2 added 115 characters in body

The following question came up in my research. I suspect that it has a slick answer, but I can't seem to find it.

Fix an integer n>=2 and a prime p. Define X(n) to be the set of primitive vectors in the Z-module Z^n and Y(n,p) to be the set of non-zero vectors "lines" in the vector space (Z/pZ)^n. Z/pZ)^n (ie the spans of non-zero vectors). There is a natural surjective map f:X(n)-->Y(n,p) ("reduce mod p")p and take the span").

Question : Does there exist a map g:Y(n,p)-->X(n) with the following two properties.

1. f(g(v))=v

1. f(g(L))=L for all v L in Y(n,p).
2. If {v_1,...,v_n} L_1,...,L_n} \subset Y(n,p) is a basis for spans the vector space (Z/pZ)^n, then {g(v_1),...,g(v_n)} g(L_1),...,g(L_n)} is a basis for the Z-module Z^n.

Of course, I expect that the answer is no except in certain simple situations (for instance, it is yes for n=p=2), but I can't seem to find a proof.

EDIT : Oops! I phrased the question incorrectly. Above is a corrected version.

1

# Lifting bases for (Z/pZ)^n to Z^n

The following question came up in my research. I suspect that it has a slick answer, but I can't seem to find it.

Fix an integer n>=2 and a prime p. Define X(n) to be the set of primitive vectors in the Z-module Z^n and Y(n,p) to be the set of non-zero vectors in the vector space (Z/pZ)^n. There is a natural surjective map f:X(n)-->Y(n,p) ("reduce mod p").

Question : Does there exist a map g:Y(n,p)-->X(n) with the following two properties.

1. f(g(v))=v for all v in Y(n,p).
2. If {v_1,...,v_n} \subset Y(n,p) is a basis for the vector space (Z/pZ)^n, then {g(v_1),...,g(v_n)} is a basis for the Z-module Z^n.

Of course, I expect that the answer is no except in certain simple situations (for instance, it is yes for n=p=2), but I can't seem to find a proof.