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],

Solutions

Expert Solution

Given function $\phi:\mathbb Q[X]\rightarrow \mathbb Q[X]$ ,

$\sum_{i=0}^n a_i X^i\mapsto\sum_{i=0}^n a_i (2X+3)^i$

We need to show that $\phi:\mathbb Q[X]\rightarrow \mathbb Q[X]$ is an automorphism. We first prove that it is a homomorphism.

Suppose $f(x),g(x)\in\mathbb Q[X]$ are given by

$f(x)=\sum_{i=0}^n a_i X^i,~~~~~g(x)=\sum_{i=0}^n b_i X^i$

Then

$\begin{align*}\phi(f+g)&=\phi\begin{pmatrix}\sum_{i=0}^n (a_i+b_i)X^i\end{pmatrix}\\ &=\sum_{i=0}^n (a_i+b_i)\begin{pmatrix}2X+3\end{pmatrix}^i\\ &=\sum_{i=0}^n a_i\begin{pmatrix}2X+3\end{pmatrix}^i+\sum_{i=0}^n b_i\begin{pmatrix}2X+3\end{pmatrix}^i\\ &=\phi(f)+\phi(g)\end{align*}$

and

$\begin{align*}\phi(fg)&=\phi\begin{pmatrix}\sum_{i=0}^{2n}X^i\sum_{j=0}^i a_jb_{i-j}\end{pmatrix}\\ &=\sum_{i=0}^{2n}\begin{pmatrix}2X+3\end{pmatrix}^i\sum_{j=0}^i a_jb_{i-j}\\ &=\begin{pmatrix}\sum_{i=0}^{n}a_i(2X+3)^i\sum_{j=0}^n b_j(2X+3)^j\end{pmatrix}\\ &=\phi(f)\phi(g)\end{align*}$

Therefore, $\phi:\mathbb Q[X]\rightarrow \mathbb Q[X]$ is a ring homomorphism.

Thus, to prove that $\phi:\mathbb Q[X]\rightarrow \mathbb Q[X]$ is automorphism, it remains to show that it is bijective. We can do this by showing that $\phi:\mathbb Q[X]\rightarrow \mathbb Q[X]$ has a two-sided inverse.

Consider the function $\psi:\mathbb Q[X]\rightarrow \mathbb Q[X]$ ,

$\sum_{i=0}^n a_i X^i\mapsto\sum_{i=0}^n a_i \begin{pmatrix}{\frac 12}X-{\frac 32}\end{pmatrix}^i$

For any element $f(x)\in\mathbb Q[X]$ , given by

$f(x)=\sum_{i=0}^n a_i X^i$

we have

$\begin{align*}\phi(\psi(f))&=\phi\begin{pmatrix}\sum_{i=0}^n \begin{pmatrix}{\frac 12}X-{\frac 32}\end{pmatrix}^i\end{pmatrix}\\ &=\phi\begin{pmatrix}\sum_{i=0}^n \sum_{j=0}^i(-1)^j\begin{pmatrix}i\\j\end{pmatrix}{\frac {3^{i-j}X^j}{2^i}}\end{pmatrix}\\ &=\sum_{i=0}^n \sum_{j=0}^i(-1)^j\begin{pmatrix}i\\j\end{pmatrix}{\frac {3^{i-j}(2X+3)^j}{2^i}}\\ &=\sum_{i=0}^n \begin{pmatrix}{\frac 12}(2X+3)-{\frac 32}\end{pmatrix}^i\\ &=\sum_{i=0}^n a_iX^i\\ &=f\end{align*}$

and

$\begin{align*}\psi(\phi(f))&=\psi\begin{pmatrix}\sum_{i=0}^n \begin{pmatrix}2X+3\end{pmatrix}^i\end{pmatrix}\\ &=\psi\begin{pmatrix}\sum_{i=0}^n \sum_{j=0}^i\begin{pmatrix}i\\j\end{pmatrix}2^{i-j}X^{i-j}3^j\end{pmatrix}\\ &=\sum_{i=0}^n \sum_{j=0}^i\begin{pmatrix}i\\j\end{pmatrix}2^{i-j}3^j\begin{pmatrix}{\frac {X}{2}}-{\frac 32}\end{pmatrix}^{i-j}\\ &=\sum_{i=0}^n \sum_{j=0}^i\begin{pmatrix}i\\j\end{pmatrix}3^j\begin{pmatrix}X-3\end{pmatrix}^{i-j}\\ &=\sum_{i=0}^n \begin{pmatrix}(X-3)+3\end{pmatrix}^i\\ &=\sum_{i=0}^n a_iX^i\\ &=f\end{align*}$

Thus, $\psi:\mathbb Q[X]\rightarrow \mathbb Q[X]$ is a two-sided inverse of $\phi:\mathbb Q[X]\rightarrow \mathbb Q[X]$ . Hence, $\phi:\mathbb Q[X]\rightarrow \mathbb Q[X]$ is an automorphism.

Colby Messinger answered 3 years ago

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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...

Solutions

Expert Solution

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