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What means a birational isomorphism from an elliptic curve to $P^1$? An elliptic curve is given by equation $y^2=x^3+p\cdot x+ q$.

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    The question is not clearly written (it is hard to understand the second sentence beginning with "But...") Anyway, there is no such thing as a birational isomorphism (or a regular isomorphism) from an elliptic curve to $\mathbf P^1$, because elliptic curves are not rational. – 2017-01-17
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    @Nefertiti I try to understand why an elliptic curve isn't birational to $P^1$. – 2017-01-17
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    OK, that wasn't clear from the question --- it sounds like you were maybe asking what is the definition of birational ismorphism. For why an elliptic curve isn't birational to $\mathbf P^1$, there are different possible answers depending on your background knowledge: you can use the ideas of genus, canonical class, topology of the underlying surface, etc. – 2017-01-17
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    @Nefertiti I read a proof of the statement: '' An elliptic curves is not rational'' and I don't understand the definition of birational/biregular isomorphism. The proof use the infinite descent method. First, I have to notice that there exists some linear transformations such that $y^2=x\cdot(x-1)\cdot(x-a)$. To second step, I suppose that the curve is rational. Then there exists $p,q,r,s$ polynomials such that $r^2\cdot q^3=s^2\cdot p\cdot(p-q)\cdot (p-aq)$. How can be deduced the last sentence? – 2017-01-17

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Let's work over an algebraically closed field $K$ of characteristic $\neq 2, 3$, so that the equation of the elliptic curve $C$ can be assumed of the form $$y^2 = x(x - 1)(x - c).$$ after some change of coordinates, as you mentioned in the comments.

Next, assume for a contradiction that there exists a birational map from $\mathbf P^1_K$ to the elliptic curve $C$. This is the same as saying that $K(t)$, the function field of $\mathbf P^1_K$, is isomorphic to the function field of $C$, which is $K(x,y)$ - the fraction field of $$K[x, y]/(y^2 - x(x-1)(x-c)).$$ Let $f, g \in K(t)$ corresponds to $x$ and $y$ under this isomorphism. Writing $f = p/q$ and $g = r/s$, for polynomials $p, q, r, s \in K[t]$, we have $$\left(\frac r s\right)^2 = \frac p q \left(\frac p q - 1\right)\left(\frac p q - c\right)$$ or, equivalently, $$q^3r^2 = s^2p(p -q)(p -cq).$$

Would you now know how to proceed?

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    Thank you! I try to understand myself. I have just one more question: Why the function field of $\mathbb{P}^1$ is $K(t)$? – 2017-01-18
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    @rafa recall that the coordinate ring of the affine line $\mathbf A^1_K $ is $K[t]$, where you should think of $t$ as a coordinate in $\mathbf A^1_K$. The function field of $\mathbf A^ 1$ is thus $K(t)$. Since $\mathbf A^1_K$ can be identified with the open dense subset $\mathbf P^1_K - \{\infty\}$ of the projective line, we must have $$K(\mathbf P_K^1) = K(\mathbf A^1_K) = K(t).$$ Again, you can think of $t$ as a coordinate in $\mathbf P^1$. – 2017-01-18
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    @AlexMacedo How do we proceed from last step. I tried to think a lot. – 2018-01-27