Let R be a commutative ring with unity. Define the nilradical N of R to be the s
ID: 2943149 • Letter: L
Question
Let R be a commutative ring with unity. Define the nilradical N of R to be the set of all nilpotent elements of R, that is elements xR for which there exists a positive integer n such that xn = 0.
(a) Show that N is an ideal of R.
(b) Show that N is equal to the intersection of all prime ideals of R.
(c) The Jacobson radical J of R is the set of all xR such that 1-xy is a unit for all y R. Show
that J is an ideal.
(d) Show that J is equal to the intersection of all maximal ideals of R.
(e) Is J a subset of N, or is N a subset of J, or neither? Explain.
Explanation / Answer
(a) Show that N is an ideal of R:
Nonempty: 0 is an element of R, and 0^n=0 for all positive integers. Thus 0 is an element of N. Thus N is nonempty.
Closure: Let x,y be elements of N. Then there exist positive integers r,s such that x^r=0 and y^s=0. R is a commutative ring, and closed under multiplication, thus xy is an element of R. Then (xy)^(rs) =x^(rs)y^(rs) (since R is a commutative ring) ==> (x^r)^s(y^s)^r ==> 0^s0^r=0. Thus (xy)^rs=0, and xy is an element of N.
Inverse: Again suppose x is an N, and x^r=0. x is an element of R, thus x^-1 is an element of R. Then (x^-1)^s=(x^s)^-1=(0)^-1=0. Thus x^-1 is an element of N.
Super-Closure: Let x be in N, and let k be an element of R. Then xk is an element of R; further (xk)^s=x^sk^s ==> (xk)^s=0*k^s =0. Thus xk is an element of N. Similar result holds for kx, since R is commutative.
Therefore, N is an ideal of R.
Bah. . .sorry I am too tired to finish this. Brain is fried. Sorry.
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