1) Let x,y∈I . There exist n,m∈N such that xn,ym∈I. Then, (x+y)n+m=k=0∑n+m(kn+m)xkyn+m−k=ymk=0∑n(kn+m)xkyn−k+xnk=n+1∑n+m(kn+m)xk−nyn+m−k∈I
Hence (x+y)∈I
For any a∈R (ax)n=anxn∈I , hence ax∈I
Therefore, I is an ideal.
2) ∀x∈I∀n∈Nxxn∈I , hence x∈I and I⊂I .
3) Let R=Z, I=(4), I={x∈Z:∃n∈N 4∣xn}={x∈Z:2∣x}=(2). So, I=I .
4) R/I×R/J=(R×R)/(I×J)
Indeed, I×J is an ideal in R×R:
∀i1,i2∈I∀j1,j2∈J i1+i2∈I,j1+j2∈J, hence (i1+i2,j1+j2)∈I×J
∀(i,j)∈I×J ∀(a,b)∈R×R ai∈I, bj∈J, (ai,bj)∈I×J
Define f:R/I×R/J→(R×R)/(I×J) by the formula f(x+I,y+J)=(x,y)+(I×J)
This definition is correct, since if x1−x2=i∈I and y1−y2=j∈J then
f(x1,y1)=f(x2+i,y2+j)=(x2+i,y2+j)+(I×J)=(x2,y2)+(I×J)=f(x2,y2)
Since I,J are ideals in R , and I×J is an ideal in R×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(ax+I,by+J)=(ax,by)+(I×J)=(a,b)f(x,y)
therefore, f is homomorphism.
f is surjective, since for any class (x,y)+I×J∈R×R/(I×J) we have (x,y)+I×J=f(x+I,y+J)
f is injective, since if f(x+I,y+J)=(0,0)+(I×J), then x∈I and y∈J and (x+I,y+J)=(0+I,0+J) is zero in f:R/I×R/J .
Therefore, f is an isomorphism.
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