To briefly put forward my question, can anyone beautifully explain me in your own view, what was the main intuition behind inventing the ideal of a ring? I want a clarified explanations in these points:

  1. Why is the name "ideal" coined?. In English 'ideal' means "One that is regarded as a standard or model of perfection or excellence." Why did people gave the name of ideal to such group?
  2. And why are the ideals not present in the case of groups?
  3. And give me a very fantastic intuition and motivation behind the ideals and what are the roles served by them in advanced mathematics.

Thanks a lot, to every one.

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    $\begingroup$ It is ideal in the sense of not real, as in the addition of ideal points ("points at infinity") when we construct the real projective plane. $\endgroup$ Sep 12, 2011 at 15:09
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    $\begingroup$ See also math.stackexchange.com/questions/362/… $\endgroup$
    – lhf
    Sep 12, 2011 at 16:06
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    $\begingroup$ thanks a lot sir,@lhf,@Andre $\endgroup$
    – IDOK
    Sep 12, 2011 at 16:09
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    $\begingroup$ It's not a good idea to accept an answer within an hour of posting a question. This means that the question will receive less exposure to others who might contribute further helpful answers. Try to wait at least a few days before accepting so that most users have a chance to see the question. Many experts visit the site less frequently, and may not browse questions that already have accepted answers. $\endgroup$ Sep 12, 2011 at 16:11
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    $\begingroup$ @Iyengar: I am very happy that you are now willing to look at simpler concepts such as the ideal class group. Thank you very much. Now you see that these considerations arise to make sense of unique factorization in algebraic number fields. May I again impose on you with yet another request? This time to please read through the proof of unique factorization of integers, and then return to algebraic integers? You naturally want to understand why this proof fails for number fields. An arithmetician moreover should know to how to prove of the fundamental theorem of arithmetic. Don't you agree? $\endgroup$
    – George
    Sep 12, 2011 at 19:56

9 Answers 9


As was already said, the term "ideal" came from Kummer's ideal numbers (more precisely, "ideal complex numbers" as Kummer was concerned with factorizations of algebraic integers which lie in the complex field). I'll try to give a brief intuition not mentioned here explicitly already.

When factoring, say, 60, you find 2 "different" factorizations: $60=15\times 4=12\times 5$. This does not contradict unique factorization in integers since you have not factored "enough": after factoring all the numbers as much as possible you obtain the unique factorization $60=2\times 2\times 3\times 5$.

However, in the context of algebraic integers this does not always hold. The famous example is in $\mathbb{Z}[\sqrt{-5}]$. There you have $6=2\times 3=(1-\sqrt{-5})\times(1+\sqrt{-5})$ but the factors are irreducible, so unique factorization fails.

Kummer's idea was that in this case as well the problem is that the factors were not factored "enough". His approach was to assume that there are better, "ideal" factors for which the unique factorization hold.

It is obvious that there is something problematic here - you need to construct such ideal numbers, prove their existence, etc. However there is also another way created by Dedekind. Dedekind defines not the ideal numbers themselves, but the sets of elements they divide. For example, instead of talking about "2" you can talk about the set $\{0,2,-2,4,-4,6,-6,\dots\}$ of even numbers in the integer - the ideal created by 2.

Dedekind noted that this concept of "being divided by" can be characterized by two properties:

  1. If some number (ideal or not) divides $a$ and $b$, it divided $a+b$.
  2. If some number (ideal or not) divides $a$, it divdes $\lambda \times a$ for all $\lambda$.

So he defined ideals using these two properties. They turned out to be just enough to prove the grand theorem that is in the base of algebraic number theory - that in Dedekind domains (and so in algebraic integer ring) there is a unique factorization of elements of ideals to products of prime ideals (this applies to elements as well, since an element can be identified with the ideal it generates).

This is quite orthogonal to the usage of ideals usually encountered in an undergrad level algebra course - where ideals pop up naturally (and more generally) as kernels of homomorphisms. But here the name "ideal" is indeed confusing.

  • $\begingroup$ I think that "ideal number" was not in the sense of "better number", but rather in the sense of "ideal" that you can still find in dictionaries: "existing in imagination or fancy only"; the "ideal numbers" did not actually exist, it's just that the other numbers behaved "as if" ideal numbers existed. $\endgroup$ Sep 12, 2011 at 16:34
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    $\begingroup$ @Gadi A: "This is quite orthogonal to the usage of ideals usually encountered in an undergrad level algebra course - where ideals pop up naturally...." -- What is the orthogonality here? The integral ideals in the ring of integers of a number field are also ideals in the sense of abstract algebra... Did you mean instead to define fractional ideal? $\endgroup$
    – George
    Sep 12, 2011 at 19:45
  • $\begingroup$ No, what I mean is that the object is exactly the same, but the way it arises is quite different (although I am sure there is a deep connection I'm missing, on the basic level the two ways of introducing ideals are not related and can be taught independently - as is done in practice) $\endgroup$
    – Gadi A
    Sep 13, 2011 at 4:14
  1. "Ideal" is meant in the sense of "not real." The story as I know it is that Kummer was interested in fixing unique factorization in number rings, and to do it he had to introduce certain "ideal numbers." Dedekind later recast these ideas in terms of ideals as we now understand them.

  2. To me ideals are kernels of ring homomorphisms. The analogue of ideals for groups is normal subgroups.

  3. Again, ideals are kernels of ring homomorphisms. In number theory, ideals appear naturally as the correct objects to use to study divisibility. After all, the statement that $a$ divides $b$ is nothing more than the statement that $b$ is contained in the ideal $(a)$ generated by $a$, or in the language of ring homomorphisms that $b = 0$ in the quotient ring $R/(a)$. So instead of studying divisibility of numbers we can study containment of ideals. For $\mathbb{Z}$ this gives us the same theory of divisibility, but for other number rings we get non-principal ideals which are precisely the correct manifestation of Kummer's "ideal numbers."

  • $\begingroup$ magnificent,elegant answer ,this is what i was expecting for,thanks a ton,and sorry for my previous Behavior,and i apologize you,But suggest me a fantastic book which covers all basic concepts(which you read,but remember my level sir) of Abstract Algebra,and an Algebraic number theory book,that has a fantastic description of Ideal classes,Class number formula and other things ,i am in debt with you,thank you anyway $\endgroup$
    – IDOK
    Sep 12, 2011 at 15:37
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    $\begingroup$ @Iyengar: Could you please be more careful with your typesetting? For example: first letter of every sentence is a capital letter; no unnecessary capitalization inside sentences, always put space after comma or full stop, put a full stop at the end of a sentence, etc, etc.. This was pointed out to you a lot times and you never listen. Next, for etiquette, would you terribly mind cutting down the number of 'sir's in your speech by some? In fact altogether avoid it if you may? $\endgroup$
    – George
    Sep 12, 2011 at 20:58

Ideals are in some sense meant to capture the notion of "multiples" of an element. They originally came in the form of Kummer's ideal numbers. Basically as mentioned above unique factorisation of elements can fail in some rings:

In $\mathbb{Z}[\sqrt{15}]$ we have $10 = 5\times 2 = (5+\sqrt{15})(5-\sqrt{15})$.

Kummer noticed that really the problem is that the ring isn't big enough. It doesn't contain the elements $\sqrt{3}$ and $\sqrt{5}$. If it did then we could explain the two different factorisations as reorderings of things in $\mathbb{Z}[\sqrt{3}, \sqrt{5}]$:

$10 = (\sqrt{5})(\sqrt{5})(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3}) = (\sqrt{5})(\sqrt{5} + \sqrt{3})(\sqrt{5})(\sqrt{5} - \sqrt{3})$

From the point of view of $\mathbb{Z}[\sqrt{15}]$ the numbers $\sqrt{3}$ and $\sqrt{5}$ don't exist so Kummer's idea was to in some sense invent them as abstract quantities, rather than having to make a ring extension.

If these abstract quantities were to be invented then they should satisfy the main rules of divisibility.

Dedekind improved on this and invented the notion of ideal of a ring. The axioms for this are exactly the axioms you would expect from divisibility (i.e. in $\mathbb{Z}$ you expect the sum of two numbers divisible by 6 to also be divisible by 6, also you expect ANY multiple of 6 to also be divisible by 6).

He then was able to prove that in some nice rings the ideals factorise uniquely into so called prime ideals...and this uniqueness is upto ordering. This takes into account what we said above, reordering prime ideals gives the two different factorisations. So the idea is not to factorise elements but to factorise the sets of multiples of elements as objects in their own right.

Define prime ideals $\mathfrak{p}_1 = \langle 5, 1+\sqrt{15}\rangle, \mathfrak{p}_2 = \langle 2, 1+\sqrt{15}\rangle, \mathfrak{p}_3 = \langle 2, 1-\sqrt{15}\rangle$.

Then we get:

$\langle 10\rangle = \langle 2\rangle \langle 5\rangle = (\mathfrak{p}_2 \mathfrak{p}_3)(\mathfrak{p}_1^2)$

$\langle 10\rangle = \langle 5+\sqrt{15}\rangle \langle 5-\sqrt{15}\rangle = (\mathfrak{p}_1\mathfrak{p}_2)(\mathfrak{p}_1\mathfrak{p}_3)$

So the two different factorisations of $10$ as an element are explained by two different orderings of the prime ideals.

How does Kummer's ideal numbers tie into all of this? Well you see we had to use ideals generated by more than one element...these are corresponding to "multiples" of elements that don't exist in the ring!

  • $\begingroup$ Ideals are not present for groups because they need two operations to be defined! $\endgroup$
    – fretty
    Mar 24, 2012 at 11:52
  • $\begingroup$ Also as for their uses in higher maths, there are many. In number theory we can study primes by studying how their associated ideals factorise in bigger rings. Ideals turn out to be the natural objects that we can quotient out by to make quotient rings...we do this very often in say algebraic geometry when trying to study geometrical solutions to polynomial equations via algebraic structures made by ideals. $\endgroup$
    – fretty
    Mar 24, 2012 at 11:55

The notion of an ideal number was introduced by Ernst Kummer in an attempt to explain and fix the failure of unique factorization in certain subrings of the complex numbers. The canonical example that students usually see is that $2\cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$.

It's fairly easy to surmise from that Wikipedia article, that working with ideal numbers was rather cumbersome. So where did the idea of considering ideals come from? I can only make an educated guess, but I think it was the same as with many other things that we now take for granted: It came from people looking at examples and eventually figuring out what it was that made them all "tick" - as it were.

So why are ideals useful? Because rings of algebraic integers in number fields are (generally) Dedekind domains where unique factorization - not in the ring itself, but rather of its ideals into its prime ideals holds.

For a long series of videos of lectures on abstract algebra ending up with some algebraic number theory about algebraic integers in imaginary quadratic number fields, there is this series of lectures from Harvard.

  • $\begingroup$ I suppose these are ideal in the sense that unique factorization holds for them. $\endgroup$
    – Srivatsan
    Sep 12, 2011 at 15:31
  • $\begingroup$ just awesome answer,everyone have presented their perspective beautifully,thanks a lot sir both for your explanation and marvelous links,i am in debt with you $\endgroup$
    – IDOK
    Sep 12, 2011 at 15:39

Another useful intuitive way of thinking about ideals, and perhaps how Dedekind first envisaged his simplified versions is in terms of sets.

(In the late 19th century, when Dedekind was working on this, Cantor's set theory was the latest thing, and no doubt informed the thinking of others just as category theory does today.)

So ask yourself, as perhaps Dedekind did initially, what is a way of expressing in terms of set inclusion the fact that an integer $a$ divides another one $b$. The answer is to consider the set $A$ of all integer multiples of $a$ and another set $B$ of all integer multiples of $b$, and then it isn't a big leap to see that $a$ divides $b$ if and only if $B$ is a subset of $A$. (Notice how the set inclusion is inverted).

To someone familiar with linear algebra (again a hot topic in the 19th century), the natural next step is to generalize multiples of a single integer to linear forms of more than one integer, for example two integers. So one might then ask when does the set $P$ of all linear forms $a p + b q$ for some fixed integers $p$, $q$ as $a$ and $b$ range over all integers contain another such set of forms c r + d s for a given pair of integers $r$, $s$.

For rational ("ordinary") integers, the answer is rather disappointing because these sets are equal to (for set $P$ typically) the set of multiples of gcd($p$, $q$). This is because $a u + b v = 1$ always has integer solutions $a$, $b$ for any given coprime pair of integers $u$, $v$. So nothing is gained.

But for rings of algebraic integers where the latter equation does not always have algebraic integer solutions in that ring, i.e. where the Euclidean algorithm fails, these linear forms do represent a proper "refinement" of the integers and the Dedekind definitions capture their relevant properties even when not thought of directly as linear forms.


For your first question, you can have a look at the corresponding Wikipedia article.

Ideals are not present in the context of groups, because they are instances of modules over a ring: an ideal of $R$ is a subgroup of $R$ that is also an $R$-module, with multiplication inherited from the one in $R$. So we really need the structure of the ring (i.e. multiplication) to be able to speak of ideals, and we do not have this extra structure for groups.

Ideals are present in so many mathematical contexts that there is no single motivation behind their definition (just as there is no single motivation behind the definition of a group). An example where you really need ideals is when you want to study how primes and factorizations behave in more general rings than just the integers. If you want to know more about this you could consider taking an introductory course in algebraic number theory. Another example where one uses ideals is in algebraic geometry. Roughly speaking, one associates to a ring a geometric space (its prime spectrum). Then the ideals of the ring correspond to subspaces of this prime spectrum.

  • $\begingroup$ good answer but purely a technical one,which is known by most of people,i was expecting the historical notion and need for introducing an Ideal and its complete background,lets wait for some beautiful answer ,anyway thanks a lot for the answer and +1 for your explanation $\endgroup$
    – IDOK
    Sep 12, 2011 at 15:32
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    $\begingroup$ @Iyengar: You are studying mathematics and there you have to go through the technical parts. If you do not like techniques, formalisms and details and want only beauty, then perhaps you are putting your efforts into the wrong direction. You should maybe look at a few gardens instead. Or take up poetry and painting, if you have any skill. $\endgroup$
    – George
    Sep 12, 2011 at 21:10

Let $K$ be an algebraic number field and let $\mathcal O_K$ be its ring of integers. A definition of an ideal of $\mathcal O_K$, which is equivalent to what is given in the earlier answers, is:

An ideal $\mathcal I \subset \mathcal O_K$ is the collection of all numbers in $\mathcal O_K$ that are divisible by a fixed algebraic integer $\alpha$, which may or may not belong to $K$.

Here an algebraic integer $\alpha$ is said to divide another algebraic integer $\beta$ if there exists an algebraic integer $\gamma$ such that $\beta = \alpha \gamma$.

Equivalence with the usual definition can be derived as a consequence of the finiteness of classnumber. It is omitted here due to the introductory nature of the question. But this point of view is worth keeping in mind, as it might give some insight to the phrase "ideal number" and how the algebraic notion in ring theory arose. This is making a little more precise what Qiaochu describes as non-principal ideals being in a sense "ideal numbers" -- they can also be thought of as some number outside the ring.


In mathematics it is very important to abstract away properties and consider equivalence relations and their associated quotients. Therefore, whenever we study some kind of structure, we wish to describe its quotients; of course, these quotients should retain the same kind of "structure" as their parent object. With certain structures it turns out that identifying certain elements is the same as identifying their difference with $0$; thus to specify an equivalence relation is to specify certain elements as being $0$. Of course, in order to preserve structure we have some rules to follow in specifying elements as being $0$.

As an example, let's derive the ideal axioms; let $a$ and $b$ be in the ideal $I$ of a ring $R$. In $R/I$, $a$ and $b$ will both be $0$; since $0 + 0 = 0$, we therefore ought to have $a + b \in I$. Similarly $c \cdot 0 = 0 = 0 \cdot c$, whence $RI = I = IR$.


Let $ R $ be a commutative ring with unity (Like in Atiyah-Macdonald, from now "ring" would mean "commutative ring with unity"). We can ask ourselves :

$\textbf{Q}$) In how many ways can we put an equivalence relation $ \sim $ on $ R $ (i.e. partition set $ R $), and put binary operations $ \oplus,\odot $ on set $ R/{\sim} $, such that "Equations in $ R $ give corresponding equations between equivalence classes in $ R/{\sim} $" that is "($a+b=c$ in $R$ implies $[a] \oplus [b] = [c]$ in $ R/{\sim} $) and ($ab=c$ in $ R$ implies $ [a] \odot [b] = [c]$ in $R/{\sim} $)" ?

[Under such $ \sim, \oplus, \odot $, notice $ (R/{\sim}, \oplus, \odot) $ automatically becomes a ring]

$\underline{\textbf{Part-1}}$ (Looking at the potential candidates for $ \sim, \oplus, \odot$)

Let $ \sim, \oplus, \odot $ be as needed. Unwrapping the constraints one by one...

$$ (1) \text{ } \sim \text{ is an equivalence relation} $$ and $$ (2) \text{ } [a]=[a'], [b]=[b'] \implies [a]\oplus [b] = [a'] \oplus [b'], [a]\odot[b] = [a']\odot [b'] $$ and $$ (3) \text{ } [a]\oplus [b] = [a+b], [a]\odot [b] = [ab]. $$

Using (3), we see (2) modifies as $ a \sim a', b \sim b' \implies a+b \sim a' + b' , ab \sim a' b'.$

Notice $ a \sim b \iff a-b \in [0] $
($ \implies $: As $ a \sim b $ and $ (-b) \sim (-b) $, we have $ a + (-b) \sim b + (-b) = 0 $.
$ \impliedby $: As $ a - b \sim 0 $ and $ b \sim b $, we have $ (a-b)+b \sim 0+b $, that is $ a \sim b $ )

Using this, constraint (1) can be rewritten as {$a - a \in [0]$; $a - b \in [0] $ implies $b-a \in [0] $; $ a-b, b-c \in [0] $ implies $ a-c \in [0] $}, which just means "$ [0] $ is a subgroup of $ (R, +) $".
Similarly constraint (2) becomes "$a-a', b-b' \in [0] $ implies $ (a+b)-(a'+b') , ab-a'b' \in [0]$", that is "$ a - a', b - b' \in [0] $ implies $ ab - a'b' \in [0] $", that is "$ x,y \in [0] $ implies $ (a' + x)(b' + y) - a'b' \in [0] $", that is "$x,y \in [0] $ implies $ a' y + b' x + xy \in [0] $", that is "$ t \in [0] $ implies $ at \in [0]$".
Constraint (3) remains the same.

So to summarise, in any such $ \sim, \oplus, \odot $ we have :

  1. $ a \sim b \iff a - b \in I $, where $ I $ is a subgroup of $ (R, +) $ satisfying $ a I \subseteq I $ for all $ a \in R $
  2. [Above condition gives $ [a] = a+I $] The operations $ \oplus, \odot $ satisfy $ (a+I) \oplus (b+I) = (a+b+I) $ and $ (a+I) \odot (b+I) = (ab + I) $.

$ \underline{\textbf{Part-2}} $ (That all such $ \sim, \oplus, \odot $ work)

Let $ I $ be a subgroup of $ R $ with $ aI \subseteq I $ for all $ a \in R $. We can readily verify $ a \sim b \overset{\text{def}}{\iff} a - b \in I $, and $ (a+I) \oplus (b+I) \overset{\text{def}}{=} (a+b+I) $, $ (a+I) \odot (b+I) \overset{\text{def}}{=} (ab+I) $ satisfy the constraints in question.

To summarise the entire discussion, equivalence relations $ a \sim b \iff a - b \in I $, arising from subgroups $ I $ of $ (R, +) $ satisfying $ a I \subseteq I $ for all $ a \in R $, are precisely the ones we were looking for. The subgroups here are traditionally called "Ideals".


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