Given a ringlet on a collection Z of values, a
collection C of its values is known as an ideal
(for this
ringlet) precisely if:
If Z has a 0, any non-empty ideal has a member a for which a +0 = a, so the first rule makes 0 a member of every non-empty ideal. (Note that, as multiplication needn't commute, we have to assert the second condition both ways round; when it does commute, it suffices to check one way round or the other.)
In a ring, the addition forms a group; that it's cancellable and associative extends inescapably to its action on any ideal C; and the first condition above (given that the addition is abelian) simplifies to saying that addition is also closed and complete on C, hence that the addition on C is a sub-group of the ring's. If Z subsumes some S satisfying the second condition, scaling members of S by the additive inverse of 1 in Z ensures S includes additive inverses for all of its members; hence merely being closed under addition suffices to make S also satisfy the first condition and thus be an ideal.
If 1 is in C, the second condition implies C = Z, which is trivially an
ideal; in particular, any sub-ringlet that is an ideal is in fact the whole
ringlet. Any ideal with an invertible member necessarily contains 1 (applying
the second condition with the invertible member in C and its inverse in Z),
hence is the whole of Z. The collections {} and {0} are also trivially
ideals. A proper
ideal is an ideal that isn't the whole
ring, {0} or {}; a ringlet is described as simple
if it
has no proper ideals. In particular, in
a division ringlet, every non-zero value is
invertible, so every division ringlet is simple.
Any intersection of ideals is an ideal. If a, b are in the intersection, C,
then they are in each of the ideals intersected, hence so are a+b and any
b−a in Z; hence these values are also in C. If a is in C then it's in
each of the ideals intersected hence so is r.a for each r in Z, hence each such
r.a is also in C. As a result, we can define the ideal
generated by
some collection C of values in a ringlet to be the intersection
of all ideals that subsume C (there is always at least one such ideal, the whole
ringlet, so this never drags intersect({}), the all-relation, into the
discussion).
Suppose we have a collection, S, of values of a ringlet R and S is closed under addition (of its members with one another) and under multiplication by arbitrary members of R. All it might lack to be an ideal is the differences in T = {x in R: s +x in S for some s in S}. Since each s in S has s +s in S, T subsumes S. When R is a ring, S subsumes T (so they are equal), but it might not in a ringlet. Given t in T, we have some s in S for which s +t is in S; if we multiply both by some r in R, we have r.s in S and r.(s +t) in S because S was closed under multiplication by R's members; but then r.s +r.t is in S and r.s is in S, so r.t is in T. For s in S and r, t in T with s +t in S and q +r in S for some q in S, we have s +q in S and (s +q) +(t +r) = (s +t) +(q +r) in S hence t +r in T. Thus T is closed under addition and under multiplication by members of R, just as S was. Finally, consider x in R for which t +x is in T, for some t in T; there are then some r, s in S for which r +t is in S and s +t +x is in S; and S was closed under addition, so (s +r +t) +x = r +(s +t +x) and s +(r +t) are in S, so x is in T. Thus T is an ideal that subsumes S; furthermore, since every member of T must be a member of any ideal that subsumes S, T is the intersection of all such ideals. Thus a collection, that almost meets the requirements for an ideal, except possibly for the differences, only needs extended to its collection of differences to get the ideal it generates.
Given two ideals C, D of a ringlet R, consider the collection S = {c +d: c in C, d in D}. As C, D are ideals, S is closed under addition and under multiplication by members of R. In a ring, this suffices to make S an ideal; in a ringlet, S's collection of differences is the ideal S generates.
Given a ringlet on a collection Z of values and a non-empty ideal C of it,
consider the (Z: :Z) relation: x ~ y
precisely if there are a, b in C for
which x+a = y+b. When x+a = y+b and y+c = z+d we obtain x+a+c = z+b+d, with a+c
and b+d in C if a, b, c, d are, so x ~ y and y ~ z implies x ~ z: the relation
is transitive. By specification, it's symmetric; and C is non-empty, so has
some member a, for which x+a = x+a for all x in Z, so the relation is
reflexive. Thus this relation is an equivalence on Z. When x ~ y and p ~ q, we
have a, b, c, d for which x+a = y+b and p+c = q+d; so (x+p)+(a+c) = (y+q)+(b+d)
with a+c and b+d in C, so x+p ~ y+q; likewise, x.p +(x.c +a.p +a.c) = y.q +(y.d
+b.q +b.d) with each parentesised expression in C, so x.p ~ y.q. Thus this
equivalence respects the addition and multiplication on Z and we can reduce Z
modulo this equivalence yet preserve the ringlet structure; so the equivalence
classes of Z under this equivalence form a ringlet, denoted Z/C and known as
the quotient
ringlet of Z modulo C. When Z's addition is complete, so is
that of Z/C, hence Z/C is a ring if Z is.
For example, for any c in Z = {naturals} with the usual
addition and multiplication, C = {n.c: n in naturals} is an ideal (as we'll
see below); and Z/C is the cyclic ringlet
of arithmetic ignoring multiples of c
, a.k.a. arithmetic modulo
c
. In this case, the addition on Z wasn't complete, but when c > 0 the
addition on Z/C is: for any n in Z there is some k in Z for which k.c > n and
there is then some p in Z for which k.c = n+p, making p an additive inverse for
n in Z/C. Thus Z/C can be a ring even when Z is only a ringlet.
When we construe the (Z: x←y; x+a = y+b for some a, b in C :Z) equivalence above as being (Z/C| |Z), i.e. reading its left values modulo the equivalence itself, it serves as a mapping (Z/C: |Z) and, by construction, is a ringlet homomorphism. Each member of its kernel is an x in Z with x+a = b for some a, b in C; and the specification of an ideal says that each such x is in C, so C subsumes the kernel; furthermore, each x in C, when added to any a in C, does indeed give b = x+a in C, so the kernel subsumes C. Thus the kernel of the natural embedding homomorphism (Z/C| |Z) is exactly C.
For any ideals C, J in Z, the image of J under the (Z/C| |Z) embedding is just J understood modulo the C-equivalence, which makes it T = {x in Z: x +b = i +a for some i in J, a, b in C}. If J subsumes C, T is just J and J respects the C-equivalence, so J's image under the (Z/C| |Z) embedding is an ideal. Otherwise, for x, y in T, with x +b = i +a, y +n = j +m for i, j in J and a, b, n, m in C, we have (x +y) +(n +b) = (i +j) +(m +a) with n +b and m +a in C and i +j in J, so x +y is also in T. For x in T with x +b = i +a for i in J, a, b in C, any r in Z gives r.x +r.b = r.i +r.a with r.i in J, while r.b and r.a are in C, so r.x is in T. Thus T's differences form an ideal that, by the structure of T, respects the C-equivalence; the image of an ideal under the (Z/C| |Z) embedding might not be an ideal in Z/C, but its collection of differences is. If Z/C is a ring, J's image is an ideal, because it's its own collection of differences.
For any ringlet homomorphism (R: f |S), consider any ideal C in R and look
at D = (C&on;f: s ←s :), the collection of members of S mapped to C. For
any a, b in D, we have f(a), f(b) in C, which is an ideal; for any s in S,
f(s.a) = f(s).f(a) is in C so s.a is in D; f(a +b) = f(a) +f(b) is in C; and,
for any s in S with a +s = b, f(a) +f(s) = f(a +s) = f(b) so f(s) is in C and s
is in D; thus D is an ideal in S. In particular, Ker(f) is always an ideal in
S. (I'm fairly sure this implies we can factor f via the natural embedding
(S/ker(f)| |S) homomorphism, when Ker(f) is non-empty.) I'll describe this as
the homomorphism pulling back
an ideal in S from one in R.
When a ringlet homomorphism is (R| f |S), i.e. it produces all of R among its outputs, consider an ideal C in S and its image D = (: r ←r :f&on;C) in R. Any member of D is f(c) for some c in C and every r in R is f(s) for some s in S, so r.f(c) = f(s.c) is in D; when we add f(c) +f(a) with a, c in C we get f(a+c) in D as a+c is in C. So consider r for which f(c) +r = f(a) with a, c in C; we have r = f(s) for some s, yielding f(c +s) = f(a), but this doesn't necessarily ensure c +s = a. When S is a ring, i.e. when the addition on S is complete, we have some s in S for which c +s = a and this s is necessarily in C with r +f(c) = f(a) = f(s +c) = f(s) +f(c) and cancellation gives r = f(s) in D, making D an ideal. Thus, for a homomorphism from one ring onto another, the image of any ideal is an ideal; but the same is not assured for ringlets or when any of the destination ring isn't an output of the homomorphism. See the last section for the special case where f is the natural projection of a ringlet onto its quotient by an ideal.
Given an ideal J of a commutative ringlet R and a value r of that ringlet, consider K = {x in R: r.x in J}. For each a, b in K, we know r.a and r.b are in J, which is an ideal; thus r.a +r.b = r.(a +b) is in J and a +b is in K; if any x in R has a +x = b then r.a +r.x = r.b with r.a and r.b in ideal J, hence r.x also in J, hence x in K. For any a in K and s in R, we have r.a in J hence r.a.s = r.(a.s) in J hence a.s in K. Thus K is an ideal. (This is a lot like the homomorphism's pull-back, even though (: r.x ←x :) isn't a homomorphism; when we come to look at principal ideals, we'll meet the analogous case for such a mapping's image of an ideal, which would run into essentially the same obstacle as just seen for a homomorphism's image of an ideal, but a partial form of additive completeness lets us work round that.)
In a ringlet R, for any c in R, the ideal generated
by {c} is known as a principal ideal (of R); I'll also
refer to it as the ideal of multiples
of c. Of course, when c = 0, this
ideal is just {0}.
The typical element of the ideal of multiples of some c is just the collection of differences of the collection of finite sums, each term of which is of form r.c.s for some r, s in the ring. This collection of finite sums is manifestly (by construction) closed under addition and under multiplication by arbitrary values of the ring; so its collection of differences is an ideal. Since every member of this is necessarily present in any ideal that contains c, this ideal must in fact be the intersection of all such ideals, i.e. the ideal generated by {c}.
When multiplication commutes, each s.c.r is s.r.c and any sum of such is just the sum, of the s.r multipliers of c, times c; so the ideal of multiples of c consists simply of the collection of differences between simple multiples of c, i.e. {r in R: s.c +r = t.c for some s, t in R}.
Let z be the polynomial power(1) = (: x ←x :) and consider the polynomials we can get from the naturals and the two generators 5.z and 3.z. The coefficient of z in any such polynomial is one of 0, 3, 5, 6, 8, or some higher natural. We inherit the multiplication and addition of polynomials over the naturals and we have all constant polynomials, in particular including 1, our multiplicative identity; so we have a ringlet. We have nothing we can add to 3.z to get 5.z; but we have 6.z that we can add to 3.(3.z) to get 3.(5.z); as 6.z isn't (for the purposes of our ringlet) a multiple of 3 (although it is a multiple of 3.z), the multiples of 3 don't form an ideal on their own, in this rather contrived ringlet. In this case, then, the principal ideal strictly subsumes the collection of multiples, as it gets 6.z as a member that isn't a multiple.
If a commutative ringlet has enough completeness
, the ideal
generated by multiples of a value is always simply the collection of that
value's multiples, without having to resort to differences to make it an
ideal. In a commutative ringlet R, I'll say that R cancels down
completions
precisely if, for a, b, c, x in R: a.c +x = b.c
implies there is a y in R for which a +y = b
. (Completeness asks for
each a, b to have an x for which a +x = b; such an x is
termed a completion for the pair
b←a; when we have a completion for (b.c)←(a.c), we're here asserting
that there must also be a completion for b←a, cancelling the .c from both
sides. So we're cancelling out a common factor from any two things between
which completion does work.) Every ring always cancels down completions, since
what it asks for is just a special case of additive completion. Helpfully,
the natural ringlet does also cancel down
completions (because its ordering is preserved under multiplication by a
non-zero).
If a commutative ringlet R cancels down completions, consider S = {r.c: r in R} as before, closed under addition and under multiplication by members of R. Each difference of its members is an x in R for which r.c +x = s.c for some r, s in R; canceling down completions then gives us a y in R for which r +y = s, whence r.c +x = r.s = r.c +y.c and cancellation gives x = y.c, thus every difference of S is in fact a member of S and S is an ideal; the ideal generated by multiples of a value is in fact just the collection of that value's multiples.
Given an ideal J of a commutative ringlet R, that cancels down completions, and a value r of the ringlet, consider the collection H = {r.x: x in J}: any two members are r.a, r.b with a, b in J. Thus a+b is in J, hence r.(a+b) = r.a +r.b is in H, so H is closed under addition; as R cancels down completions, any x with r.a +x = r.b gives us a y in R for which a +y = b; as a and b are in J, which is an ideal, this implies y is also in J; as x +r.a = r.b = r.(y +a) = r.y +r.a we have x = r.y with y in J so x is in H; finally, given s in R and r.a in H, we have a in J, which is an ideal, so s.a in J, so r.(s.a) = s.(r.a) is in H. Thus H is an ideal.
Given an ideal C of a commutative ringlet R and a value s of the ringlet, consider the collection S = {r.s +c: r in R, c in C}. It is manifestly closed under addition and under multiplication by members of R, so its collection of differences is an ideal. Furthermore, even if R hadn't been commutative, for each c in C, we have c+c in C and 1 in R, so s +c and (s +c) +c = s +(c +c) in S, so c is in S's collection of differences, which thus subsumes C.
If a non-trivial (specifically: not {0}) commutative ringlet R is simple, consider any non-zero s in R; let S be the ideal generated by its multiples, {r in R: s.u +r = s.v for some u, v in R}. As R is simple, S is {}, {0} or R; as 1 and 1+1 are in R, 1.s +s = (1+1).s implies non-zero s in S, so S = R and 1 is in S. Thus there are some u, v in R for which u.s +1 = v.s; if R cancels down completions, there is then some y in R for which u +y = v and 1 = y.s, so s has a multiplicative inverse. Thus, in a non-trivial simple commutative ringlet that cancels down completions, the non-zero multiplication forms a group; it is both complete and cancellable, so we have a division ringlet. (In particular, since rings naturally cancel down completions, a non-trivial simple commutative ring is a field.)
A prime ideal of a ringlet R is a proper ideal C for
which r, s in R with r.s in C
implies r in C or s in C
. (Note
that this is just the converse of the part of the definition of an ideal that
requires that, for r, s in R, r in C or s in C
implies r.s in C.) A
value, in a ringlet, that generates (as its principal ideal) a prime ideal is
known as a prime. Since the ideal it generates is proper, so
not the whole of R, no prime is invertible.
There is some diversity as to whether mathematicians consider 0 and/or 1 to be prime. By requiring prime ideals to be proper, I've ruled them out as primes; one could as readily skip this requirement, making only modest changes to the rest of the analysis, to allow 0 and 1 to be prime. I prefer to have these two non-prime; they are special enough in their own ways, as it is.
When C is a prime ideal in R, R/C has no proper zero-divisors; so consider any ideal C of a ringlet R for which R/C has no proper zero-divisors. If r, s in R have r.s in C then their equivalence classes in R/C have zero product, hence one of r, s is equivalent to the zero in R/C; but it's then an x for which x+a = b for some a, b in C and the specification of an ideal implies it's in C. Thus C is prime. So an ideal C is prime in a ringlet R precisely if R/C has no proper zero-divisors.
An integral ringlet is a commutative descalable ringlet; being descalable (i.e. a common non-zero factor can be cancelled out through an equation) implies an absence of proper zero-divisors (i.e. if r.s = 0, one of r, s was zero). In a commutative ringlet R, if R/C is an integral ringlet for some ideal C, then C is prime. A ring is descalable precisely if it lacks zero-divisors so, in a commutative ring R, an ideal is prime precisely if R/C is an integral domain.
An ideal C of a ringlet R is described
as maximal precisely if it isn't R and: R subsumes J
subsumes C with J an ideal in R
implies J in {R, C}
. In this case,
R/C is simple (as is R/R); it has no proper ideals, as any ideal in R/C pulls
back, via the natural projection of R onto R/C, to an ideal of R that subsumes
C.
Conversely, suppose a ringlet R has a non-empty ideal C for which R/C has no proper ideals. The natural projection of R onto R/C projects each ideal that subsumes C onto an ideal; the only candidates here are {}, {0} and R/C itself. So consider any ideal J in R that subsumes C. If it projects onto {} or {0} then C subsumes it and J = C. Otherwise, J projects onto all of R/C so every member of R is C-equivalent to some member of J, i.e. for each r in R there is some j in J for which j +a = r +b for some a, b in C. Since J subsumes C, a and b are in J, so j +a is in J and r +b and b in J implies r in J also, so J is all of R. Thus, when R/C has no proper ideals, any ideal of R that subsumes C is either C or R, so C is maximal.
Now consider any maximal ideal: is it prime ? In a ringlet R, suppose
an ideal C is maximal and consider any r, s in R for which r.s is in C. The
ideal S generated by T = {s.t +c: t in R, c in C} necessarily subsumes C (see
last section), hence is either C or R; if it is C, then s is in C; otherwise, it
is R, of which 1 is a member. When S is just T's collection of differences (as
arises in the commutative case), this gives us s.u +a +1 = s.v +b for some u, v
in R, a, b in C; multiplying by r gives (r.s.u +r.a) +r = r.s.v +r.b, in which
r.s is given to be in C, so r.s.u and r.s.v are in C; a, b are in C, so r.a and
r.b are in C; thus r.s.u +r.a and r.s.v +r.b are in C and r is the difference
between them, hence also in C. Thus, at least in a commutative ringlet, r.s in
C implies either s in C or r in C
, i.e. C is prime, provided it's
proper. Thus proper maximal ideals are prime in commutative ringlets.
Recall that, in a non-trivial simple commutative ringlet that cancels down completions, the non-zero multiplication forms a group. This implies that, in a commutative ringlet R, for any non-trivial maximal ideal C, the non-zero multiplication on R/C forms a group. (If R/C is a ring, e.g. because R is, it is then a field.)
Now consider a ringlet R, a proper ideal C and a non-empty collection B of proper ideals, each of which subsumes C, within which: for all a, b in B, either a subsumes b or b subsumes a. Now consider A = unite(B), the union of all of these proper ideals. Clearly A subsumes C, since each member of B does; and, as each member of B is a proper ideal, 1 is not in any member of B (since an ideal of which 1 is a member is the whole ringlet, so not proper), hence not in the union. Now consider any x, y in A; as A is unite(B), there are a, b in B for which x is in a and y is in b; as a, b are in B, one of them subsumes the other, hence x and y are both members of the one that subsumes, which is an ideal, so x +y is also a member and thus a member of our union A. Likewise if r in R has x +r = y for x in a, y in b with a, b in B; whichever of a, b subsumes the other has both x and y as members, hence also r, so r is in A. For any x in A, there's some ideal a in B with x in a so, for each r in R, r.x is in a, hence in A. Thus A is an ideal; and, as it doesn't have 1 as a member and does subsume the proper ideal C, it's a proper ideal.
The preceding, taken with Zorn's lemma
(a variant on
the axiom of choice), proves that every proper ideal of a ringlet R is subsumed
by some maximal ideal of R, using subsumes
as partial ordering on D =
{proper ideals of R that subsume C}; B was an arbitrary chain
in D and
has an uppper bound in D. This being true for every chain in D, Zorn asserts
that there is some maximal (for this ordering on D) element of D; the
specification of D and its ordering, subsumes, then makes each such maximal-in-D
element a (ringlet) maximal ideal. I am, however, somewhat unwilling to make
Zorn's assumption, or assume any other equivalent of the (transfinite) axiom of
choice. Fortunately, we can avoid the need for Zorn in some cases, notably when
the collection of proper divisors of any given member of our ringlet is finite
– as happens for {naturals}, for example.
We can likewise use is a proper multiple of
as a partial order on
members of our ringlet and look at chains of values, each of which is a proper
multiple of all those later in the chain. Each value has an associated
principal ideal, so the chain of multiples induces a chain of ideals, each of
which subsumes those that come before it, so that the union is necessarily an
ideal also. This shall prove useful when we
come to factorise in terms of primes.
Written by Eddy.