Ring Isomorphisms

By Arun Iyer on Tuesday, November 05, 2002 - 05:55 am:

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

By Demetres Christofides on Tuesday, November 05, 2002 - 08:56 am:

Definition: Two rings

R

W

are isomorphic if there is a bijection

f : R \to S

such that (for all

a

b

R

) the following hold

ϕ (a + b) = ϕ (a) + ϕ (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) $ϕ (a b) = ϕ (a) ϕ (b)$

[Similar comments to (i)]
(iii) $ϕ (1_{R}) = 1_{S}$

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

[Now that I think of it (iii) is not necessary but never mind]

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 $a^{3} + a + 1 = 0$ in $R$ and suppose there is an isomorphism $f : R \to S$ then

$0 = ϕ (0) = ϕ (a^{3} + a + 1) = ϕ (a^{3}) + ϕ (a) + ϕ (1) = [ϕ (a {)]}^{3} + ϕ (a) + (1)$ . (*)

Now if $b^{3} + b + 1 = 0$ does not hold for any $b$ in $S$ then it means that (*) cannot be satisfied so the two rings are not isomorphic.

Demetres

By Arun Iyer on Tuesday, November 05, 2002 - 09:28 am:

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

love arun