684: Polynomials Ring over Integral Domain Is Integral Domain

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description/proof of that polynomials ring over integral domain is integral domain

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Topics

About: ring

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of integral domain.
• The reader knows a definition of polynomials ring over commutative ring.

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Target Context

• The reader will have a description and a proof of the proposition that any polynomials ring over any integral domain is an integral domain.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the integral domains }\}$
$R[x]$: $=\text{ the polynomials ring over }R$
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Statements:
$R[x]$: $\in \{\text{ the integral domains }\}$
//

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2: Natural Language Description

For any integral domain, $R$, the polynomials ring over $R$, $R[x]$, is an integral domain.

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3: Proof

Whole Strategy: Step 1: see that $R[x]$ is a nonzero commutative ring; Step 2: prove the qualification that the multiplication of any 2 elements being 0 implies that one of the elements is 0.

Step 1:

As $R$ is nonzero, $R[x]$ is nonzero, because any nonzero element of $R$ is contained in $R[x]$ as the constant.

$R[x]$ is a commutative ring, as is shown in Note for the definition of polynomials ring of commutative ring.

Step 2:

Let us prove that for each $p(x),p^{\prime }(x)\in R[x]$ such that $p(x)p^{\prime }(x)=0$, $p(x)=0$ or $p^{\prime }(x)=0$.

Step 2 Strategy: Step 2-1: suppose that $p(x)$ is with certain coefficients with n-degree with $0\le n$ and $p(x)\ne 0$ and $p^{\prime }(x)$ is with certain coefficients with m-degree, and see that $p^{\prime }(x)=0$; Step 2-2: suppose that $p^{\prime }(x)$ is with certain coefficients with m-degree with $0\le m$ and $p^{\prime }(x)\ne 0$ and $p(x)$ is with certain coefficients with n-degree, and see that $p(x)=0$.

In fact, Step 2-2 is not necessary by the known fact that $R[x]$ is commutative, but we do it anyway for a fun.

Step 2-1:

Let us suppose that $p(x)=p_n{x}^n+...+p_0$ with $p_n\ne 0$ and $p^{\prime }(x)=p_m^{\prime }{x}^m+...+p_0^{\prime }$. $p_n$ may be $p_0$ when $n=0$. The supposition for $p^{\prime }(x)$ means that $p^{\prime }(x)$ is equal to or smaller than $m$ degree, which does not lose any generality.

$p_n{p}_m^{\prime }$ has to be $0$, because that is the coefficient of the $n+m$ degree, which implies that $p_m^{\prime }=0$, because $p_n\ne 0$ and $R$ is an integral domain.

Then, $p_n{p}_{m-1}^{\prime }$ has to be $0$, because that is the coefficient of the $n+m-1$ degree, which implies that $p_{m-1}^{\prime }=0$, because $p_n\ne 0$ and $R$ is an integral domain.

And so on, after all, $p_n{p}_0^{\prime }$ has to be $0$, because that is the coefficient of the $n$ degree, which implies that $p_0^{\prime }=0$, because $p_n\ne 0$ and $R$ is an integral domain.

So, all the coefficients of $p^{\prime }(x)$ are $0$, so, $p^{\prime }(x)=0$.

Step 2-2:

Let us suppose that $p^{\prime }(x)=p_m^{\prime }{x}^m+...+p_0^{\prime }$ with $p_m^{\prime }\ne 0$ and $p(x)=p_n{x}^n+...+p_0$. $p_m^{\prime }$ may be $p_0^{\prime }$ when $m=0$. The supposition for $p(x)$ means that $p(x)$ is equal to or smaller than $n$ degree, which does not lose any generality.

$p_n{p}_m^{\prime }$ has to be $0$, because that is the coefficient of the $n+m$ degree, which implies that $p_n=0$, because $p_m^{\prime }\ne 0$ and $R$ is an integral domain.

Then, $p_{n-1}{p}_m^{\prime }$ has to be $0$, because that is the coefficient of the $n-1+m$ degree, which implies that $p_{n-1}=0$, because $p_m^{\prime }\ne 0$ and $R$ is an integral domain.

And so on, after all, $p_0{p}_m^{\prime }$ has to be $0$, because that is the coefficient of the $m$ degree, which implies that $p_0=0$, because $p_m^{\prime }\ne 0$ and $R$ is an integral domain.

So, all the coefficients of $p(x)$ are $0$, so, $p(x)=0$.

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References

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