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enter image description here I am reading from Topics in Galois theory by Serre,

How this lemma implies that $Q(E)=Q(X)(T_{1},\cdots,T_{n})$

I looked through Hartshorne's Book. Corollary 4.5 says that two varieties $X,Y$ are birationally equivalent if and only if their function field are isomorphic (as $K$ algebra, where $K$ is the base field)

So using this, we get $Q(E)=Q(X\times W)$. Now what ?

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Compute $\mathbb{Q}(X\times W)$. It's clear that $\mathbb{Q}[X\times W]\cong \mathbb{Q}[X]\otimes\mathbb{Q}[W]$, and $\mathbb{Q}[W]\cong \mathbb{Q}[T_1,\cdots,T_n]$ where $n=\dim W$. So $\mathbb{Q}[X\times W]\cong \mathbb{Q}[X]\otimes\mathbb{Q}[T_1,\cdots,T_n]\cong \mathbb{Q}[X][T_1,\cdots,T_n]$. Taking the field of fractions, we get that $\mathbb{Q}(X\times W)\cong \mathbb{Q}(X)(T_1,\cdots,T_n)$.

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    What is the map for this isomorphism, $ \mathbb{Q}[X]\otimes\mathbb{Q}[T_1,\cdots,T_n]\cong \mathbb{Q}[X][T_1,\cdots,T_n]$ and If I am not wrong by $Q[X]$ you are denoting the coordinate ring of the variety $X$. – 2017-02-09
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    $\mathbb{Q}[X]$ is indeed the coordinate ring of $X$. The map is the natural one, sending $f\otimes p(T_1,\cdots,T_n)\mapsto f\cdot p(T_1,\cdots,T_n)$ and extending by linearity. – 2017-02-09