathbb{Z}[\sqrt{-5}]$), neither of which is $1$ or $-1$. The idea was that there were these "ideal numbers" $\alpha$, $\beta$, and $\gamma$, which had the property that $\alpha\beta=1+\sqrt{-5}$, $\alpha\gamma=1-\sqrt{-5}$, $\alpha^2 = 2$, and $\beta\gamma=3$. Instead of "inventing" these numbers, Dedekind considered a collection of all numbers which would be "the multiples of $\alpha$" if $\alpha$ was actually an element of $\mathbb{Z}[\sqrt{-5}]$. A necessary and sufficient condition for a collection of elements of $\mathbb{Z}[\sqrt{-5}]$ to qualify as "all multiples of $x$" (where $x$ was either an ideal number of an actual number) was that the collection be nonempty, closed under sums and differences, and that if $a$ was in the collection of $r$ was in $\mathbb{Z}[\sqrt{-5}]$, then $ra$ was in the collection. They were "ideals" because they were playing the role of ideal numbers (here, "ideal" means existing in fancy or imagination). This generalized to any ring contained in a finite extension of $\mathbb{Q}$ and containing all integral elements (roots of monic polynomials with integer coefficients), called the "number field case"; later Dedekind and Weber developed a similar theory in what is called the "function field" case. Later, when rings were abstracted and generalized from the number field and function field cases by Artin and by Noether, the conditions were extended but they were still based in the notion that the ideals corresponded to "all multiples of" an ideal number in the number field.

Now, there is no problem with the fact that sometimes these collections are everything: this happens whenever the collection satisfies the conditions and contains a unit. Some collections do, some don't. In fact, that is one way in which one can test to see if an element is a unit: if the ideal it generates is the whole ring, then it is a unit (this works in the case of commutative rings with unit, not in more general rings). This actually makes sense because everything is a multiple of a unit.

For rings like $\mathbb{Z}$, and to some extent number rings as above, every ideal consists exactly of "all the multiples of $x$" for some $x$ (though in the number ring, the $x$ may be an "ideal number"). For more general (commutative) rings (with unit), if you have a family of elements $x_1,\ldots,x_n$, then the elements of the ideal generated by $x_1,\ldots,x_n$ are exactly those elements that are "$R$-linear combinations" of $x_1,\ldots,x_n$; that is, all elements that can be expressed as $r_1x_1+\cdots+r_nx_n$ for some $r_1,\ldots,r_n\in R$; this includes all multiples of each $x_i$, as well as other elements. If one of the $x_i$ is a unit, then since everything is a multiple of a unit, then everything is in the ideal in question. There are other kinds of ideals (not "finitely generated", as you start dealing with more and more complicated rings).

Now, on to the second question. As it happens, Dedekind also noted that the conditions on the set (nonempty, closed under sums and differences, and closed under products by elements of the ring) were exactly the condition one needed in order to be able to do "congruence modulo" the multiples of $x$, so he also called them "modules" (because you could do modular arithmetic with them). This is essentially the same as talking about quotients.

But one can get to ideals differently if you are interested in quotients. This is essentially the same thing I did in this answer about normal subgroups. Suppose you want to define an equivalence relation $\sim$ on the ring $R$ so that you can make $R/\sim$ into a ring using the operations $[a]+[b]=[a+b]$ and $[a][b]=[ab]$ (where $[a]$ is the equivalence class of $a$ under $\sim$). It turns out that the only way this can work is if $\sim$ is a subring of $R\times R$ (when we consider $\sim$ as a collection of ordered pairs of elements of $R$, hence a subset of $R\times R$), and that if you let $I$ be the collection of all elements of $R$ that are equivalent to $0$, then $I$ is an ideal of $R$ and the equivalence relation $\sim$ is $a\sim b\Longleftrightarrow a-b\in I$. You can try proving the theorems in the abovementioned answer for rings instead of groups, they are essentially the same arguments.

Again, this has no problem with regards to having units in the ideal: this is what happens if your equivalence relation is just "everything is related to everything", or if your quotient is the one element ring; these are perfectly valid (if somewhat boring) equivalence relations and quotients.

But: in a commutative ring with $1$, an ideal is the whole ring if and only if it contains a unit. If there are no units, then it is not the whole ring. A ring is said to be simple if the only ideals are the zero ideal and the whole ring; for commutative rings, the only simple rings are the fields and the zero ring. In the noncommutative case there are simple rings that are not division rings: for example, the set of $n\times n$ matrices over any simple ring (for instance, any field) is simple for every $n$.

Note also that while a proper ideal necessarily does not contain a unit, it is not true that any subset that does not contain a unit is an ideal. The collection of all numbers that are either multiples of $2$ or of $3$ in $\mathbb{Z}$ contains no units, but is not an ideal (it is not even a subgroup).

share|improve this answer
    
Thank you! Your answer has helped a lot to clarify what ideals are for me. Also, is the fact that everything is a multiple of a unit (since multiplying by $ax^{-1}$ will take us to $a$ if our unit is $x$) the reason why it is called a unit, similar to the fact that everything is a multiple of 1? –  Harry Stern Dec 29 '10 at 0:31
    
(also, I would love to read your paper but unfortunately I don't have access to jstor) –  Harry Stern Dec 29 '10 at 0:31
    
@Harry Stern: I confess I do not know off the top of my head the reason for calling them "units". In fact, in a commutative ring, a unit is an element that divides $1$. Of course, if $y$ divides $x$, then every multiple of $x$ is a multiple of $y$. If $u$ divides $1$ (that is, if $u$ is a unit), then every multiple of $1$ is a multiple of $u$. Since everything is a multiple of $1$, then everything is a multiple of $u$. As to the paper, most math departments (at least in the U.S.) have subscriptions to the Monthly, you might be able to get your hands on a hard copy. –  Arturo Magidin Dec 29 '10 at 0:34
    
@Harry Stern: (Unfortunately, I am away from the office and my local machine has been turned off for a few days, so I cannot even offer to send you a version of the paper by e-mail right now) –  Arturo Magidin Dec 29 '10 at 0:40

One way to think of an ideal as being the set of multiples of the (possibly non-existent) g.c.d of all the elements that contain it. (I am thinking here of the case of a commutative ring with $1$, so that distinctions between left, right, and two-sided don't matter.)

In the integers, for example, any set of elements has a g.c.d. with all the reasonable properties that you could want, and furthermore if $\{a_i\}_{i \in I}$ is a set of integers, then the ideal generated by the $a_i$ is in fact principal, and its generator is a g.c.d. of this collection. In a more general ring, g.c.d.s don't necessarily exist, or even if they do, they don't have all the properties that they do in the integers. So rather than trying to work with g.c.d.s, we can introduce ideals, which generally have better properties (or, rather, do many of the same jobs in more general rings that g.c.d.s and their set of multiples do in the context of the integers).

With this in mind, one sees (a) why ideals are the natural kernels of quotient maps: think about the case of the integers, working {\em modulo} $n$ means setting all multiples of $n$ equal to zero; (b) why an ideal with a unit will be the trivial ideal (i.e. the whole ring): because the g.c.d. of any set containing a unit will have to be $1$.

share|improve this answer

Ideals are not "unitless subsets", in the sense that there are many unitless subsets that aren't ideals. Proper ideals, indeed, contain no units, but they are also closed under addition, subtraction and multiplication by arbitrary ring elements.

I'm not sure why you're confused about the case of "multiples of n". In the ring $\mathbb{Z}$ of integers, the proper ideals are precisely the multiples of some fixed integer. Indeed, there are many other "unitless subsets" in $\mathbb{Z}$, but they are not ideals.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.