Thread: Vector space - proof or counterexample?

Hi everyone,

Let V be a vector space over Z_p (p a prime number), then every nonempty subset of V that is closed under addition is a subspace of V.

Any ideas?
Thanks

To prove that a subset of a vector space over any field is a sub-space you need only prove that it is closed under addition and scalar multiplication. You want to prove that if the field is then "closed under addition" is sufficient- that "closed under addition" implies "closed under scalar multiplication". Can you show that, for n in , nv is just v added to itself n times?

let S be such a set.

for any k in Z_p we have either:

k = 0, or k = 1+1+...+1 (k times).

let u be any element of S. if k ≠ 0, then ku = (1+1+..+1)u = u+u+...+u (k times), which is in S, since S is closed under addition.

if k = 0, we have 0 = 1+1+...+1 (p times), so that 0u = (1+1+...+1)u = u+u+...+u (p times), so 0u = 0 is also in S.

thus S is closed under scalar multiplication. since S is a non-empty subset of V closed under vector addition and scalar multiplication, it is a subspace of V.

(a bit of a blurb as to why this magic works: in Z_n, we get a "ring structure" for free: since 1 generates Z_n as an abelian group, we can use "the law of exponents", to define:

km = m(k1) = (1+1+...+1) + (1+1+...+1) +...+ (1+1+...+1) (where each term in parentheses has "k" ones, and we have "m" such groupings)

[[-if we write such a cyclic group multiplicatively, the use of the "laws of exponents" becomes clearer: 0 becomes e, and k becomes a^k (a being a generator), so we get a 2nd operation

(not the usual group operation) (a^k)*(a^m) := (a^k)^m = a^km, which has identity a-]]

if n = p, a prime, then all of the non-zero elements of Z_p are units, so we have a field, again, just from the fact that we have a cyclic group of order p. in such a group, the additive structure "creates" the multiplicative structure).

Perfect, thanks!