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In the affine chart $z=1$, the equation is $y^2=x^2(x+1)$, irreducible because $x+1$ is not a square in $k(x)$. The partial derivatives show the unique singular point is $P=[0:0:1]$; the point at infinity is smooth.

A line $y=tx$ through $P$ meets the curve again atThus $t\mapsto [t^2-1:t(t^2-1):1]$ is birational, with inverse $t=y/x$ away from $P$. The incidence closure has one point over every smooth $Q$ and two points over $P$, corresponding to $t=\pm 1$. This is the normalization of a nodal curve and separates the two branches.

Dominated by projective space does not imply being projective space. A smooth quadric surface $Y\simeq {\mathbb{P}}^1\times {\mathbb{P}}^1$ is rational and hence dominated by ${\mathbb{P}}^2$, but is not any ${\mathbb{P}}^k$; for example its Picard rank is $2$, not $1$.

Its homology and chain formulas areFor Lefschetz fixed-point theorem, argue contrapositively: after barycentric subdivision, a fixed-point-free map has a simplicial approximation such that no simplex meets its image. Every diagonal coefficient on chains is then zero, so $L(f)=0$.

The antipodal map on $S^n$ has degree $(-1{)}^{n+1}$. For even $n$ this is $-1$, whereas the identity has degree $1$, so they are not homotopic.

If the tangent field $v$ were nowhere zero, thenwould stay on the sphere because $x\cdot v(x)=0$, and would homotope the identity to the antipodal map. This is impossible for even $n$, so $v$ has a zero.

A primitive root modulo $N$ is a unit of order $\phi (N)$; a nonzero quadratic residue modulo an odd prime $p$ is a square.

If a primitive root $g$ modulo $p$ were $x^2$, then its order would divide $(p-1)/2$, the order of the subgroup of squares, contradicting ${\mathrm{ord}}_p(g)=p-1$.

Modulo $11$, $2$ has order $10$, so the primitive roots are $2^k$ for $\gcd (k,10)=1$:Finally $2^{10}=1024\equiv 56\not\equiv 1(\mathrm{mod}121)$, so the lifting criterion makes $2$ primitive modulo $121$. It is plainly the smallest possible positive primitive root.