Answer to Question #184205 in Abstract Algebra for Rohit

Question

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

Answer 1

1) Let $x, y\in\sqrt{I}$ . There exist $n,m\in \mathbb{N}$ such that $x^n,y^m\in I$. Then, $(x+y)^{n+m}=\sum\limits_{k=0}^{n+m}\binom{n+m}{k}x^ky^{n+m-k}=y^m\sum\limits_{k=0}^{n}\binom{n+m}{k}x^ky^{n-k}+x^n\sum\limits_{k=n+1}^{n+m}\binom{n+m}{k}x^{k-n}y^{n+m-k}\in I$

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

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

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

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

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

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

Indeed, $I\times J$ is an ideal in $R\times R$:

$\forall i_1,i_2\in I \forall j_1,j_2\in J\ i_1+i_2\in I, j_1+j_2\in J$, hence $(i_1+i_2,j_1+j_2)\in I\times J$

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

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

This definition is correct, since if $x_1-x_2=i\in I$ and $y_1-y_2=j\in J$ then

$f(x_1,y_1)=f(x_2+i,y_2+j)=(x_2+i,y_2+j)+(I\times J)=(x_2,y_2)+(I\times J)=f(x_2,y_2)$

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

$f(x_1+x_2+I,y_1+y_2+J)=(x_1,y_1)+(x_2,y_2)+(I\times J)=f(x_1+I,y_1+J) + f(x_2+I,y_2+J)$

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

therefore, f is homomorphism.

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

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

Therefore, f is an isomorphism.