| Theorem 01: Given a ring
R then 0*x = x*0 = 0 for any x in R.
Proof
| 0*x = (0+0)*x |
by definition of the additive identity
(0) |
| 0*x = 0*x +
0*x |
by the distributive property |
| 0*x -(0*x) = 0*x + 0*x - (0*x) |
since the additive inverse always
exists in a Ring |
| 0 = 0*a |
by the definition of the additive inverse |
The proof for a*0 follows analogously. |
| Theorem 02: Given x and
y in a ring R then x*(-y) = (-x)*y = -(x*y).
Proof
| (-x)*y + x*y = (-x + x)* y |
by the distributive property |
| (-x)*y + x*y = (0)*y |
by the definition of the additive inverse |
| (-x)*y + x*y = 0 |
by Theorem 1 |
| (-x)*y = - (x*y) |
by the definition of the additive inverse |
The proof for x*(-y) follows analogously. |
| Theorem 03: Given x and
y in a ring R then (-x)*(-y) = (x*y).
Proof
The proof for x*(-y) follows analogously. |
| Theorem 04: F(S) (read as F
adjoin S) is the smallest field containing both F and S.
Proof
Define X to be the smallest field containing both F and
S.
By the definition of adjoin the smallest field containing
both F and S must also contain F(S), i.e. X contains F(S). Since
a finite intersection of fields is also a field F(S) must be a field.
Since X contains F(S) and F(S) is also a field, X must equal F(S). |
Theorem 05: If S=S1 S2
then F(S) = F(S1)(S2) |
| Theorem 06: ILet E be s subset of F which
is in turn a subset of G then the degree
of G over E is equal to the degree of G over F time the degree of F over
E, i.e. [G:E]=[G:F][F:E]. |