When an affine algebraic scheme Spec(R) can be written as the “categorical union” of two closed proper subschemes Spec(R/I) and Spec(R/J), we say that it is reducible; the term “categorical union” indicates the smallest closed subscheme Spec(R/K) of Spec(R) such that Spec(R/I)(A) Spec(R/J)(A) is contained in Spec(R/K)(A) for every ring A in Spec. Assuming the result that the solutions in finite rings determine the ideal, we see that reducibility is equivalent to the condition that (0) = I capJ with neither I nor J the zero ideal. Thus we can define a ring R to be irreducible if (0) cannot be written as the intersection of two non-zero ideals; an ideal I in R is irreducible if R/I is irreducible. From the Noetherian-ness of the ring R it follows that the zero ideal is a finite intersection of irreducible ideals.

There are some problems with this decomposition. First of all there is the lack of
unique-ness. Secondly, the condition that an ideal is irreducible seems somewhat difficult
to characterise in terms of elements. Finally, (and very importantly) the notion of
categorical irreducibility seems too strong a condition to impose. For example, even the
scheme Spec(k[x,y]/(x^{2},xy,y^{2}) is reducible with this definition; this seems rather
counter-intuitive.

Another way to decide if a scheme is irreducible is as follows. Let f and g be non-zero functions with fg = 0, then f = 0 and g = 0 would define a decomposition of the space unless f (or g is nilpotent). This corresponds to the condition that (0) is a primary ideal; an ideal Q is primary if whenever fg lies in Q and g does not lie in Q, some power of f lies in Q.

1.1. Irreducibles are primary. Let R be an irreducible ring and a be in R. If a is not
nilpotent then R_{a} = R[T]/(aT - 1) is a non-zero ring. The kernel J of the natural
homomorphism R R_{a} is a proper ideal. Let J = (x_{1},...,x_{r}) (since R is Noetherian)
and let n be such that a^{n}x_{i} = 0 for all i. If y is an element of R such that a^{m}y = 0 for
some m, then y lies in J and so x^{n}y = 0. Now, suppose that c lies in Ra^{n} Ann(a); we
have c = da^{n} and ca = da^{n+1} = 0. But then d lies in J and so c = da^{n} = 0. Since we have
written (0) as an intersection of two ideals and Ra^{n} is non-zero we see that
Ann(a) = 0. So we have shown that every non-nilpotent element of R is not a
zero-divisor. Let P be the ideal consisting of nilpotent elements. If xy lies in P then
x^{n}y^{n} is zero for some n, so either x^{n} = 0 or y is a zero divisor, hence nilpotent.
Thus P is a prime ideal. Moreover, if xy = 0 and x does not lie in P then x is
not nilpotent and so we must have y = 0. In other words, (0) is a P-primary
ideal.

1.2. Irredundant primary decomposition. If P is a prime ideal in R, then the
intersection of finitely many P-primary ideals in R is again a P-primary ideal. Thus,
given an expression _{i}I_{i} = (0) of (0) as an intersection of irreducible ideals we can bunch
together all the I_{i}’s which are primary for the same prime ideal. This way we obtain an
expression _{j}Q_{j} = (0), where Q_{j} are P_{j}-primary for some finite collection of distinct
prime ideals P_{j} of R. Now, we can further assume that (0) _{jk}Q_{k} for any j else we
can shorten the above expression. Such an expression of (0) as an intersection of primary
ideals that cannot be shortened further is called an irredundant primary decomposition
of (0).
Consider the ring R = k[x,y] = k[X,Y ]/(X^{2},XY ). We see that (0) is the intersection
of the ideals RxRy^{n} for any n. This shows that we cannot home for uniqueness of the
primary decomposition. However, we will see that some form of uniqueness can be
proved.

1.3. Annihilators which are primary. Let Q_{i} = (0) be an irredundant primary
decomposition of (0) and let P_{i} be the prime ideal which is the radical of Q_{i}. By
irredundancy, the intersection _{jk}Q_{j} is non-zero; let c be an element of this
intersection. Now suppose xy lies in Ann(c) so that xyc = 0. Thus xyc lies in Q_{k}, if x
does not lie in P_{k} then yc lies in Q_{k} and as a multiple of c it lies in _{jk}Q_{j} as well. But
then yc lies in (0) = _{j}Q_{j} and so is 0; in other words y lies in Ann(c). We
have shown that Ann(c) is P_{j}-primary. It is also clear that Ann(c) contains
Q_{j}.
For any d not in Ann(c) consider the proper ideal Ann(dc). If xy lies in this ideal then
xyd lies in Ann(c); so if x is not in P_{j}, then yd lies in Ann(c) (by the P_{j}-primality of this
ideal) and so y lies in Ann(dc). In other words Ann(dc) is also P_{j}-primary. On the other
hand, if b lies in P_{j} there is a positive integer n such that b^{n+1} lies in Ann(c) but b^{n} does
not. Then b lies in Ann(b^{n}c). Thus the maximal ideal of the form Ann(dc) must contain
and thus equal P_{j}. We have therefore shown that prime ideals that occur in an
irredundant primary decomposition of (0) in the ring R are all annihilators of some
element of R.

1.4. Associated primes. The associated primes of a ring R are the prime ideals P
which are of the form Ann(c) for some c in R. Suppose P is such a prime. Now consider
an irredundant primary decomposition _{j}Q_{j} = (0). Among the Q_{j} let A be the collection
of those j for which c does not lie in Q_{j} and let I be their intersection. If d lies in I then
dc lies in all the Q_{j} and so must be 0; hence d lies in Ann(c) = P; so I P. Consider
some Q_{j} which does not contain c; let P_{j} be the radical of Q_{j}. If d lies in P = Ann(c)
then dc = 0 lies in Q_{j} and c does not lie in Q_{j} so d must lie in P_{j}. In other words P is
contained in the intersection _{jA}P_{j}. Since P is prime it contains the radical of I
which is the intersection _{jA}P_{j}. Thus P must be equal to one of the primes
P_{j}.
We have shown that the associated primes are precisely the primes that occur as the
radicals of the primary ideals in an irredundant primary decomposition.

1.5. Minimal primes. A prime ideal in R that is not contained in any other prime
ideal is called a minimal prime of R. Let _{j}Q_{j} be an irredundant primary decomposition
of (0) in R. The radical of (0) consists of all nilpotents and is the thus the
intersection of the radicals P_{j}’s of the Q_{j}’s. Since the nil radical of a ring R is
contained in all the prime ideals of R it follows that the minimal primes in
R are the minimal elements in the collection of the P_{j}’s. The minimal prime
ideals of a ring R are the minimal elements in the finite collection of associated
primes of the ring R. The non-minimal associated primes are called embedded
primes.
Let P_{k} be an associated prime in R and let Q be the collection of all c such that there
is some d not in P such that dc = 0. For any irredundant primary decomposition
_{j}Q_{j}, we see that Q is contained in Q_{k}. The ideal I_{k} = _{jk}Q_{j} is non-zero and
its radical is _{jk}P_{j}. If P_{k} is a minimal prime then it does not contain the
latter ideal and hence it does not contain I_{k} either. It follows that there is a d
that lies in I_{k} but does not lie in P_{k}. For any c in Q_{k}, the element dc lies in
I_{k} Q_{k} = (0) so is 0. Thus we see that Q = Q_{k}. The primary ideal in an irredundant
primary decompostion that corresponds to a minimal prime in R is uniquely
determined.

1.6. Zero-divisors in a ring. From the description given above we see that the
associated prime ideals consist of zero-divisors. Conversely, if xy = 0 and x does not lie in
an associated prime P_{j} then y must lie in the corresponding Q_{j}. So if x does not lie
in any associated prime P_{j} then y lies in all the Q_{j}’s and so must be 0. The
collection of zero-divisors in R is thus precisely the union of all its associated
primes.

1.7. Some geometry. We have seen that the primary decomposition is related to the decomposition of Spec(R) into irreducibles components. If we ignore for the moment, the distinction between irreducible components which differ only in their “nilpotent thickenings”, then the primary decomposition precisely corresponds to the decomposition into irreducible components. A zero-divisor c should be seen as a function that vanishes on some components but not others; the components on which it does not vanish are defined by the vanishing of those functions d such that dc is identically the zero function; that is, Ann(c). We have seen above that for any component, there is a function c such that Ann(c) is a primary ideal associated to that component; this is a function that vanishes on all components except the chosen one. This interprepretation makes sense unless the corresponding prime ideal is embedded (i. e. non-minimal). How can a function vanish on a component but apparently not on a closed subset? What can and does happen is that the function vanishes on the larger component but is nilpotent and non-zero on the embedded component which is contained in this larger component.

The primes associated with a component consist of functions that are nilpotent on that component. The above description also leads to characterisation of functions c that vanish on a non-embedded component as those for which there is function d which is not nilpotent on this component such that dc identically vanishes.

Finally, while dealing with primary decomposition, it is easy (but erroneous) to loose track of the elements of R which are not zero-divisors. We will see these elements acquire importance in a later section.