If F is a field, R a nonzero ring, and f: F >> R a surjective homomorphism, prove that f is an isomorphism.

we will assume Ker f = {0} or Ker f = F

when Ker f = 0 ( nothing to prove )

but when Ker f = F

f ( x ) = 0 ,  ∀x ∈ F

as ⇒Im F = 0

but when F = R and R ≠ 0 , it proves that ker f = {0}

and this proves that f is an isomorphism

Step-by-step explanation:

F been a subjective homomorphism shows that f is one-one

Therefore to prove that f is an isomorphism

we will assume Ker f = {0} or Ker f = F

when Ker f = 0 ( nothing to prove )

but when Ker f = F

f ( x ) = 0 ,  ∀x ∈ F

as ⇒Im F = 0

but when F = R and R ≠ 0 , it proves that ker f = {0}

and this proves that f is an isomorphism

yes

step-by-step explanation:

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y