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For a divisor $D$ on a smooth projective curve $X$, defineThe Riemann-Roch theorem stateswhere $K_X$ is the divisor of a nonzero rational differential, $g$ is the genus of $X$, and $\mathrm{deg}D$ is the sum of the divisor coefficients because the ground field is algebraically closed.

Under the coordinate change $u=1/x$ and $v=y/x^n$, we haveHenceThis is a polynomial, and its value at zero is the nonzero leading coefficient of $f$. Thus the even-degree hyperelliptic model has two points $P_{+},P_{-}$ above $u=0$, distinguished by the two values of $v$.

Consider the rational differentialAt a finite branch point $x=a$, square-freeness gives a local parameter $y$, with $x-a$ a nonzero constant times $y^2$ to first order. Hence $dx/y$ is regular and nonzero there. It is also regular and nonzero at finite unramified points.

At infinity,soBecause $v(P_{\pm })\ne 0$ and $u$ is a local parameter at each point, $\omega$ has a zero of order $n-2$ at both. Therefore the canonical divisor of an even-degree hyperelliptic curve isand

Since $\mathrm{deg}{K}_X=2g-2$, it follows that

The function $x$ has a simple pole at each of $P_{+}$ and $P_{-}$. ConsequentlyThese $n-1$ functions are linearly independent. Riemann-Roch with $D=K_X$ gives $\ell (K_X)=g=n-1$, so

With respect to this basis, the canonical map of a hyperelliptic curve isIt identifies $(x,y)$ with $(x,-y)$ for general $x$, and hence is not an embedding.

If $X$ embedded in ${\mathbb{P}}^2$, its smooth plane image would have degree $d\ge 4$: degrees one and two have genus zero, and degree three has genus one, whereas $g=n-1\ge 2$. But the canonical map of a smooth plane curve of degree $d\ge 4$ is an embedding by adjunction, contradicting the preceding calculation. Therefore

The universal coefficient theorem for homology and flatness of $\mathbb{Q}$ giveThus, ifwith $T_i$ torsion, then $H_i(X;\mathbb{Q})\cong {\mathbb{Q}}^{b_i}$.

For a finite triangulable space, its Euler characteristic isIf $K$ is a finite triangulation and $f_i$ is its number of $i$-simplices, then the Euler-Poincare formula isTo prove it over $\mathbb{Q}$, write $Z_i=\ker {\partial }_i$ and $B_i=\mathrm{im}{\partial }_{i+1}$. Rank-nullity and $H_i=Z_i/B_i$ giveAfter multiplying by $(-1{)}^i$ and summing, the two boundary sums cancel, leaving the claimed equality.

The standard simplex ${\Delta }^3$ is contractible, so