Unique Automorphism of R[x] - A Hint for Invertible a & b

[mathmari]

Hey!

Let $R$ be a commutative ring with unity and $a,b\in R$ with $a$ invertible.

I want to show that the mapping $x\rightarrow ax+b$ defines a unique automorphism of $R[x]$ that is idempotent in $R$. Could you give me a hint what I am supposed to do? (Wondering)

 

[Deveno]

How does your text/notes define "idempotent in $R$?"

 

[mathmari]

How does your text/notes define "idempotent in $R$?"

It sends each element to itself.

 

[Deveno]

First, show it is an ring-homomorphism.

Next, show it is bijective (this is where you will need the fact that $a$ is a unit).

Since constant polynomials (the embedding of $R$ in $R[x]$) have no powers of $x$ at all, it is clear that this mapping is thus idempotent on $R$.

 

[mathmari]

First, show it is an ring-homomorphism.

Let $f(x)=ax+b$ and $p(x)\in R[x]$ a polynomial. Then we have the mapping $\phi : p(x)\mapsto (p\circ f)(x)$.

We have that $$\phi (p_1+p_2)=(p_1+p_2)\circ f$$ Is this equal to $p_1\circ f+p_2\circ f$ ? (Wondering)

We also have that $$\phi (p_1\cdot p_2)=(p_1\cdot p_2)\circ f$$ Is this equal to $(p_1\circ f)(\cdot (p_2\circ f)$ ? (Wondering)

Next, show it is bijective (this is where you will need the fact that $a$ is a unit).

$$\phi (p_1(x))=\phi (p_2(x))\Rightarrow (p_1\circ f)(x)=(p_2\circ f)(x)\Rightarrow p_1(f(x))=p_2(f(x))$$
Do we conclude from that that $p_1=p_2$ ? (Wondering) To show that the mapping is onto, we have to show that for each $q\in R[x]$ there is a $p\in R[x]$ such that $\phi (p)=q$, right? (Wondering)
So, for each $q$ there is a $p$ such that $p\circ f=q$.

Could you give me a hint how we could show that? (Wondering)

 

[Deveno]

Let $f(x)=ax+b$ and $p(x)\in R[x]$ a polynomial. Then we have the mapping $\phi : p(x)\mapsto (p\circ f)(x)$.

We have that $$\phi (p_1+p_2)=(p_1+p_2)\circ f$$ Is this equal to $p_1\circ f+p_2\circ f$ ? (Wondering)

Is it? Let's write:

$p(x) = \sum\limits_{i = 0}^n c_ix^i$
$q(x) = \sum\limits_{i = 0}^n d_ix^i$

then $\phi(p(x)+q(x)) = (p+q)(f(x)) = \sum\limits_{i = 0}^n (c_i+d_i)(ax + b)^i$

$= \sum\limits_{i = 0}^n c_i(ax + b)^i + \sum\limits_{i = 0}^n d_i(ax + b)^i$

$= p(f(x)) + q(f(x)) = \phi(p(x)) + \phi(q(x))$

We also have that $$\phi (p_1\cdot p_2)=(p_1\cdot p_2)\circ f$$ Is this equal to $(p_1\circ f)(\cdot (p_2\circ f)$ ? (Wondering)

Do we? You try this one.

$$\phi (p_1(x))=\phi (p_2(x))\Rightarrow (p_1\circ f)(x)=(p_2\circ f)(x)\Rightarrow p_1(f(x))=p_2(f(x))$$
Do we conclude from that that $p_1=p_2$ ? (Wondering)

Well, that's what we want to prove-but just stating that isn't a proof. You need to use explicit expressions for $p_1(x), p_2(x)$.

To show that the mapping is onto, we have to show that for each $q\in R[x]$ there is a $p\in R[x]$ such that $\phi (p)=q$, right? (Wondering)
So, for each $q$ there is a $p$ such that $p\circ f=q$.

Could you give me a hint how we could show that? (Wondering)

First, you might try to find a pre-image for $x$ and a constant polynomial $c$.

 

[mathmari]

Let's write:

$p(x) = \sum\limits_{i = 0}^n c_ix^i$
$q(x) = \sum\limits_{i = 0}^n d_ix^i$

then $\phi(p(x)+q(x)) = (p+q)(f(x)) = \sum\limits_{i = 0}^n (c_i+d_i)(ax + b)^i$

$= \sum\limits_{i = 0}^n c_i(ax + b)^i + \sum\limits_{i = 0}^n d_i(ax + b)^i$

$= p(f(x)) + q(f(x)) = \phi(p(x)) + \phi(q(x))$

Ah ok... I see... (Nerd)

Do we? You try this one.

Let
$p(x) = \sum\limits_{i = 0}^n c_ix^i$
$q(x) = \sum\limits_{i = 0}^n d_ix^i$

then $\phi(p(x)q(x)) = (pq)(f(x)) = \sum\limits_{i = 0}^n h_i(ax + b)^i$, where $h_i=a_{j_1}b_{j_2}$ with $j_1+j_2=i$

$\Rightarrow \phi(p(x)q(x)) =\left (\sum\limits_{i = 0}^n c_i(ax + b)^i \right )\cdot \left ( \sum\limits_{i = 0}^n d_i(ax + b)^i\right )$

$= p(f(x)) q(f(x)) = \phi(p(x)) \phi(q(x))$

I am not really sure if this is correct... Especially, the part $\phi(p(x)q(x)) =\left (\sum\limits_{i = 0}^n c_i(ax + b)^i \right )\cdot \left ( \sum\limits_{i = 0}^n d_i(ax + b)^i\right )$.

(Wondering)

 

[Deveno]

Ah ok... I see... (Nerd)

Let
$p(x) = \sum\limits_{i = 0}^n c_ix^i$
$q(x) = \sum\limits_{i = 0}^n d_ix^i$

then $\phi(p(x)q(x)) = (pq)(f(x)) = \sum\limits_{i = 0}^n h_i(ax + b)^i$, where $h_i=a_{j_1}b_{j_2}$ with $j_1+j_2=i$

$\Rightarrow \phi(p(x)q(x)) =\left (\sum\limits_{i = 0}^n c_i(ax + b)^i \right )\cdot \left ( \sum\limits_{i = 0}^n d_i(ax + b)^i\right )$

$= p(f(x)) q(f(x)) = \phi(p(x)) \phi(q(x))$

I am not really sure if this is correct... Especially, the part $\phi(p(x)q(x)) =\left (\sum\limits_{i = 0}^n c_i(ax + b)^i \right )\cdot \left ( \sum\limits_{i = 0}^n d_i(ax + b)^i\right )$.

(Wondering)

That's the right idea, so do a very careful calculation of the coefficient of $(ax + b)^k$ (for an arbitrary, but fixed $k$ in-between $0$ and $2n$) in your product:$\left (\sum\limits_{i = 0}^n c_i(ax + b)^i \right )\cdot \left ( \sum\limits_{i = 0}^n d_i(ax + b)^i\right )$

Also, remember polynomials may have different degrees, so we may have to "pad out with 0's" to be able to use the same $n$.

 

[mathmari]

That's the right idea, so do a very careful calculation of the coefficient of $(ax + b)^k$ (for an arbitrary, but fixed $k$ in-between $0$ and $2n$) in your product:$\left (\sum\limits_{i = 0}^n c_i(ax + b)^i \right )\cdot \left ( \sum\limits_{i = 0}^n d_i(ax + b)^i\right )$

Also, remember polynomials may have different degrees, so we may have to "pad out with 0's" to be able to use the same $n$.

We have the following:
$$\phi(p(x)q(x)) = (pq)(f(x)) = \sum\limits_{i = 0}^{2n} \sum\limits_{j=0}^ic_jd_{i-j}(ax + b)^i$$
Why is this equal to $\left (\sum\limits_{i = 0}^n c_i(ax + b)^i \right )\cdot \left ( \sum\limits_{i = 0}^n d_i(ax + b)^i\right )$ ? (Wondering)

 

[Deveno]

We have the following:
$$\phi(p(x)q(x)) = (pq)(f(x)) = \sum\limits_{i = 0}^{2n} \sum\limits_{j=0}^ic_jd_{i-j}(ax + b)^i$$
Why is this equal to $\left (\sum\limits_{i = 0}^n c_i(ax + b)^i \right )\cdot \left ( \sum\limits_{i = 0}^n d_i(ax + b)^i\right )$ ? (Wondering)

As I suggested before, compare the coefficients of $(ax + b)^k$ in each expression.

 

[mathmari]

As I suggested before, compare the coefficients of $(ax + b)^k$ in each expression.

The coefficient of $(ax+b)^k$ is $$\sum_{j=0}^kc_jd_{k-j}$$ or not? (Wondering)

 

[Deveno]

The coefficient of $(ax+b)^k$ is $$\sum_{j=0}^kc_jd_{k-j}$$ or not? (Wondering)

Lol, don't tell me, convince yourself that it is, in *both* expressions. Try it for $k = 0,1,2$ until you get the hang of how to show it for ANY $k$.

 

[mathmari]

We have the following:
$$\sum_{i=0}^{2n}\sum_{j=0}^ic_jd_{i-j}(ax+b)^i=\sum_{i=0}^{2n}\left (c_0d_i+c_1d_{i-1}+\dots +c_id_0\right )(ax+b)^i \\ =\sum_{i=0}^{2n}[c_0d_i(ax+b)^i+c_1d_{i-1}(ax+b)^i+\dots +c_id_0(ax+b)^i] \\ =\sum_{i=0}^{2n}c_0d_i(ax+b)^i+\sum_{i=0}^{2n}c_1d_{i-1}(ax+b)^i+\dots +\sum_{i=0}^{2n}c_id_0(ax+b)^i$$
right? (Wondering)

We have that $c_i=d_i=0, i>n$.
So, the last equation is equal to $$c_0\sum_{i=0}^nd_i(ax+b)^i+c_1(ax+b)\sum_{i=0}^nd_{i-1}(ax+b)^{i-i}+\dots +\left (\sum_{i=0}^nc_i(ax+b)^i\right ) d_0$$
or not? (Wondering)

Is this equal to $$\left (\sum_{i=0}^nc_i(ax+b)^i\right )\left (\sum_{i=0}^nd_i(ax+b)^i\right )$$ ? (Wondering)

 

[Deveno]

Use the distributive law.

 

[mathmari]

Use the distributive law.

At which point exactly? I got stuck right now... (Wondering)

 

[Deveno]

We want to show that $\phi$ is multiplicative.

Now:

$\left(\sum\limits_{i = 0}^m c_ix^i\right)\left(\sum\limits_{i = 0}^n d_ix^i\right) = \sum\limits_{i = 0}^{m+n} h_ix^i$

where $h_i = \sum\limits_{j+k = i} c_jd_k$

Thus $\phi\left[\left(\sum\limits_{i = 0}^m c_ix^i\right)\left(\sum\limits_{i = 0}^n d_ix^i\right) \right]$

$= \phi\left(\sum\limits_{i = 0}^{m+n} h_ix^i\right) = \phi\left(\sum\limits_{i = 0}^{m+n} \left(\sum\limits_{j+k = i} c_jd_k\right)x^i\right)$

$= \sum\limits_{i= 0}^{m+n} \phi\left(\left(\sum\limits_{j+k = i}c_jd_k\right)x^i\right)$

(since we already showed $\phi$ is additive)

$= \sum\limits_{i = 0}^{m+n}\left(\sum\limits_{j+k = i}c_jd_k\right)(ax + b)^i$

(by our definition of $\phi$)

$= \left(\sum\limits_{i = 0}^m c_i(ax + b)^i\right)\left(\sum\limits_{i = 0}^n d_i(ax + b)^i\right)$

(to see this, temporarily replace "$(ax + b)$" with "$u$", the polynomial multiplication works out the same)

$ = \left(\phi\left(\sum\limits_{i = 0}^m c_ix^i\right)\right)\left(\phi\left(\sum\limits_{i = 0}^n d_ix^i\right)\right)$, QED.

 

[mathmari]

We want to show that $\phi$ is multiplicative.

Now:

$\left(\sum\limits_{i = 0}^m c_ix^i\right)\left(\sum\limits_{i = 0}^n d_ix^i\right) = \sum\limits_{i = 0}^{m+n} h_ix^i$

where $h_i = \sum\limits_{j+k = i} c_jd_k$

Thus $\phi\left[\left(\sum\limits_{i = 0}^m c_ix^i\right)\left(\sum\limits_{i = 0}^n d_ix^i\right) \right]$

$= \phi\left(\sum\limits_{i = 0}^{m+n} h_ix^i\right) = \phi\left(\sum\limits_{i = 0}^{m+n} \left(\sum\limits_{j+k = i} c_jd_k\right)x^i\right)$

$= \sum\limits_{i= 0}^{m+n} \phi\left(\left(\sum\limits_{j+k = i}c_jd_k\right)x^i\right)$

(since we already showed $\phi$ is additive)

$= \sum\limits_{i = 0}^{m+n}\left(\sum\limits_{j+k = i}c_jd_k\right)(ax + b)^i$

(by our definition of $\phi$)

$= \left(\sum\limits_{i = 0}^m c_i(ax + b)^i\right)\left(\sum\limits_{i = 0}^n d_i(ax + b)^i\right)$

(to see this, temporarily replace "$(ax + b)$" with "$u$", the polynomial multiplication works out the same)

$ = \left(\phi\left(\sum\limits_{i = 0}^m c_ix^i\right)\right)\left(\phi\left(\sum\limits_{i = 0}^n d_ix^i\right)\right)$, QED.

Ah ok... I see... (Thinking)

Next, show it is bijective (this is where you will need the fact that $a$ is a unit).

To show that $\phi$ is 1-1

I have done the following:

Let $p_1(x)=\sum_{i=0}^nc_ix^i$ and $p_2(x)=\sum_{i=0}^md_ix^i$ and let $n>m$ then
$$\phi (p_1(x))=\phi (p_2(x))\Rightarrow (p_1\circ f)(x)=(p_2\circ f)(x)\Rightarrow p_1(f(x))=p_2(f(x)) \\ \Rightarrow \sum_{i=0}^nc_if(x)^i=\sum_{i=0}^md_if(x)^i\\ \Rightarrow \sum_{i=0}^nc_if(x)^i=\sum_{i=0}^nd_if(x)^i \text{ with } d_i=0 \text{ for } i=m+1, \dots , n \\ \Rightarrow \sum_{i=0}^n (c_i-d_i)f(x)^i=0 \\ \Rightarrow c_i=d_i, \forall i=1, \dots , n$$
This means that $p_1=p_2$, or not? (Wondering)

First, you might try to find a pre-image for $x$ and a constant polynomial $c$.

Why do we have to do that? (Wondering)

 

[Deveno]

Ah ok... I see... (Thinking)


To show that $\phi$ is 1-1

I have done the following:

Let $p_1(x)=\sum_{i=0}^nc_ix^i$ and $p_2(x)=\sum_{i=0}^md_ix^i$ and let $n>m$ then
$$\phi (p_1(x))=\phi (p_2(x))\Rightarrow (p_1\circ f)(x)=(p_2\circ f)(x)\Rightarrow p_1(f(x))=p_2(f(x)) \\ \Rightarrow \sum_{i=0}^nc_if(x)^i=\sum_{i=0}^md_if(x)^i\\ \Rightarrow \sum_{i=0}^nc_if(x)^i=\sum_{i=0}^nd_if(x)^i \text{ with } d_i=0 \text{ for } i=m+1, \dots , n \\ \Rightarrow \sum_{i=0}^n (c_i-d_i)f(x)^i=0 \\ \Rightarrow c_i=d_i, \forall i=1, \dots , n$$
This means that $p_1=p_2$, or not? (Wondering)

Yes, a polynomial is the 0-polynomial if and only if all of its coefficients are 0.

Why do we have to do that? (Wondering)

It just makes things easier, since we already know $\phi$ is a ring-homomorphism.

 

[mathmari]

It just makes things easier, since we already know $\phi$ is a ring-homomorphism.

And how could we find a pre-image of $x$ ? I got stuck right now... (Wondering)

 

[Deveno]

And how could we find a pre-image of $x$ ? I got stuck right now... (Wondering)

Suppose $ay + b = x$, what is $y$? Why does $a$ have to be a unit to find $y$?

 

[mathmari]

Suppose $ay + b = x$, what is $y$? Why does $a$ have to be a unit to find $y$?

$y$ is then equal to $a^{-1}(x-b)$. So that $a^{-1}$ exists $a$ must be a unit, right? (Wondering)

For each $p$ we set $q(x)=p(a^{-1}(x-b))$ and then we have that $\phi (q(x))=(q\circ f)(x)=p(a^{-1}(f(x)-b))=p(a^{-1}(ax+b-b))=p(x)$. That means that $\phi$ is onto, right? (Wondering)

 

[Deveno]

Yes, because every element of $R[x]$ has a pre-image under $\phi$.

Thus $\phi$ is an automorphism.

Convince yourself that $\phi(f(x)) = f(2x)$ is *not* an automorphism of $\Bbb Z[x]$, even though the image is isomorphic to the domain.

 

[mathmari]

Convince yourself that $\phi(f(x)) = f(2x)$ is *not* an automorphism of $\Bbb Z[x]$, even though the image is isomorphic to the domain.

To which domain is the image isomorphic to? (Wondering)

To see why it is not an automorphism we have to check if $\phi$ satisfies the conditions $\phi (f_1+f_2)=\phi (f_1)+\phi (f_2)$ and $\phi (f_1f_2)=\phi (f_1)\phi (f_2)$, right? (Wondering)

 

[Deveno]

To which domain is the image isomorphic to? (Wondering)

To see why it is not an automorphism we have to check if $\phi$ satisfies the conditions $\phi (f_1+f_2)=\phi (f_1)+\phi (f_2)$ and $\phi (f_1f_2)=\phi (f_1)\phi (f_2)$, right? (Wondering)

The image is isomorphic to $\Bbb Z[x]$, but it is *not* all of $\Bbb Z[x]$ (you might want to spend a little time to see how that is even possible).

It *is* a ring-homomorphism, just not an automorphism.