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

#### Peter

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

I tried to first deal with the rings [TEX] \mathbb{R}[x]/<x^2> [/TEX] and [TEX] \mathbb{R}[x]/<x^2 +1> [/TEX] 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 [TEX] \mathbb{R}[x]/<x^2> [/TEX] where [TEX] \mathbb{R}[x] [/TEX] is the ring of polynomials with real co-efficients.

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

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

[TEX] g(x) = q(x)(x^2) + r(x) [/TEX] where r(x) = 0 or r(x) has degree less than 2.

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

Thus [TEX] g(x) + <x^2> = q(x)(x^2) + r(x) + <x^2> [/TEX]

= [TEX] r(x) + <x^2> [/TEX] since the ideal [TEX] <x^2> [/TEX] absorbs the term [TEX] q(x)(x^2) + r(x) [/TEX]

= [TEX] ax + b + <x^2> [/TEX]

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

Now, by a similar argument we can demonstrate that

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

which makes the two rings [TEX] \mathbb{R}[x]/<x^2> [/TEX] and [TEX] \mathbb{R}[x]/<x^2 +1> [/TEX] 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 [TEX] \mathbb{R}[x] [/TEX] is a Euclidean Domain ... so the same argument as above does not apply to

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

How do we rigorously demonstrate that

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

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

Peter

#### Fernando Revilla

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which makes the two rings [TEX] \mathbb{R}[x]/<x^2> [/TEX] and [TEX] \mathbb{R}[x]/<x^2 +1> [/TEX] 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
[TEX] \mathbb{Z}[x]/<x^2> = \{ ax + b + <x^2> | a, b \in \mathbb{Z} \} [/TEX] and
[TEX] \mathbb{Z}[x]/<x^2 +1> = \{ ax + b + <x^2 + 1> | a, b \in \mathbb{Z} \} [/TEX]
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 $p(x)-q(x)=h(x)(x^2+1)$.

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#### Peter

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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 $p(x)-q(x)=h(x)(x^2+1)$.

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$$\displaystyle (ax+b)(a'x+b') x^2+1)$$

How can we carry out Euclidean Divisions in $$\displaystyle \mathbb{Z}[x]$$ when it is not a Euclidean Domain?

Can you clarify? Am i misunderstanding something?

Peter

#### Fernando Revilla

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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$$\displaystyle (ax+b)(a'x+b') x^2+1)$$

How can we carry out Euclidean Divisions in $$\displaystyle \mathbb{Z}[x]$$ 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.

#### Peter

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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][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]<[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main][/FONT]

But then how do we rigorously demonstrate that

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Can you help?

Peter

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#### Fernando Revilla

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But then how do we rigorously demonstrate that [FONT=MathJax_AMS]Z[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]<[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]>[FONT=MathJax_Main]=[/FONT][/FONT][FONT=MathJax_Main]{[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]<[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]∈[/FONT][FONT=MathJax_AMS]Z[/FONT][FONT=MathJax_Main]}[/FONT] 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. $[p(x)]=[a_0+a_1x]$.

#### Peter

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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 $$\displaystyle \mathbb{Z}[x]/<x^2>$$ very clear.

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

Can someone please clarify this for me?

Peter

#### Fernando Revilla

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MHB Math Helper
But maybe I did not ask a general enough question because I cannot see how the abve would work for $$\displaystyle \mathbb{Z}[x]/<x^2 + a >$$ where $$\displaystyle a \in \mathbb{Z}$$ - 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.

#### Peter

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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

#### Peter

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Take a particular example:

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

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

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

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

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

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

and

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

So if we put $$\displaystyle b_1 = 3$$ and $$\displaystyle b_0 = 7$$ then LHS = RHS and we have verified (1)

Generalise: (not quite sure here)

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

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

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

Thus

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

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

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

Peter

#### Fernando Revilla

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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 $f(x)-(\alpha x+\beta)=(x^2+1)(b_{n-2}x^{n-2}+\ldots+b_0)$.

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#### Peter

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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,