Question

# Show that the map Q[X] → Q[X], sum^{n}{i=0}(a_nX^i) → sum^{n}{i=0}(a_n(2X + 3)^i) , is an automorphism...

Show that the map Q[X] → Q[X], sum^{n}{i=0}(a_nX^i) → sum^{n}{i=0}(a_n(2X + 3)^i) , is an automorphism of Q[X],

Let ϕ:Q[x]→Q[x] be given by ϕ(f(x))=f(2x+3). First note that if φ:Q→Q is given by φ(x)=2x+3, we can re-write ϕ as ϕ(f)=f∘φ.

We want to see if ϕ is an automorphism.

Firstly, we check that ϕ is well defined. Clearly φ is a polynomial, and the composition of polynomials is a polynomial, so ϕ is well defined.

Now we want to check that it preserves the structure. Let f,g∈Q[x]. Then

ϕ((f+g)(x))=(f+g)(2x+3)=f(2x+3)+g(2x+3)=ϕ(f(x))+ϕ(g(x))

Similarly,

ϕ((fg)(x))=(fg)(2x+3)=f(2x+3)g(2x+3)=ϕ(f(x))ϕ(g(x))

Finally it suffices to show that ϕ is bijective.

First we show ϕ is injective.

Let f,g∈Q[x] and suppose ϕ(f)=ϕ(g).

Then f∘φ=g∘φ,

so f=f∘φ∘φ−1=g∘φ∘φ−1=g (as we can easily see that φ is invertible.)

Similarly, we can show that ϕ is surjective.

Thus, ϕ is an automorphism.

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