I am having trouble with proving or disproving Ring Isomorphisms.

Is it necessary that if two (groups/rings) are isomorphic then they have same properties???

One more thing ... in my book ... they have given "for disproving a ring isomorphism just find an equation which makes sense in both rings,which is solvable in one and not in the other"

Why is this so??

love arun

(i) f(a+b)=f(a)+f(b)

[Note that the first addition is in the ring R and the second in S (we should write for example +_R etc but this would result in a horrible notation). This means that if you have two elements in R then it doesn't matter if you add them in R and then map the result to S, or if you first map them to S and then add the results]
(ii) f(a b)=f(a)f(b)

[Similar comments to (i)]
(iii) f(1_R)=1_S

[From (ii) you have f(a)=f(a)f(1) where you deduce that either f(1) is 1 or 0. (iii) is there to ensure that it is not 0]

Saying that two rings are isomorphic means that 'they have the same properties' e.g.

If multiplication is commutative in R then it is in S

If an element has a (multiplicative) inverse in R then its image also has one in S.

If both rings are finite then they have the same number of elements

etc.

What your book is saying is the following:

Suppose a³+a+1=0 in R and suppose there is an isomorphism f:R® S then

0=f(0)=f(a³+a+1)=f(a³)+f(a)+f(1)=[f(a)]³+f(a)+(1). (*)

Now if b³+b+1=0 does not hold for any b in S then it means that (*) cannot be satisfied so the two rings are not isomorphic.

Demetres

Thanks for the clarification Demetres....
i think i understood everything u have said

love arun