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Given the canonical map $\mathbb{C}[x] \to \mathbb{C}[x,y]/(y^2 - x(x-1)(x-2))$, how can I compute the norm of a $\mathbb{C}(x)$-algebra morphism $$ \text{Frac}\left( \frac{\mathbb{C}[x,y]}{(y^2 - x(x-1)(x-2))} \right) \xrightarrow{\cdot r} \text{Frac}\left( \frac{\mathbb{C}[x,y]}{(y^2 - x(x-1)(x-2))} \right) $$ for some $r \in \mathbb{C}(x)$? I am trying to understand how to construct example computations of proper pushforward for maps of curves using the formula at the bottom of page 9 of http://www.cmi.ac.in/~asengupta/Intersection_Theory.pdf

Thank you, Mohan. Given $$f + g\cdot y \in \text{Frac}\left( \frac{\mathbb{C}[x,y]}{(y^2 - x(x-1)(x-2))} \right)$$ multiplying this element by $y$ gives $$ f\cdot y + g\cdot x(x-1)(x-2) $$ hence the matrix of the morphism is given by $$ \begin{bmatrix} 0 & x(x-1)(x-2) \\ 1& 0 \end{bmatrix} $$ giving a norm of $-x(x-1)(x-2)$.

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    The field in question is a two dimensional vector space over $\mathbb{C}(x)$ generated by $1,y$. So, multiplication by $r$ is just the diagonal matrix $[r,r]$ and thus the determinant, which is the norm is just $r^2$. The more interesting case is $r$ in the bigger filed and I will let you check its norm by writing down the $2\times 2$ matrix. – 2017-02-06
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    You still have confusion about writing the matrix using the basis. You should not have a $y$ in the norm, since norm must be in $\mathbb{C}(x)$. So, carefully check. – 2017-02-06

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As Mohan points out in the comments, you have a small mistake in your calculation. Let $A = \mathbb{C}[x,y]/(y^2 - x(x-1)(x-2))$ and $K = \operatorname{Frac}(A)$. The matrix of the multiplication-by-$y$ map \begin{align*} \lambda_y: K &\to K\\ g(x,y) &\mapsto y g(x,y) \end{align*} with respect to the basis $1,y$ is \begin{align*} \begin{bmatrix} 0 & x(x-1)(x-2) \\ 1 & 0 \end{bmatrix} \end{align*} and the norm is the determinant of this matrix. Note that this only gives you the norm of $y$: to compute the norm of a general element $g + hy$ you can either repeat the computation or use the fact that $\lambda: K \to \operatorname{End}(K)$ is a $\mathbb{C}(x)$-algebra homomorphism.

Here's a slightly different argument you could make. The extension $K/\mathbb{C}(x)$ is of degree $2$, hence is Galois. Thus an equivalent definition of the norm of $\alpha \in K$ is the product of all Galois conjugates of $\alpha$. The conjugate of $g + hy$ is simply $g - hy$, so $$ N_{K/\mathbb{C}(x)}(g + hy) = (g + hy)(g - hy) = g^2 - h^2 y^2 = g^2 - h^2 x(x-1)(x-2) \, . $$

Compare this to the norm of a quadratic number field: if $K = \mathbb{Q}[\sqrt{D}]$ for $D$ square-free, then for $\alpha = a + b\sqrt{D}$, the norm of $\alpha$ is \begin{align*} N(\alpha) = \alpha \sigma(\alpha) = (a + b\sqrt{D})(a - b\sqrt{D}) = a^2 - b^2D \end{align*} where $\sigma$ is the nontrivial element of $\operatorname{Gal}(\mathbb{Q}(\sqrt{D})/\mathbb{Q})$.