Step by step solution to "if f:R→R is a homomorphism and p is..."

Bryant Miranda

Answered question

2022-02-15

f : R \to R

is a homomorphism and p is an integer-coefficient polynomial, then:

f (p (r)) = p (f (r)) \forall r \in R

Vaughn Bradley

Beginner2022-02-16Added 16 answers

We have

f (1) = f (1 + 0) = f (1) + f (0) so f (0) = 0

.
By induction on

n \in N

we have

f (\sum_{i = 1}^{n} r_{i}) = \sum_{i = 1}^{n} f (r_{i}) and f (\prod_{i = 1}^{n} s_{i}) = \prod_{i = 1}^{n} f (s_{i})

.
In particular when

r_{i} = 1

for each i then

f (n) = f (\sum_{i = 0}^{n} 1) = f (\sum_{i = 1}^{n} r_{i}) = \sum_{i = 1}^{n} f (r_{i}) =

\sum_{i = 1}^{n} f (1) = \sum_{i = 1}^{n} 1 = n

.
And so if

n \in N

then

0 = f (0) = f (n + (- n)) = f (n) + f (- n) = n so + f (- n)

f (- n) = - n

f (v) = v

for every integer v.
And if

s_{i} = x

for each i then

f (x^{n}) = f (\prod_{i = 1}^{n} x) = f (\prod_{i = 1}^{n} s_{i}) =

= \prod_{i = 1}^{n} f (s_{i}) = \prod_{i = 1}^{n} f (x) = {f (x)}^{n}

.
Let

p (x) = v_{0} + \sum_{i = 1}^{n} v_{i} x^{i}

where each

v_{i} \in Z

. For

i > 0

let each

v_{i} x^{i} = r_{i}

and we have

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