Prove the equivalence of a) $ V $ is a point; b) $ \Gamma(V)\simeq k $ and c) dim$ _k \Gamma(V)<\infty $ Let $ k $ be an arbitrary field, and $ V\subseteq \mathbb{A}^n(k) $ a non-empty variety. I want to show that the following are equivalent:
a) $ V $ is a point;
b) $ \Gamma(V)\simeq k $;
c) dim$ _k \Gamma(V)<\infty $.
Can someone help me to show that $ c) \Rightarrow a) $? Also, is my proof of $ a) \Rightarrow b) $ correct? 
I prove $ a) \Rightarrow b) $ like this: For any variety $ V $ and for any $ P\in V $, consider the function
\begin{align*}
\phi:\mathcal{O}_P(V)&\to k\\
f&\mapsto f(P).
\end{align*}
Since $ k\subseteq \mathcal{O}_P(V) $, we see that $ \phi $ is surjective by considering the constant functions in $ \mathcal{O}_P(V) $. Hence, by the First Isomorphism Theorem, 
$$\mathcal{O}_P(V)/\text{ker}(\phi)\cong k.      (1)$$
Actually, ker$ (\phi)=m_P(V) $ by definition, but that is not so important for us here(?). Assume now that $ V=P $. Then, (by Proposition 2 of chapter 2.4 in Algebraic Curves by Fulton), 
$$ \Gamma(P)=\Gamma(V)=\bigcup_{P\in V}\mathcal{O}_P(V)=\bigcup_{P\in P}\mathcal{O}_P(V)=\mathcal{O}_P(V) $$
where I have used the fact that $ \mathcal{O}_P(V) $ is only defined for $ P\in V $. Hence
$$ \text{ker}(\phi)=\{ f\in \Gamma(P):f(P)=0 \} $$
witch is zero since only functions in $ I(P) $ vanish modulo $ I(P) $. Hence (1) yields
$$ \Gamma(V)=\Gamma(P)/\{0\}=\mathcal{O}_P(V)/\text{ker}(\phi)\cong k. $$
I prove $ b) \Rightarrow c) $ like this: Since $ \Gamma(V)\simeq k $, we have that dim$ _k \Gamma(V)=\text{dim}_k(k)=1<\infty $.
 A: Assume toward a contradiction that there is $y\in \Gamma(V)$ such that $y\not \in k$. If $1,y,y^2\cdots$ are independent then the dimension of $\Gamma(V)$ is $\infty$. If they are dependent over $k$ then for some polynomial $f$, (assumed to be of lowest degree)
$$f(y)=0$$ identically on $V$. 
And now 
$$f(y)=\prod (y-\alpha_i)$$ where $\alpha_i$ are algebraic over $k$. Since $V$ is non empty there is $P\in V$ and we have 
$$\prod (y(P)-\alpha_i)=0$$ 
this implies that for some $j$, 
$$y(P)-\alpha_j=0,$$ so $\alpha_j\in k$ and $f$ is reducible. Since $V$ is irreducible of the factors of $f$, say $g$ will satisfy $g(y)=0$ on $V$ and this contradicts the choice of $f$. 
It follows that $\Gamma(V)=k$. And thus that $I(V)$ is a maximal ideal so $V$ is one point.  
A: Yes your proof is correct. 
(Also probably your definition of a variety is an irreducible algebraic set, since the union of two points verify $\dim_k(\Gamma(V)) = 2$.)
For your implication, consider a variety $V$ with strictly positive dimension and $x_1, \dots, x_r \in V$ distinct. It's easy to find $f_1, \dots, f_r$ with $f_i$ with $f_i(x_j) = \delta_{ij}$. In particular $\dim_k \Gamma(V) \geq r$, and $r$ was arbitrary.
Edit : just in case, for finding the $f_i$, consider a linear form $\ell_{ij}$ such that $\ell_{ij}(x_i) = 1$ and $\ell_{ij}(x_j) = 0$. Now you can verify that $f_i(x) = \prod_{j \neq i}\ell_{ij}(x)$ verify $f_i(x_j) = \delta_{ij}$.
