Isomorphism question

Hi,

For a finite field isomorphism can one use the result that two fields of order p are isomorphic or does one have to construct a specific bijective map?

Thanks for any help.

(edited 8 years ago)

Reply 1

Gregorius

14

Original post by DennisC

Hi,

I'm trying to show a subring,

R= \{aT+bI: a,b \in mod5\}

, (T is a matrix in mod 5 and I is the 2x2 Identity if mod 5) of the set of 2x2 matrices mod5 is isomorphic to GF(25). I know that this subring has 25 elements the same as GF(25). Is it then sufficent to show that the subring has multiplicative inverses for all non-zero elements, to show that it is isomorphic, or do I have to define a mapping and show it's a bijective homomorphism still?

If it's the latter I'm not really sure what type of mapping to use...

If the former why is this approach ok to use?

Thanks for any help.

I presume that you are trying to show that R is isomorphic as a field to GF(25) = GF(5^2). If so, then recall that a finite field of order q is unique up to isomorphism. Hence if you show that R is a field, you are done. However, if you are doing this as an exercise for a course, it would be as well to check that you are allowed to assume the result on uniqueness of fields! Otherwise you had better do it the hard way (finding an explicit isomorphism).

Reply 2

DennisC

OP

Original post by Gregorius

I presume that you are trying to show that R is isomorphic as a field to GF(25) = GF(5^2). If so, then recall that a finite field of order q is unique up to isomorphism. Hence if you show that R is a field, you are done. However, if you are doing this as an exercise for a course, it would be as well to check that you are allowed to assume the result on uniqueness of fields! Otherwise you had better do it the hard way (finding an explicit isomorphism).

I thought that to be the case thanks. If I do have to do it the had way, do you have any insights into what type of mapping to use, I was initally thinking something like the Frobenius mapping, but not really sure it would work...