Bezout needs algebraic closure: an imaginary conic has no real point

Given: AC The Axiom of Choice, the imaginary conic $C=V(x_0^2+x_1^2+x_2^2)$ and the line $L=V(x_1)$, first over the ordered field $\mathbb{R}$ The reals form a totally ordered field and then over the algebraically closed field $\mathbb{C}$: the complex field construction $\mathbb{C}=\mathbb{R}[x]\mathrm{/}(x^2+1)$ is a field, every element is uniquely $a+bi$, and every nonzero element has inverse $(a-bi)\mathrm{/}(a^2+b^2)$ and the root theorem Fundamental theorem of algebra: every nonconstant complex polynomial has a complex root give the latter assertion.

The derivatives of the quadratic are $2x_0,2x_1,2x_2$. Any repeated irreducible factor would divide all three, impossible since these coordinates have no common nonconstant factor; thus the quadratic is square-free, as is the linear form $x_1$. Their degrees are two and one. Over $\mathbb{C}$ they define curves in the convention of Plane projective curves and their components. Over $\mathbb{R}$, $C(\mathbb{R})$ and $L(\mathbb{R})$ mean the rational zero sets of these homogeneous equations; the degrees refer to the equations (or their geometric curves after extension to $\mathbb{C}$), and no degree is assigned to the empty real point set.

A point $[a_0:a_1:a_2]\in L$ has $a_1=0$ and satisfies the conic equation exactly when $a_0^2+a_2^2=0$. In an ordered field a nonzero square is positive by trichotomy and closure of the positive cone Ordered field, so over $\mathbb{R}$ the sum of the two squares can be zero only if $a_0=a_2=0$, impossible for a projective point; over $\mathbb{C}$ the solutions are $[1:0:i]$ and $[1:0:-i]$ (or $[i:0:1]$, $[-i:0:1]$) Evaluation and roots of a polynomial in a commutative target ring, An algebraically closed field: every nonconstant polynomial has a root in the field.

Over the algebraically closed field, the Bezout identity gives ${\sum }_p{I}_p(C,L)=2\cdot 1=2$ Bezout's theorem for plane projective curves. At $p=[1:0:i]$ the gradients of $x_0^2+x_1^2+x_2^2$ and $x_1$ are $(2,0,2i)$ and $(0,1,0)$, both nonzero, so both curves are smooth at $p$ with tangent lines $x_0+i{x}_2=0$ and $x_1=0$, which are distinct; hence $I_p(C,L)=1$ Transversal smooth curves meet with multiplicity one, Multiplicity one characterises smooth points with a unique tangent, and symmetrically at the conjugate point Intersection multiplicity dominates the product of multiplicities, with equality for separated tangent cones, Local intersection multiplicity of two plane curves. The two points are conjugate, with non-real coordinates, and are not $\mathbb{R}$-rational: at either point the ratio $x_2\mathrm{/}{x}_0=\pm i$ is non-real, since a real square cannot equal $-1$ $\mathbb{C}=\mathbb{R}[x]\mathrm{/}(x^2+1)$ is a field, every element is uniquely $a+bi$, and every nonzero element has inverse $(a-bi)\mathrm{/}(a^2+b^2)$, Ordered field.