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For real closed fields this is fairly easy.

First show that for any infinite cardinal k there are 2^k nonisomorphic linear orders of cardinality k

For example if X is a subset of k let A_x be Q+2+Q if x is in k and Q+3+Q if x is not in X. Let L_X be the sum of the A_x for x in k. It is easy to see that L_X is isomorphic to L_Y if and only if X=Y.

If F is a real closed and x and y are infinite element of R we say that x and y are comparable if and only if there are natural numbers m and n such that x is less than y^m and y is less than x^n. The ordering of R induces a linear order L_R of the comparability classes, which we call the ladder of R.

Suppose L is a linear order. Let F be the real algebraic numbers. Let R_L be the real closure of the transcendental extension of the real algebraic numbers F(x_l:l\in L) ordered such that if i is less than j then x_i^n is less than x_j for all n. It's not hard to show that the ladder of R_L is isomorphic to L.

Thus if we start with nonisomorphic orders A and B then the fields R_A and R_B will be nonisomorphic.

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For real closed fields this is fairly easy.

First show that for any infinite cardinal k there are 2^k nonisomorphic linear orders of cardinality k

For example if X is a subset of k let A_x be Q+2+Q if x is in k and Q+3+Q if x is not in X. Let L_X be the sum of the A_x for x in k. It is easy to see that L_X is isomorphic to L_Y if and only if X=Y.

If F is a real closed and x and y are infinite element of R we say that x and y are comparable if and only if there are natural numbers m and n such that x

Suppose L is a linear order. Let F be the real algebraic numbers. Let R_L be the real closure of the transcendental extension of the real algebraic numbers F(x_l:l\in L) ordered such that if i is less than j then x_i^n is less than x_j for all n. It's not hard to show that the ladder of R_L is isomorphic to L.

Thus if we start with nonisomorphic orders A and B then the fields R_A and R_B will be nonisomorphic.