is a mathematical
term for a self-isomorphism
in some category
. More concretely,
an automorphism of a group G
is a group homomorphism
which is an isomorphism
an automorphism of a ring R
is a bijective ring homomorphism
The collection of all automorphisms of a given object (a ring, group,
vector space, metric space, topological space, graph,...) form a group
with binary operation given by composition of functions, and the
neutral element being the identity function that fixes every element
of the object. This group can be thought of as a
symmetry group for the object.
To illustrate the concept,
we consider an example that is important in Galois theory.
If L is a field extension of K.
a function f:L-->L is a K-automorphism iff
f is a ring homomorphism.
f is a bijection
f(k)=k for all k in K
Here are some examples.
- Well first of all we always
have the identity 1L:L-->L defined by
1L(a)=a for all a in L.
- C is a field extension of
R. We have seen
one R-automorphisms of C already, the identity,
but there is another one. Complex conjugation
gives a map bar:C-->C defined by
bar(a+ib)=a-ib (for real numbers a,b). It is not
difficult to check that this is indeed an R-automorphism
The question becomes are there any others? The answer is no. To see this
suppose that f is some R-automorphism
of C. Think about f(i). It is a complex number, what can it
be? Well i2+1=0,
and so f(i2+1)=f(0)=0. (The last equality is because
f is a ring homomorphism.) But using that f is a ring
homomorphism twice more we get that
f(i2+1)=f(i)2+f(1). Finally, one more
application tells us that f(1)=1.
Putting it altogether we have
f(i)2+1=0. In other words f(i) is a square
root of -1 and so that tells us that f(i) is either
i or -i.
But once we know f(i) we know f.
This is because a typical complex number looks like
a+bi, for real numbers a,b.
Now f(a+bi)=f(a)+f(b)f(i) (since f is a
homomorphism) and since f fixes real numbers,
It follows that f is either the identity or complex conjugation.