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Distinct lines in the projective plane are in general position when no three are concurrent. Their union is a reduced plane curve with exactly $\left(\genfrac{}{}{0ex}{}{d}{2}\right)$ singular points, all ordinary pairwise crossings, if there are $d$ lines. An explicit family is $y=t_{i}x+t_i^{2}z$ for distinct complex $t_i$: the intersection of the $i$th and $j$th lines is $[-t_i-t_j:-t_i{t}_j:1]$, and substituting it into a third equation gives $-(t_k-t_i)(t_k-t_j)\ne 0$.

Solutions to all three questions of the University of Oxford Part B algebraic curve paper, sat on 6 June 2023. Source: Oxford's released examination PDF, compilation pages 218–220; paper pages 1–3. The solutions use the paper's labels. Curves in Question 1 are understood to be reduced plane curves; the qualification concerning lines in 2(a), and the unnecessary restriction on the half-period values in 3(d), are discussed where they arise.

Let $d_i=\mathrm{deg}{C}_i$. Under the reduced plane curve convention, the distinct irreducible components have ${\sum }_i{d}_i=d$. Away from intersections with other components, a point of $C_i$ is nonsingular on the union exactly when it is nonsingular on $C_i$. For example, a local equation of the union is the equation of $C_i$ multiplied by a nonvanishing factor. At an intersection of two components, differentiating their product makes every first partial derivative vanish. Therefore

Part (b) bounds the number of singular points of $C_i$ by $d_i(d_i-1)/2$. For each distinct pair of irreducible components, Bézout's theorem bounds the number of distinct intersections by $d_i{d}_j$. The cardinality of a union is at most the sum of the cardinalities, even if some points have been counted several times. Hence

The reduced convention matters. If one instead allows the equation $L^2=0$ as a nonreduced curve and uses its equation to test singularity, every point on its supporting line has vanishing first partial derivatives. A finite bound would then be false.

For a complex reduced plane curve of degree $d$, the number $s$ of distinct singular points is at most $d(d-1)/2$. For an irreducible polynomial $F$ of degree $d>1$, $F_x$ is nonzero, since a binary homogeneous polynomial over $\mathbb{C}$ factors into linear factors. It has no common factor with $F$. Each singular point contributes at least two to the intersection with the polar curve $F_x=0$, so Bézout's theorem gives $2s\le d(d-1)$.

For an irreducible curve with $d>1$, one can improve this to $s\le d(d-1)/2-1$. Place a nonsingular point at $p=[1:0:0]$. The Euler theorem for homogeneous functions gives $F_x(p)=0$ and $\nabla F_x(p)=(d-1)\nabla F(p)$, so the polar curve is nonsingular and tangent to the curve at $p$. This consumes at least two further units of intersection multiplicity. For a reducible curve with component degrees $d_i$, add the irreducible bounds and at most $d_i{d}_j$ intersections per component pair. The sum is $d(d-1)/2$ because $\sum d_i=d$.