Do field homomorphisms preserve characteristic

[icantadd]

Homework Statement

Given two fields F,E with different characteristic. Prove or disprove the following statement: "Field homomorphisms between fields of different characteristic cannot exist"

Homework Equations

T : F1 --> F2 is a field homomorphism if
1) T(a+b) = T(a) + T(b)
2) T(ab) = T(a)T(b)
3) T(1) = 1
4) T(0) = 0.

The Attempt at a Solution

Intuition says no...

All field homorphisms are injective. So T:F --> E where F has bigger order than E cannot exist. On the other hand, if E has bigger order than F, F must contain an isomorphic copy of E.

Hmm, not sure where to go from here. Here is my attempt... Suppose we do have a hom from F to E where char E is bigger than char F. Then by the fundamental homomorphism theorem, F/kerT is isomomorphic to E. However since T is injective the kernel is trivial. Therefore F is isomorphic to E contradicting the assumption of different characteristic...

 

[Dick]

Homework Statement

Given two fields F,E with different characteristic. Prove or disprove the following statement: "Field homomorphisms between fields of different characteristic cannot exist"

Homework Equations

T : F1 --> F2 is a field homomorphism if
1) T(a+b) = T(a) + T(b)
2) T(ab) = T(a)T(b)
3) T(1) = 1
4) T(0) = 0.

The Attempt at a Solution

Intuition says no...

All field homorphisms are injective. So T:F --> E where F has bigger order than E cannot exist. On the other hand, if E has bigger order than F, F must contain an isomorphic copy of E.

Hmm, not sure where to go from here. Here is my attempt... Suppose we do have a hom from F to E where char E is bigger than char F. Then by the fundamental homomorphism theorem, F/kerT is isomomorphic to E. However since T is injective the kernel is trivial. Therefore F is isomorphic to E contradicting the assumption of different characteristic...

That is nonsense. Look up the definition of 'field characteristic'. Read it several times. Then look at your requirement T(1)=1.

 

[icantadd]

Right read the definitions

$$T(1_{F1}) = 1_{F2}$$
Thus,
$$T(n1_{F1}) = nT(1_{F1}) = n1_{F2}$$
Thus suppose char(F1) = m,
$$T(m1_{F1}) = T(0) = 0 = mT(1_{F1}) = m1_{F2}$$
Therefore, char(F2) <= m.
Suppose p < m satisfies $$p1_{F2} = 0$$.
Then,
$$T^{-1}(p1_{F2}) = T(0) = 0 = T^{-1}(p1_{F2}) = pT^{-1}(1_{F2}) = p1_{F1}$$
Contradicting the minimality of m. Therefore, char(F2) = m.

 

[icantadd]

P.s. thank you!

 

[Dick]

Right read the definitions

$$T(1_{F1}) = 1_{F2}$$
Thus,
$$T(n1_{F1}) = nT(1_{F1}) = n1_{F2}$$
Thus suppose char(F1) = m,
$$T(m1_{F1}) = T(0) = 0 = mT(1_{F1}) = m1_{F2}$$
Therefore, char(F2) <= m.
Suppose p < m satisfies $$p1_{F2} = 0$$.
Then,
$$T^{-1}(p1_{F2}) = T(0) = 0 = T^{-1}(p1_{F2}) = pT^{-1}(1_{F2}) = p1_{F1}$$
Contradicting the minimality of m. Therefore, char(F2) = m.

Much better. You're welcome!