TITLE: How do I prove this field homomorphism is an isomorphism?
QUESTION [3 upvotes]: The question is as follows. Let $F$ be a finite field with unit $1$ not equal to zero. Let the function $f: F \to F$ be given by $f(x) = x^3$, where the $\operatorname{char}(F) = 3$. Prove it is a ring isomorphism.

I can prove it is a ring homomorphism by showing $f(xy) = (xy)^3 = x^3y^3 = f(x)f(y)$ by commutativity. Similarly, since the characteristic is three,
$$(x+y)^3 = x^3 + y^3$$
$$f(a+b) = f(a) + f(b)$$
$$f(1) = 1,\ \ \textrm{and}\ \  f(0) = 0$$
Now, how do I show $f$ is a bijection?

REPLY [4 votes]: It's actually not an isomorphism in general, as Zev explained. But in the case that $F$ is finite it is:
It is a homomorphism, as you have shown, and any homomorphism of fields is injective. But also, any injective function from a finite set to itself is a bijection, hence $f$ is an isomorphism.


TITLE: Find the length of the common chord
QUESTION [1 upvotes]: "Two circles with centres C1 and C2 and radius $6$ cm and $8$ cm respetively cut each other at right angles.
Find the length of the common chord."
I tried it but could not get to the answer.
I am not getting that circles cut each other at right angles.
So do we need to make a tangent that will intersect at right angle?
Any suggestions welcome.
Thanks in advance.

REPLY [0 votes]: Find out the radical axis of the two circles. This is nothing but the common chord of the two circles. Substitute for $x$ or $y$ in any of the equation of the circle from the radical axis. You will get two points of intersection, and then you can apply the distance formula you will get the length of the common chord.