Why is $F(x,y)$ isomorphic to $F(x)(y)$?

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Here $F$ is a field, $F(x,y)$ is the field of rational functions in $x$ and $y$ with coefficients in $F$, and $F(x)(y)$ is the field of rational functions in $y$ with coefficients in $F(x)$, the field of rational functions in $x$ with coefficients in $F$.

I pretty much see intuitively why this is true, but it bugs me that I don't really see it formally. How can rigurously I prove this?

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It is just a formal game:

Consider the inclusion $F(X) \hookrightarrow F(X,Y)$. The universal property of the polynomial ring gives us a homomorphism $$F(X)[Y] \to F(X,Y), Y \mapsto Y$$

, which extends the inclusion above. This map is injective, since $Y$ is not algebraic over $F(X)$ by definition. Then the universal property of the quotient field tells us that this map factors over the quotient field, hence we get a map $$F(X)(Y) \to F(X,Y)$$

As a map of fields, it is clearly injective, and for the sake of surjectivity, we only have to note that $X$ and $Y$ are contained in the image.