Hi everyone,

Let V be a vector space over Z_{p}(p a prime number), then every nonempty subset of V that isclosed under additionis a subspace of V.

Any ideas?

Thanks

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- May 12th 2012, 08:42 AMloui1410Vector 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 - May 12th 2012, 09:05 AMHallsofIvyRe: Vector space - proof or counterexample?
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 $\displaystyle Z_p$ then "closed under addition" is sufficient- that "closed under addition" implies "closed under scalar multiplication". Can you show that, for n in $\displaystyle Z_p$, nv is just v added to itself n times? - May 12th 2012, 09:13 AMDevenoRe: Vector space - proof or counterexample?
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). - May 12th 2012, 09:30 AMloui1410Re: Vector space - proof or counterexample?
Perfect, thanks! :)