How do I show that this morphism from a smooth curve to $\mathbb{P}^m$ is an isomorphism onto it's image with image being smooth? This is a problem from Silverman, Arithmetic of Elliptic Curves, Chapter 3 (basically an AG problem).
Assume $K$ is algebraically closed and $C$ is a smooth curve of genus $g$, point $P \in C$, and $n \geq 2g+1$ an integer. We choose a basis of $L(n(P))$, the vector space associated with a divisor $D$, $\{f  \in K(C)^{×} | div(f) \geq -D\} \cup \{0\}$
A basis be ${f_0, f_1,...f_m}$ and let's define a map,$ \phi: [f_0, ...f_m]:C \to \mathbb{P}^m$
Then we have to show 
(1) image of $C$ under map $ \phi : C \to C'$ is a curve and I was able to show this by showing how $\phi$ is a morphism and takes closed sets to closed sets and preserves irreducibility. 
(2) degree of map is $1$.
(3) $C'$ is smooth and $ \phi: C \to  C'$ is an isomorphism.
I want to show first that $C'$ is smooth, if I'm able to do that then since $\phi$ is a morphism it'll follow that it's also an isomorphism. But I am clueless how to go on about it. 
Another way we can do is show $\phi$ is injective (that's all left to show it's an isomorphism) but then again how is $C'$ is smooth? And how do I show $\phi$ is injective in the first place?
Any hint/help is appreciated!
 A: Since $n\geq 2g+1$, $nP$ is an very ample divisor. And these properties follow. You can check this at Hartshorne's Algebraic Geometry, chapter IV - section 3.
However I'll try to make it simple.
Suppose that these exist $Q_1,Q_2\in C$ such that $\phi(Q_1)=\phi(Q_2)$. This would imply that every element of $L(nP)$ that vanishes at one of these points must vanish on the other i.e. $L(nP-Q_1) = L(nP- Q_1-Q_2)$.  However, by the Riemann-Roch theorem together with $n\geq 2g+1$ shows that $l(nP-Q_1) = n-g$ and $l(nP-Q_1-Q_2) = n-g-1$ (See corollary 5.5 right after Silverman states the Riemann-Roch theorem).
Low let us see why $C'=\phi(C)$ is smooth. Pick a point $Q'=\phi(Q) \in C'$. We have $C'$ is singular at $Q'$  if and only if a hyperplane intersects $C'$ at $Q'$ with multiplicity at least two. Since the pre-image of these hylperplanes are given by the elements of $L(nP-Q)$ we have that $C'$ is singular if and only if  $L(nP-Q) = L(nP-2Q)$. This is not possible, again by Riemann-Roch.
Let me try to improve the assertion "$C'$ is singular at $Q'$ iff $L(nP-Q) = L(nP-2Q)$".
First, the multiplicity $m_{Q'}(C')$ is defined by the minimum of the intersection multiplicities of $C'$ with an hyperplane at $Q'$. We have that $m_{Q'}(C')=1$ iff $C'$ is regular at $Q'$. 
Let $H$ be a hyperplane given by $\sum_{j=0}^n a_jx_j = 0$. Then $g= \sum_{j=0}^n a_jf_j \in L(nP)$. Conversely, given $g =\sum_{j=0}^n a_jf_j\in L(nP)$, we build the hyperplane $H=\{\sum_{j=0}^n a_jx_j = 0\}$. We have that $$Q'\in H \Leftrightarrow g(Q) =0 \Leftrightarrow g\in L(nP -Q)$$
The vanishing order of $g$ at $Q$ is exactly the intersection multiplicity between $C'$ and $H$ at $Q'$. Hence, if $Q'$ is a singular point for $C'$, $m_{Q'}(C')\geq 2$ and any $g\in  L(nP -Q)$ vanishes at $Q$ with order at least two. This means that $L(nP -Q) \subset L(nP-2Q)$ and the other inclusion is obvious.
