Polynomial Rings - Z[x]/(x^2) and Z[x^2 + 1>

[Math Amateur]

I am trying to get a good understanding of the structure of the rings $$\mathbb{Z}[x]/<x^2>$$ and $$\mathbb{Z}[x]/<x^2 +1>$$.

I tried to first deal with the rings $$\mathbb{R}[x]/<x^2>$$ and $$\mathbb{R}[x]/<x^2 +1>$$ as they seemed easier to deal with ... my thinking ... and my problems are as follows: (Would really appreciate clarification)Following an example I found in Gallian (page 257), first consider $$\mathbb{R}[x]/<x^2>$$ where $$\mathbb{R}[x]$$ is the ring of polynomials with real co-efficients.

Then $$\mathbb{R}[x]/<x^2> = \{ g(x) + <x^2> | g(x) \in \mathbb{R}[x] \}$$

But $$\mathbb{R}[x]$$ is a Euclidean Domain and hence possesses a Division Algorithm, so we may write:

$$g(x) = q(x)(x^2) + r(x)$$ where r(x) = 0 or r(x) has degree less than 2.

so we can write r(x) = ax + b where a, b $$\in \mathbb{R}$$

Thus $$g(x) + <x^2> = q(x)(x^2) + r(x) + <x^2>$$

= $$r(x) + <x^2>$$ since the ideal $$<x^2>$$ absorbs the term $$q(x)(x^2) + r(x)$$

= $$ax + b + <x^2>$$

Thus $$\mathbb{R}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{R} \}$$

Now, by a similar argument we can demonstrate that

$$\mathbb{R}[x]/<x^2 +1> = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{R} \}$$

which makes the two rings $$\mathbb{R}[x]/<x^2>$$ and $$\mathbb{R}[x]/<x^2 +1>$$ look to have the same structure?

One of my questions is how exactly are these two ring structures different?

A second worry is that the above demonstration works because $$\mathbb{R}[x]$$ is a Euclidean Domain ... so the same argument as above does not apply to

$$\mathbb{Z}[x]$$ because $$\mathbb{Z}[x]$$ is not a Euclidean Domain and hence we cannot use the Division algorithm.

How do we rigorously demonstrate that

$$\mathbb{Z}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{Z} \}$$ and$$\mathbb{Z}[x]/<x^2 +1> = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{Z} \}$$

Can someone please help clarify the above problems and issues?

Peter

 

[Fernando Revilla]

which makes the two rings $$\mathbb{R}[x]/<x^2>$$ and $$\mathbb{R}[x]/<x^2 +1>$$ look to have the same structure?

One of my questions is how exactly are these two ring structures different?

The operations are not the same. For example, in $\mathbb{R}/<x^2>$ the product is is $[ax+b][a'x+b']=[cx+d]$ where $cx+d$ is the remainder of the eucldean division $(ax+b)(a'x+b'):x^2$ and in $\mathbb{R}/<x^2+1>$ the remainder of the eucldean division $(ax+b)(a'x+b'):(x^2+1)$. By the way, $\mathbb{R}/<x^2+1>$ is naturally isomorphic to $\mathbb{C}$.

How do we rigorously demonstrate that
$$\mathbb{Z}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{Z} \}$$ and
$$\mathbb{Z}[x]/<x^2 +1> = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{Z} \}$$

Try $p(x)\sim q(x)$ iff there exists $h(x)\in\mathbb{Z}[x]$ such that $p(x)-q(x)=h(x)x^2$ (or $p(x)-q(x)=h(x)(x^2+1)$.

 

[Math Amateur]

The operations are not the same. For example, in $\mathbb{Z}/<x^2>$ the product is is $[ax+b][a'x+b']=[cx+d]$ where $cx+d$ is the remainder of the eucldean division $(ax+b)(a'x+b'):x^2$ and in $\mathbb{Z}/<x^2+1>$ the remainder of the eucldean division $(ax+b)(a'x+b'):(x^2+1)$. By the way, $\mathbb{Z}/<x^2+1>$ is naturally isomorphic to $\mathbb{C}$.



Try $p(x)\sim q(x)$ iff there exists $h(x)\in\mathbb{Z}[x]$ such that $p(x)-q(x)=h(x)x^2$ (or $p(x)-q(x)=h(x)(x^2+1)$.

Thanks Fernando, most helpful.You write:

"in $\mathbb{Z}/<x^2>$ the product is is $[ax+b][a'x+b']=[cx+d]$ where $cx+d$ is the remainder of the eucldean division $(ax+b)(a'x+b'):x^2$ and in $\mathbb{Z}/<x^2+1>$ the remainder of the eucldean division$$ (ax+b)(a'x+b') x^2+1)$$How can we carry out Euclidean Divisions in $$ \mathbb{Z}[x]$$ when it is not a Euclidean Domain?

Can you clarify? Am i misunderstanding something? :(

Peter

 

[Fernando Revilla]

Thanks Fernando, most helpful.You write:

"in $\mathbb{Z}/<x^2>$ the product is is $[ax+b][a'x+b']=[cx+d]$ where $cx+d$ is the remainder of the eucldean division $(ax+b)(a'x+b'):x^2$ and in $\mathbb{Z}/<x^2+1>$ the remainder of the eucldean division$$ (ax+b)(a'x+b') x^2+1)$$How can we carry out Euclidean Divisions in $$ \mathbb{Z}[x]$$ when it is not a Euclidean Domain?

Can you clarify? Am i misunderstanding something? :(

Peter

Of course, it is a typo, I meant $\mathbb{R}$ instead of $\mathbb{Z}$, I've already corrected it.

 

[Math Amateur]

Of course, it is a typo, I meant $\mathbb{R}$ instead of $\mathbb{Z}$, I've already corrected it.

OK thanks Fernando, but then I am still left with the puzzle;

I can see your arguments regarding [FONT=MathJax_AMS]R[FONT=MathJax_Main][[FONT=MathJax_Math]x[FONT=MathJax_Main]][FONT=MathJax_Main]/[FONT=MathJax_Main]<[FONT=MathJax_Math]x[FONT=MathJax_Main]2[FONT=MathJax_Main]>[FONT=MathJax_Main]

But then how do we rigorously demonstrate that

[FONT=MathJax_AMS]Z[FONT=MathJax_Main][[FONT=MathJax_Math]x[FONT=MathJax_Main]][FONT=MathJax_Main]/[FONT=MathJax_Main]<[FONT=MathJax_Math]x[FONT=MathJax_Main]2[FONT=MathJax_Main]>[FONT=MathJax_Main]=[FONT=MathJax_Main]{[FONT=MathJax_Math]a[FONT=MathJax_Math]x[FONT=MathJax_Main]+[FONT=MathJax_Math]b[FONT=MathJax_Main]+[FONT=MathJax_Main]<[FONT=MathJax_Math]x[FONT=MathJax_Main]2[FONT=MathJax_Main]>[FONT=MathJax_Main]|[FONT=MathJax_Math]a[FONT=MathJax_Main],[FONT=MathJax_Math]b[FONT=MathJax_Main]∈[FONT=MathJax_AMS]Z[FONT=MathJax_Main]} given that we cannot use the Division Algorithm

Can you help?

Peter[FONT=MathJax_AMS]Z[FONT=MathJax_Main][[FONT=MathJax_Math]x[FONT=MathJax_Main]][FONT=MathJax_Main]/[FONT=MathJax_Main]<[FONT=MathJax_Math]x[FONT=MathJax_Main]2[FONT=MathJax_Main]>[FONT=MathJax_Main]=[FONT=MathJax_Main]{[FONT=MathJax_Math]a[FONT=MathJax_Math]x[FONT=MathJax_Main]+[FONT=MathJax_Math]b[FONT=MathJax_Main]+[FONT=MathJax_Main]<[FONT=MathJax_Math]x[FONT=MathJax_Main]2[FONT=MathJax_Main]>[FONT=MathJax_Main]|[FONT=MathJax_Math]a[FONT=MathJax_Main],[FONT=MathJax_Math]b[FONT=MathJax_Main]∈[FONT=MathJax_AMS]Z[FONT=MathJax_Main]}

 

[Fernando Revilla]

But then how do we rigorously demonstrate that [FONT=MathJax_AMS]Z[FONT=MathJax_Main][[FONT=MathJax_Math]x[FONT=MathJax_Main]][FONT=MathJax_Main]/[FONT=MathJax_Main]<[FONT=MathJax_Math]x[FONT=MathJax_Main]2[FONT=MathJax_Main]>[FONT=MathJax_Main]=[FONT=MathJax_Main]{[FONT=MathJax_Math]a[FONT=MathJax_Math]x[FONT=MathJax_Main]+[FONT=MathJax_Math]b[FONT=MathJax_Main]+[FONT=MathJax_Main]<[FONT=MathJax_Math]x[FONT=MathJax_Main]2[FONT=MathJax_Main]>[FONT=MathJax_Main]|[FONT=MathJax_Math]a[FONT=MathJax_Main],[FONT=MathJax_Math]b[FONT=MathJax_Main]∈[FONT=MathJax_AMS]Z[FONT=MathJax_Main]} given that we cannot use the Division Algorithm

If $p(x)=a_nx^n+\ldots+a_2x^2+a_1x+a_0\in\mathbb{Z}[x]$, then $p(x)-(a_0+a_1x)=x^2h(x)$ with $h(x)\in\mathbb{Z}[x]$, so $p(x)+(x^2)=(a_0+a_1x)+(x^2)$ i.e. $[p(x)]=[a_0+a_1x]$.

 

[Math Amateur]

If $p(x)=a_nx^n+\ldots+a_2x^2+a_1x+a_0\in\mathbb{Z}[x]$, then $p(x)-(a_0+a_1x)=x^2h(x)$ with $h(x)\in\mathbb{Z}[x]$, so $p(x)+(x^2)=(a_0+a_1x)+(x^2)$ i.e. $[p(x)]=[a_0+a_1x]$.

Thanks again Fernando, that has made the case for $$ \mathbb{Z}[x]/<x^2>$$ very clear.

But maybe I did not ask a general enough question because I cannot see how the abve would work for $$ \mathbb{Z}[x]/<x^2 + a >$$ where $$ a \in \mathbb{Z} $$ - since then it seems that you would have to use the Division Algorithm.

Can someone please clarify this for me?

Peter

 

[Fernando Revilla]

But maybe I did not ask a general enough question because I cannot see how the abve would work for $$ \mathbb{Z}[x]/<x^2 + a >$$ where $$ a \in \mathbb{Z} $$ - since then it seems that you would have to use the Division Algorithm.

Can someone please clarify this for me?

I give you a hint. Suppose $p(x)\in\mathbb{Z}[x]$ has degree $\geq 2$, for example $p(x)=3x^3+7x^2-6x+5$. We have to proof that there exists $\alpha x+\beta\in\mathbb{Z}[x]$ such that $$p(x)-(\alpha x +\beta)=h(x)(x^2+1)\mbox { for some }h(x)\in\mathbb{Z}[x]\qquad (*)$$ The polynomial $h(x)$ has necessarily the form $h(x)=b_1x+b_0$. Now, identify coefficients in $(*)$ and you'll easily verify that there exists $\alpha x +\beta=-9x-2$. Try to generalize but please, show some work before asking.

 

[Math Amateur]

I give you a hint. Suppose $p(x)\in\mathbb{Z}[x]$ has degree $\geq 2$, for example $p(x)=3x^3+7x^2-6x+5$. We have to proof that there exists $\alpha x+\beta\in\mathbb{Z}[x]$ such that $$p(x)-(\alpha x +\beta)=h(x)(x^2+1)\mbox { for some }h(x)\in\mathbb{Z}[x]\qquad (*)$$ The polynomial $h(x)$ has necessarily the form $h(x)=b_1x+b_0$. Now, identify coefficients in $(*)$ and you'll easily verify that there exists $\alpha x +\beta=-9x-2$. Try to generalize but please, show some work before asking.

Thanks Fernando.

Just working on this now ... but please permit one question ... as I am working ...

You write:

"The polynomial $h(x)$ has necessarily the form $h(x)=b_1x+b_0$..."

Why is this the case?

Peter

 

[Math Amateur]

Take a particular example:

Take $$ p(x) \in \mathbb{Z}[x] $$ to be

$$ p(x) = 3x^3 + 7x^2 -6x + 5 $$

Then p(x0 is a member of the coset $$ \alpha x + \beta $$ if

$$ p(x) - (\alpha x + \beta ) = (b_1 x + b_0) (x^2 + 1 $$ ... (1)

If we assume $$ (\alpha x + \beta ) $$ = (-9x - 2) then we have

LHS of (1) = $$ 3x^3 + 7x^2 + 3x + 7 $$

and

RHS of (1) = $$ b_1 x^3 + b_0 x^2 = b_1 x + b_0 $$

So if we put $$ b_1 = 3 $$ and $$ b_0 = 7 $$ then LHS = RHS and we have verified (1)Generalise: (not quite sure here)

If p(x) is a member of $$ <x^2 + 1> $$ then

$$ p(x) - ( \alpha x + \beta ) = (b_0 + b_1 x) (x^2 + 1) $$

This is true if $$ \alpha = b_1 $$ and $$ \beta = b_0 $$

Thus

$$ p(x) + <x^2 + 1> = ( \alpha x + \beta) + <x^2 + 1> $$

i.e $$ <p(x)> = (\alpha x + \beta )$$

Is that OK? (I am not sure of the genalisation)

Peter

 

[Fernando Revilla]

Is that OK? (I am not sure of the genalisation)

Yes, it is OK. You can generalize using following theorem:

Theorem. Let $D$ be an integral domain, $f(x), g(x)\in D[x]$ with $g(x)\neq 0$. Suppose that $\mbox{deg }g(x)=m$ with leading coefficiet $b_m$. Then, there exist $k\in\mathbb{N}$, $q(x),r(x)\in D[x]$ with $\mbox{deg }r(x)<\mbox{deg }g(x)$ such that: $$b_m^kf(x)=q(x)g(x)+r(x)$$ Now, apply this theorem for $D=\mathbb{Z}$, $g(x)=x^2+1$ (so $b_m=1$) and you'll obtain

$f(x)=q(x)(x^2+1)+\alpha x+\beta$ i.e. $f(x)+(x^2+1)=\alpha x +\beta +(x^2+1)$
​

Alternatively to that theorem, choose a general polynomial $f(x)=a_nx^n+\ldots+a_1x+a_0$ and identify coefficients in $f(x)-(\alpha x+\beta)=(x^2+1)(b_{n-2}x^{n-2}+\ldots+b_0)$.

 

[Math Amateur]

Yes, it is OK. You can generalize using following theorem:

Theorem. Let $D$ be an integral domain, $f(x), g(x)\in D[x]$ with $g(x)\neq 0$. Suppose that $\mbox{deg }g(x)=m$ with leading coefficiet $b_m$. Then, there exist $k\in\mathbb{N}$, $q(x),r(x)\in D[x]$ with $\mbox{deg }r(x)<\mbox{deg }g(x)$ such that: $$b_m^kf(x)=q(x)g(x)+r(x)$$ Now, apply this theorem for $D=\mathbb{Z}$, $g(x)=x^2+1$ (so $b_m=1$) and you'll obtain

$f(x)=q(x)(x^2+1)+\alpha x+\beta$ i.e. $p(x)+(x^2+1)=\alpha x +\beta +(x^2+1)$
​

Alternatively to that theorem, choose a general polynomial $f(x)=a_nx^n+\ldots+a_1x+a_0$ and identify coefficients in $f(x)-(\alpha x+\beta)=(x^2+1)(b_{n-2}x^{n-2}+\ldots+b_0)$.

Fernando,

My thanks for all your helpful guidance

Peter