Question

Show that if F_1 and F_2 are free modules over a ring with the

invariant dimension property then rank (F_(1 )⨁ F_2) = rank F_1 +
rank F_2.

Answer 1

If F_1 and F_2 are free modules over a ring with the invariant dimension property, then the rank of their direct sum F_1⨁F_2 is equal to the sum of their ranks.

Step-by-step explanation:

Let F_1 and F_2 be free modules over a ring.

The rank of a free module is defined as the number of basis elements in the module.

We need to show that if F_1 and F_2 have the invariant dimension property, then the rank of their direct sum F_1⨁F_2 is equal to the sum of their ranks, i.e., rank(F_1⨁F_2) = rank(F_1) + rank(F_2).

Since F_1 and F_2 have the invariant dimension property, their ranks are the same as their dimensions.

Let m be the rank of F_1 and n be the rank of F_2. This means F_1 is isomorphic to a free module of rank m and F_2 is isomorphic to a free module of rank n.

Then, F_1⨁F_2 is isomorphic to a free module of rank m + n.

Since the rank of a module is equal to the dimension of the module, we can conclude that rank(F_1⨁F_2) = rank(F_1) + rank(F_2), which is m+n = m+n