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Theorem 1. f(x,y)=yn+a1(x)yn-1+…+an-1(x)y+an(x) (∈R[x,y]),
C={(x,y)∈R2| f(x,y)=0} and
only an algebraic singular point of C exists on the line x=x"(⊂R2). ⇒
The singular point (x",y") is an even point when you see C as a graph.(1982(?))
Proof. Let y1(x),…,yn(x) be the algebraic functions defined by f(x,y)=0.
∃s,∃r(>0), |x-x"|≤r ⇒y1(x),…,ys(x)∈R,
(i≠k,1≤i,k≤s ⇒ yi(x")≠yk(x"),yi(x")≠y")
And ∃t, x"-r≤x≤x" ⇒
ys+1(x),…,ys+t(x)∈R, ys+1(x")=…=ys+t(x")=y".
ys+t+1(x)∈C-R,…ys+t+2u(x)∈C-R, ys+t+i+u(x) is the conjugate of ys+t+i(x).
And ∃t", x"≤x≤x"+r ⇒ ys+1(x),…,ys+t"(x)∈R, ys+1(x")=…=ys+t"(x")=y",
ys+t"+1(x)∈C-R,…,ys+t"+2u"(x)∈C-R, ys+t"+i+u"(x) is the conjugate of ys+t"+i(x).
n=s+t+2u=s+t"+2u". t+t"≡t+t"+2s+2u+2u"=2n≡0 (mod 2).
t+t" curves join at (x",y"). (x",y") is an even point.♦
Theorem 2. The real algebraic curve is union of unicursal curves.
proof. Let C": g(z,y)=am(y)zm+…+a1(y)z+a0(y) (ai(y)∈R[y]) be a given algebraic curve.
Put x=z+rye (r∈R, r≠0)
and g(z,y)=g(x-rye,y)=r"yme+deg(am(y))+(the terms whose degrees about y
are less than me)=f(x,y) (r"∈R).
Let (zi,yi) (1≤i≤N) be the algebraic singular points of C" in R2.
(xi,yi) (xi=zi+ryie, 1≤i≤N) are the algebraic singular points of C:f(x,y)=0 in R2.
If r is little enough,
zi≠zk ⇒ xi-xk=zi-zk+r(yie-yke)≠0 (1≤i,k≤N)
If zi=zk and yi≠yk, xi-xk=r(yie-yke)≠0.
It never ocurres that yi≠yk and xi=xk.
The (xi,yi)'s are even points by theorem 1.
Take such an r.
Let M be a number which is larger than the |xi|'s.
Let y1(x),…,yn(x) be the algebraic functions defined by f(x,y)=0.
∃t, |x|≥M ⇒ y1(x),…,yt(x)∈R, yt+1(x),…,yt+2u(x)∈C-R.
yt+i+u(x) is the conjugate of yt+i(x). t+2u=n,
i≠k ⇒ yi(x)≠yk(x).
The t real curves join at the point x=+∞. t≡n (mod 2)
Similarly, t" real curves join at the point x=-∞. t"≡n (mod 2)
C is union of unicursal curves. C" is homeomorphic to C by
C∋(x,y) → (z,y)=(x-rye,y)∈C".♦
I noticed the above in connection with a problem submitted to
"Suugaku seminar (November 1982)" by Dr. Kako.
I heard that Gauss had used theorem 2 in the first proof of the
fundamental theorem of the algebra of his.
Is this useful to think the 16th problem of Hilbert?
C={(x,y)∈R2| f(x,y)=0} and
only an algebraic singular point of C exists on the line x=x"(⊂R2). ⇒
The singular point (x",y") is an even point when you see C as a graph.(1982(?))
Proof. Let y1(x),…,yn(x) be the algebraic functions defined by f(x,y)=0.
∃s,∃r(>0), |x-x"|≤r ⇒y1(x),…,ys(x)∈R,
(i≠k,1≤i,k≤s ⇒ yi(x")≠yk(x"),yi(x")≠y")
And ∃t, x"-r≤x≤x" ⇒
ys+1(x),…,ys+t(x)∈R, ys+1(x")=…=ys+t(x")=y".
ys+t+1(x)∈C-R,…ys+t+2u(x)∈C-R, ys+t+i+u(x) is the conjugate of ys+t+i(x).
And ∃t", x"≤x≤x"+r ⇒ ys+1(x),…,ys+t"(x)∈R, ys+1(x")=…=ys+t"(x")=y",
ys+t"+1(x)∈C-R,…,ys+t"+2u"(x)∈C-R, ys+t"+i+u"(x) is the conjugate of ys+t"+i(x).
n=s+t+2u=s+t"+2u". t+t"≡t+t"+2s+2u+2u"=2n≡0 (mod 2).
t+t" curves join at (x",y"). (x",y") is an even point.♦
Theorem 2. The real algebraic curve is union of unicursal curves.
proof. Let C": g(z,y)=am(y)zm+…+a1(y)z+a0(y) (ai(y)∈R[y]) be a given algebraic curve.
Let e be an odd number which is larger than max (deg(ai(y))).
0≤i≤mPut x=z+rye (r∈R, r≠0)
and g(z,y)=g(x-rye,y)=r"yme+deg(am(y))+(the terms whose degrees about y
are less than me)=f(x,y) (r"∈R).
Let (zi,yi) (1≤i≤N) be the algebraic singular points of C" in R2.
(xi,yi) (xi=zi+ryie, 1≤i≤N) are the algebraic singular points of C:f(x,y)=0 in R2.
If r is little enough,
zi≠zk ⇒ xi-xk=zi-zk+r(yie-yke)≠0 (1≤i,k≤N)
If zi=zk and yi≠yk, xi-xk=r(yie-yke)≠0.
It never ocurres that yi≠yk and xi=xk.
The (xi,yi)'s are even points by theorem 1.
Take such an r.
Let M be a number which is larger than the |xi|'s.
Let y1(x),…,yn(x) be the algebraic functions defined by f(x,y)=0.
∃t, |x|≥M ⇒ y1(x),…,yt(x)∈R, yt+1(x),…,yt+2u(x)∈C-R.
yt+i+u(x) is the conjugate of yt+i(x). t+2u=n,
i≠k ⇒ yi(x)≠yk(x).
The t real curves join at the point x=+∞. t≡n (mod 2)
Similarly, t" real curves join at the point x=-∞. t"≡n (mod 2)
C is union of unicursal curves. C" is homeomorphic to C by
C∋(x,y) → (z,y)=(x-rye,y)∈C".♦
I noticed the above in connection with a problem submitted to
"Suugaku seminar (November 1982)" by Dr. Kako.
I heard that Gauss had used theorem 2 in the first proof of the
fundamental theorem of the algebra of his.
Is this useful to think the 16th problem of Hilbert?