Define \[ F:\mathbb C^3\to\mathbb C^3, \qquad F(u,v,w)=(u+v+w,uv+vw+uw,uvw). \] Show that \(F\) is onto but not one-to-one.
Solution
1. Fix an arbitrary point\[ (a,b,c)\in\CC^3. \]The polynomial\[ p(t) \coloneqq t^3-at^2+bt-c \]has three complex roots \(u,v,w\), counted with multiplicity.
By the fundamental theorem of algebra, the degree-\(3\) polynomial \(p\) factors over \(\CC\) as \[ p(t) =(t-u)(t-v)(t-w) \] for some \(u,v,w\in\CC\).
2. The roots from step 1 satisfy\[ u+v+w=a, \]\[ uv+uw+vw=b, \]and\[ uvw=c. \]
Expanding the factorization from step 1 gives \[ (t-u)(t-v)(t-w) = t^3-(u+v+w)t^2 +(uv+uw+vw)t -uvw. \] Compare coefficients with \[ t^3-at^2+bt-c. \]
3. The map \(F\) is onto.
For the arbitrary point \((a,b,c)\) chosen in step 1, step 2 gives \[ F(u,v,w) = (a,b,c). \] Thus every point of \(\CC^3\) has a preimage.
4. The map \(F\) is invariant under permutations of its three inputcoordinates.
Each component of \[ F(u,v,w) = (u+v+w,\ uv+vw+uw,\ uvw) \] is a symmetric polynomial in \(u,v,w\).
5. The map \(F\) is not one-to-one.
The points \[ (1,2,3) \qquad\text{and}\qquad (2,1,3) \] are distinct, but step 4 gives \[ F(1,2,3) = F(2,1,3). \] Explicitly, both images are \[ (6,11,6). \]
6. Q.E.D.
Step 3 proves surjectivity and step 5 proves noninjectivity.