Radical of an ideal explained
of a
commutative ring is another ideal defined by the property that an element
is in the radical
if and only if some power of
is in
. Taking the radical of an ideal is called
radicalization. A
radical ideal (or
semiprime ideal) is an ideal that is equal to its radical. The radical of a
primary ideal is a
prime ideal.
This concept is generalized to non-commutative rings in the semiprime ring article.
Definition
The radical of an ideal
in a
commutative ring
, denoted by
or
, is defined as
\sqrt{I}=\left\{r\inR\midrn\inI \hbox{forsome} n\in\Z+\right\},
(note that
).Intuitively,
is obtained by taking all roots of elements of
within the
ring
. Equivalently,
is the preimage of the ideal of
nilpotent elements (the
nilradical) of the
quotient ring
(via the natural map
). The latter
proves that
is an ideal.
[1] If the radical of
is finitely generated, then some power of
is contained in
. In particular, if
and
are ideals of a
Noetherian ring, then
and
have the same radical if and only if
contains some power of
and
contains some power of
.
If an ideal
coincides with its own radical, then
is called a
radical ideal or
semiprime ideal.
Examples
of integers.
of integer multiples of
is
(the evens).
is
.
is
.
- In general, the radical of
is
, where
is the product of all distinct prime factors of
, the largest
square-free factor of
(see
Radical of an integer). In fact, this generalizes to an arbitrary ideal (see the Properties section).
I=\left(y4\right)\subseteq\Complex[x,y]
. It is trivial to show
(using the basic property but we give some alternative methods: The radical
corresponds to the
nilradical
of the quotient ring
R=\Complex[x,y]/\left(y4\right)
, which is the
intersection of all prime ideals of the quotient ring. This is contained in the
Jacobson radical, which is the intersection of all
maximal ideals, which are the
kernels of
homomorphisms to
fields. Any ring homomorphism
must have
in the kernel in order to have a well-defined homomorphism (if we said, for example, that the kernel should be
the composition of
\Complex[x,y]\toR\to\Complex
would be
, which is the same as trying to force
). Since
is algebraically closed, every homomorphism
must factor through
, so we only have to compute the intersection of
\{\ker(\Phi):\Phi\in\operatorname{Hom}(R,\Complex)\}
to compute the radical of
We then find that
Properties
This section will continue the convention that I is an ideal of a commutative ring
:
- It is always true that , i.e. radicalization is an idempotent operation. Moreover,
is the smallest radical ideal containing
.
is the intersection of all the
prime ideals of
that contain
and thus the radical of a prime ideal is equal to itself. Proof:
On one hand, every prime ideal is radical, and so this intersection contains
. Suppose
is an element of
that is not in
, and let
be the set \left\{rn\midn=0,1,2,\ldots\right\}
. By the definition of
,
must be disjoint from
.
is also multiplicatively closed. Thus, by a variant of Krull's theorem, there exists a prime ideal
that contains
and is still disjoint from
(see Prime ideal). Since
contains
, but not
, this shows that
is not in the intersection of prime ideals containing
. This finishes the proof. The statement may be strengthened a bit: the radical of
is the intersection of all prime ideals of
that are
minimal among those containing
.
- Specializing the last point, the nilradical (the set of all nilpotent elements) is equal to the intersection of all prime ideals of
[2] This property is seen to be equivalent to the former via the natural map
, which yields a
bijection
:
defined by
u\colonJ\mapstoJ/I=\lbracer+I\midr\inJ\rbrace.
[3] [4]
in a ring
is radical if and only if the
quotient ring
is
reduced.
- The radical of a homogeneous ideal is homogeneous.
- The radical of an intersection of ideals is equal to the intersection of their radicals:
\sqrt{I\capJ}=\sqrt{I}\cap\sqrt{J}
.
- The radical of a primary ideal is prime. If the radical of an ideal
is maximal, then
is primary.
is an ideal,
. Since prime ideals are radical ideals,
for any prime ideal
.
be ideals of a ring
. If
are comaximal, then
are comaximal.
[5]
be a
finitely generated module over a
Noetherian ring
. Then
where
is the
support of
and
is the set of
associated primes of
.
Applications
The primary motivation in studying radicals is Hilbert's Nullstellensatz in commutative algebra. One version of this celebrated theorem states that for any ideal
in the
polynomial ring
over an
algebraically closed field
, one has
\operatorname{I}(\operatorname{V}(J))=\sqrt{J}
where
\operatorname{V}(J)=\left\{x\inkn\midf(x)=0forallf\inJ\right\}
and
\operatorname{I}(V)=\{f\ink[x1,x2,\ldotsxn]\midf(x)=0forallx\inV\}.
Geometrically, this says that if a
variety
is cut out by the
polynomial equations
, then the only other polynomials that vanish on
are those in the radical of the ideal
.
Another way of putting it: the composition
\operatorname{I}(\operatorname{V}(-))=\sqrt{-}
is a
closure operator on the set of ideals of a ring.
See also
References
- Book: Atiyah. Michael Francis. Michael Atiyah. 1994. Addison-Wesley. Ian G. . Macdonald. Ian G. Macdonald. Introduction to Commutative Algebra. 0-201-40751-5. Reading, MA.
- Book: Eisenbud, David . David Eisenbud . Commutative algebra with a view toward algebraic geometry . New York . . . 150 . 1995 . 1322960 . 0-387-94268-8 . none.
Notes and References
- Here is a direct proof that
is an ideal. Start with
with some powers
. To show that
, we use the binomial theorem (which holds for any commutative ring):
style(a+b)n+m-1
\binom{n+m-1}{i}aibn+m-1-i.
For each
, we have either
or
. Thus, in each term
, one of the exponents will be large enough to make that factor lie in
. Since any element of
times an element of
lies in
(as
is an ideal), this term lies in
. Hence
, and so
.To finish checking that the radical is an ideal, take
with
, and any
. Then
, so
. Thus the radical is an ideal.
- For a direct proof, see also the characterisation of the nilradical of a ring.
- Book: Aluffi, Paolo. Algebra: Chapter 0. AMS. 2009. 978-0-8218-4781-7. 142.
- This fact is also known as fourth isomorphism theorem.
- Proof: implies
.