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'' $  A $
in a commutative associative ring  $  R $''

The set of all elements  $  b \in R $
some power of which is contained in  $  A $. 
This set is denoted by  $  \sqrt A $. 
It is an [[Ideal|ideal]] in  $  R $; 
moreover,  $  \sqrt A \supset A $
and  $  \sqrt {\sqrt A } = \sqrt A $.

A generalization of this idea is that of the radical of a submodule. Let  $  M $
be a [[Module|module]] over  $  R $
and let  $  N $
be a submodule of  $  M $. 
The radical of the submodule  $  N $
is the set of all elements  $  a \in R $
such that  $  a  ^ {n} M \subset  N $
for some integer  $  n $(
in general, depending on  $  a $). 
The radical of a submodule is an ideal in  $  R $.

====Comments====
Consider the quotient ring  $  R/A $
and the natural quotient homomorphism  $  \pi :  R \rightarrow R/A $. 
The radical of  $  A $
is the inverse image of the nil radical (cf. [[Nil ideal|Nil ideal]]) of  $  R/A $.

Radical ideal is the phrase sometimes used to denote an ideal that is equal to its radical.

Let  $  k $
be an algebraically closed field. To each ideal  $  A \subset  R [ X _ {1} \dots X _ {n} ] $
one associates the algebraic set  $  V( A) \subset  k  ^ {n} $, 
$  V( A) = \{ {a = ( a _ {1} \dots a _ {n} ) \in k  ^ {n} } : {f( a) = 0  \textrm{ for  all  }   f \in A } \} $. 
The Hilbert Nullstellensatz says that  $  \{ {g \in R } : {g( x) = 0  \textrm{ for  all  }   x \in V( A) } \} = \sqrt A $. 
Thus there is in this setting a bijective correspondence between radical ideals and algebraic sets.

In the setting of affine schemes (cf. [[Affine scheme|Affine scheme]])  $   \mathop{\rm Spec} ( R) $
this takes the following form. To each ideal  $  A \subset  R $
one associates the closed subspace  $   \mathop{\rm Spec} ( R/A) = V( A) = \{ {\mathfrak p } : {\mathfrak p  \textrm{ is  a  "prime"  ideal  such  that  }   \mathfrak p \supset A } \} $. 
Conversely, to each closed subspace  $  V \subset   \mathop{\rm Spec} ( R) $
one associates the ideal  $  I( V)= \{ {f \in R } : {f \in \mathfrak p  \textrm{ for  all  }   \mathfrak p \in V } \} $. 
Then again  $  IV( A) = \sqrt A $
because  $  \sqrt A $
is the intersection of all prime ideals containing  $  A $, 
and, again,  $  I $
and  $  V $
set up a bijective correspondence between radical ideals and closed subsets of  $   \mathop{\rm Spec} ( R) $. 
The difference with the setting of the Nullstellensatz is that in that case only prime ideals of the form  $  ( X _ {1} - a _ {1} \dots X _ {n} - a _ {n} ) $
are considered.

====References====
<table><TR><TD valign="top">[a1]</TD> <TD valign="top">  P.M. Cohn,   "Algebra" , '''II''' , Wiley  (1977)  pp. Sects. 11.2, 11.10</TD></TR><TR><TD valign="top">[a2]</TD> <TD valign="top">  A. Grothendieck,   "Eléments de géométrie algébrique. I: Le language des schémas"  ''Publ. Math. IHES'' , '''20'''  (1960)  pp. 80</TD></TR></table>