The zig-zag trick

This is the zig-zag trick. The inspiration is a post from a friend’s blog. The “trick” is just the proof of the statement in the post: “ring homomorphisms out of the rationals that agree on the integers must agree everywhere.”

Let f,g: ℚ  ──> R be two ring homomorphisms such that f|         
                                                        ℤ
= g|  Now, let x ∈ℚ be arbitrary. Thus, ∃y,z ∈ℤ 
    ℤ
              z
such that x = ─. Then, using the properties of rings and ring
              y
homomorphisms, we have:
                                                    2
         z      zy           1            1       zy    1
f(x) = f(─) = f(──) = f(zy)f(──) = g(zy)f(──) = g(───)f(──) =
         y       2            2            2       y     2
                y            y            y             y 

                              z    2   1
                            g(─)g(y )f(──) =
                              y         2
                                       y 

                          2
  z     2   1       z    y       z          z          z*1       z
g(─) f(y )f(──) = g(─) f(──) = g(─)f(1) = g(─)g(1) = g(───) = g(─) =
  y          2      y     2      y          y           y       y
            y            y 

                                  g(x)


But, since x was arbitrarily chosen, we must have f = g.