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Radical of An Ideal
In commutative ring theory, a branch of mathematics, the radical of an idealI is an ideal such that an element x is in the radical if and only if some power of x is in I. (Taking the radical is called radicalization.) A radical ideal (or semiprime ideal) is an ideal that is equal to its own radical. The radical of a primary ideal is a prime ideal.
This concept is generalized to noncommutative rings in the Semiprime ring article.
Definition
The radical of an ideal I in a commutative ringR, denoted by Rad(I) or ${\sqrt {I}}$, is defined as
(Note that $I\subset {\sqrt {I}}$.)
Intuitively, ${\sqrt {I}}$ is obtained by taking all roots of elements of I within the ring R. Equivalently, ${\sqrt {I}}$ is the pre-image of the ideal of nilpotent elements (the nilradical) in the quotient ring$R/I$. The latter shows ${\sqrt {I}}$ is itself an ideal.^{[Note 1]}
If the radical of I is finitely generated, then some power of ${\sqrt {I}}$ is contained in I.^{[1]} In particular, if I and J are ideals of a noetherian ring, then I and J have the same radical if and only if I contains some power of J and J contains some power of I.
If an ideal I coincides with its own radical, then I is called a radical ideal or semiprime ideal.
The radical of the ideal 4Z of integer multiples of 4 is 2Z.
The radical of 5Z is 5Z.
The radical of 12Z is 6Z.
In general, the radical of mZ is rZ, where r is the product of all distinct prime factors of m, the largest square-free factor of m (see radical of an integer). In fact, this generalizes to an arbitrary ideal (see the Properties section).
Consider the ideal $I=(y^{4})\subset \mathbb {C} [x,y].$ It is trivial to show ${\sqrt {I}}=(y)$ (using the basic property ${\sqrt {I^{n}}}={\sqrt {I}}$), but we give some alternative methods.^{[clarification needed]} The radical ${\sqrt {I}}$ corresponds to the nilradical${\sqrt {0}}$ of the quotient ring $R=\mathbb {C} [x,y]/(y^{4}),$ 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 morphism $R\to \mathbb {C}$ must have $y$ in the kernel in order to have a well-defined morphism (if we said, for example, that the kernel should be $(x,y-1)$ the composition of $\mathbb {C} [x,y]\to R\to \mathbb {C}$ would be $(x,y^{4},y-1)$ which is the same as trying to force $1=0$). Since $\mathbb {C}$ is algebraically closed, every morphism $R\to \mathbb {F}$ must factor through $\mathbb {C} ,$ so we only have the compute the intersection of $\{\ker(\Phi ):\Phi \in {\text{Hom}}(R,\mathbb {C} )\}$ to compute the radical of $(0).$ We then find that ${\sqrt {0}}=(y)\subset R.$
Properties
This section will continue the convention that I is an ideal of a commutative ring R:
It is always true that ${\sqrt {\sqrt {I}}}={\sqrt {I}}$, i.e. radicalization is an idempotent operation. Moreover, ${\sqrt {I}}$ is the smallest radical ideal containing I.
${\sqrt {I}}$ is the intersection of all the prime ideals of R that contain I, 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 ${\sqrt {I}}$. Suppose r is an element of R which is not in ${\sqrt {I}}$, and let S be the set $\{r^{n}\mid n=0,1,2,...\}$. By the definition of ${\sqrt {I}}$, S must be disjoint from I. S is also multiplicatively closed. Thus, by a variant of Krull's theorem, there exists a prime ideal P that contains I and is still disjoint from S. (see prime ideal.) Since P contains I, but not r, this shows that r is not in the intersection of prime ideals containing I. This finishes the proof. The statement may be strengthened a bit: the radical of I is the intersection of all prime ideals of R that are minimal among those containing I.
Specializing the last point, the nilradical (the set of all nilpotent elements) is equal to the intersection of all prime ideals of R.
An ideal I in a ring R is radical if and only if the quotient ringR/I 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: ${\sqrt {I\cap J}}={\sqrt {I}}\cap {\sqrt {J}}$.
The radical of a primary ideal is prime. If the radical of an ideal I is maximal, then I is primary.^{[2]}
If I is an ideal, ${\sqrt {I^{n}}}={\sqrt {I}}$. Since prime ideals are radical ideals, ${\sqrt {P^{n}}}=P$ for any prime ideal P.
Let I, J be ideals of a ring R. If ${\sqrt {I}},{\sqrt {J}}$ are comaximal, then $I,J$ are comaximal.^{[Note 2]}
Let M be a finitely generated module over a noetherian ring R. Then
$\operatorname {V} (J)=\{x\in k^{n}\ |\ f(x)=0{\mbox{ for all }}f\in J\}$
and
$\operatorname {I} (S)=\{f\in k[x_{1},x_{2},\ldots x_{n}]\ |\ f(x)=0{\mbox{ for all }}x\in S\}.$
Geometrically, this says that if a varietyS is cut out by the polynomial equations $f_{1}=0,\ldots ,f_{r}=0$, then the only other polynomials which vanish on S are those in the radical of the ideal $(f_{1},\ldots ,f_{r})$.
Another way of putting it: The composition $\operatorname {I} (\operatorname {V} (-))={\sqrt {-}},$ is a closure operator on the set of ideals of a ring.
^Here is a direct proof. Start with a, b ∈ ${\sqrt {I}}$ with some powers a^{n}, b^{m} ∈ I. To show that a+b ∈ ${\sqrt {I}}$, we use the binomial theorem (which holds for any commutative ring):
For each i, we have either i ≥ n or n+m−1−i ≥ m. Thus, in each term a^{i}b^{n+m−1−i}, one of the exponents will be large enough to make that factor lie in I. Since any element of I times an element of R lies in I (as I is an ideal), this term lies in I. Hence (a+b)^{n+m−1} ∈ I, and a+b ∈ ${\sqrt {I}}$.
To finish checking that the radical is an ideal, take a ∈ ${\sqrt {I}}$ with a^{n} ∈ I, and any r ∈ R. Then (ra)^{n} = r^{n}a^{n} ∈ I, so ra ∈ ${\sqrt {I}}$. Thus the radical is an ideal.