3
$\begingroup$

I need to determine if the following statement is true or false, if it's true, I need to prove it, else I need to give a counterexample:

Let $V$ be a vector space over $\mathbb{Z}_p$ when $p$ is a prime, so each subset (not empty) of $V$ that has closure under addition is a sub-vectorspace of $V$.

What I have tried:

I am not sure if this statement is true, but if it is I thought about using a Theorem that says: Any subset of vector space $V$ which is not empty is a sub-vectorspace of $V\Leftrightarrow V$ has closure under addition and multiplication(with scalars).$\alpha*v=(1+1+1+1...)v$ and because the subset has closure under addition, it will be under multiplication too.

$\endgroup$
1
  • 4
    $\begingroup$ Yes, that looks fine. You are also using the fact that $-1 \equiv (p-1)$ as a scalar. $\endgroup$ Dec 3, 2013 at 16:11

1 Answer 1

1
$\begingroup$

Since the field is $\mathbb{F}_p$, scalar multiplication is just addition: $\lambda x=x+\cdots +x$, $\lambda\in \{0,1,\ldots ,p-1\}$. Now $(V.+)$ is an abelian group. If $U$ is a nonempty subset closed under addition, then $(U,+)$ is a subgroup of $(V,+)$, because $x-y=x+(p-1)y$. By the above remark, it is a sub-vectorspace of $V$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .