Question #184205

For an ideal I of a commutative ring R, define

√I {x belongs to R |x^n belongs to I |for some n ∈N}. Show that

i) √I is an ideal of R.

ii) I ⊆ √I.

iii) I ≠ √I in some cases.

b) Is R/I×R/J=R×R/I×J ,for any two ideals I and J of a ring R ? Give reasons for your

answer


1
Expert's answer
2021-05-07T09:15:08-0400

1) Let x,yIx, y\in\sqrt{I} . There exist n,mNn,m\in \mathbb{N} such that xn,ymIx^n,y^m\in I. Then, (x+y)n+m=k=0n+m(n+mk)xkyn+mk=ymk=0n(n+mk)xkynk+xnk=n+1n+m(n+mk)xknyn+mkI(x+y)^{n+m}=\sum\limits_{k=0}^{n+m}\binom{n+m}{k}x^ky^{n+m-k}=y^m\sum\limits_{k=0}^{n}\binom{n+m}{k}x^ky^{n-k}+x^n\sum\limits_{k=n+1}^{n+m}\binom{n+m}{k}x^{k-n}y^{n+m-k}\in I

Hence (x+y)I(x+y)\in \sqrt{I}

For any aR (ax)n=anxnIa\in R\ (ax)^n=a^nx^n\in I , hence axIax\in \sqrt{I}

Therefore, I\sqrt{I} is an ideal.


2) xInNxxnI\forall x\in I\,\forall n\in\mathbb{N}x\, x^n\in I , hence xIx\in \sqrt{I} and III\subset \sqrt{I} .


3) Let R=ZR=\mathbb{Z}, I=(4)I=(4), I={xZ:nN 4xn}={xZ:2x}=(2)\sqrt{I}=\{x\in \mathbb{Z}: \exists n\in\mathbb{N}\ 4|x^n\}=\{x\in \mathbb{Z}: 2|x\}=(2). So, III\ne\sqrt{I} .


4) R/I×R/J=(R×R)/(I×J)R/I\times R/J=(R\times R)/(I\times J)

Indeed, I×JI\times J is an ideal in R×RR\times R:

i1,i2Ij1,j2J i1+i2I,j1+j2J\forall i_1,i_2\in I \forall j_1,j_2\in J\ i_1+i_2\in I, j_1+j_2\in J, hence (i1+i2,j1+j2)I×J(i_1+i_2,j_1+j_2)\in I\times J

(i,j)I×J (a,b)R×R aiI, bjJ, (ai,bj)I×J\forall (i,j)\in I\times J\ \forall (a,b)\in R\times R\ ai\in I,\ bj\in J,\ (ai,bj)\in I\times J


Define f:R/I×R/J(R×R)/(I×J)f:R/I\times R/J\to(R\times R)/(I\times J) by the formula f(x+I,y+J)=(x,y)+(I×J)f(x+I,y+J)=(x,y)+(I\times J)


This definition is correct, since if x1x2=iIx_1-x_2=i\in I and y1y2=jJy_1-y_2=j\in J then

f(x1,y1)=f(x2+i,y2+j)=(x2+i,y2+j)+(I×J)=(x2,y2)+(I×J)=f(x2,y2)f(x_1,y_1)=f(x_2+i,y_2+j)=(x_2+i,y_2+j)+(I\times J)=(x_2,y_2)+(I\times J)=f(x_2,y_2)

Since I,JI,J are ideals in RR , and I×JI\times J is an ideal in R×RR\times R , we have

f(x1+x2+I,y1+y2+J)=(x1,y1)+(x2,y2)+(I×J)=f(x1+I,y1+J)+f(x2+I,y2+J)f(x_1+x_2+I,y_1+y_2+J)=(x_1,y_1)+(x_2,y_2)+(I\times J)=f(x_1+I,y_1+J) + f(x_2+I,y_2+J)

f(ax+I,by+J)=(ax,by)+(I×J)=(a,b)f(x,y)f(ax+I,by+J)=(ax, by)+(I\times J)=(a,b)f(x,y)

therefore, f is homomorphism.


f is surjective, since for any class (x,y)+I×JR×R/(I×J)(x,y)+I\times J\in R\times R/(I\times J) we have (x,y)+I×J=f(x+I,y+J)(x,y)+I\times J=f(x+I,y+J)


f is injective, since if f(x+I,y+J)=(0,0)+(I×J)f(x+I,y+J)=(0,0)+(I\times J), then xIx\in I and yJy\in J and (x+I,y+J)=(0+I,0+J)(x+I,y+J)=(0+I,0+J) is zero in f:R/I×R/Jf:R/I\times R/J .

Therefore, f is an isomorphism.


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