683: Polynomials Ring over Commutative Ring

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definition of polynomials ring over commutative ring

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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: Note

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Starting Context

• The reader knows a definition of ring.

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Target Context

• The reader will have a definition of polynomials ring over commutative ring.

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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:
$R$: $\in \{\text{ the commutative rings }\}$
$\ast R[x]$: $=\{\sum _{j\in \mathbb{N}}{p}_j{x}^j|p_j\in R\text{ such that only a finite number of }{p}_j\text{ s are not zero }\}$, with the addition and the multiplication specified below, $\in \{\text{ the commutative rings }\}$
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Conditions:
$\mathrm{\forall }\sum _{j\in \mathbb{N}}{p}_j{x}^j,\sum _{j\in \mathbb{N}}{p}_j^{\prime }{x}^j\in R[x](\sum _{j\in \mathbb{N}}{p}_j{x}^j+\sum _{j\in \mathbb{N}}{p}_j^{\prime }{x}^j=\sum _{j\in \mathbb{N}}(p_j+p_j^{\prime })x^j)$
$\wedge$
$\mathrm{\forall }\sum _{j\in \mathbb{N}}{p}_j{x}^j,\sum _{j\in \mathbb{N}}{p}_j^{\prime }{x}^j\in R[x](\sum _{j\in \mathbb{N}}{p}_j{x}^j\sum _{j\in \mathbb{N}}{p}_j^{\prime }{x}^j=\sum _{j\in \mathbb{N}}\sum _{l\in \{0,...,j\}}{p}_l{p}_{j-l}^{\prime }{x}^j)$
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Let us understand the logical structure of the construction: 1st, $\sum _{j\in \mathbb{N}}{p}_j{x}^j$, denoted also as $p(x)=p_n{x}^n+...+p_0$, in the definition of $R[x]$ is a single element, not the sum of the multiple elements: $\sum$ or $+$ is just a symbol used inside the expression of a single element and does not mean 'addition' (in fact, it cannot be understood as 'addition', because 'addition' has not been defined yet); 2nd, the addition of elements of $R[x]$, $+$, is defined; but after all, the single element equals the sum, because addition is defined to cause so.

So, $+$ in $p(x)=p_n{x}^n+...+p_0$ and $+$ in $p(x)+p^{\prime }(x)$ are really different.

Using the same symbol with the 2 meanings may seem confusing (it is indeed confusing), but of course, the same symbol is used in order to intentionally allow confusions: confusing the 2 meanings is harmless and in fact convenient: if the 1st meaning was denoted like $p(x)=p_n{x}^n@...@p_0$, we would have to write conversions like $p_n{x}^n@...@p_0=p_n{x}^n+...+p_0$, but we just write $p_n{x}^n+...+p_0$ and interpret it conveniently because the both interpretations are valid.

$\sum _{j\in \mathbb{N}}{p}_j{x}^j\sum _{k\in \mathbb{N}}{p}_k^{\prime }{x}^k=\sum _{j\in \mathbb{N}}\sum _{k\in \mathbb{N}}{p}_j{p}_k^{\prime }{x}^{j+k}$ is obviously true, where the right hand side is the sum of the elements.

Obviously, only a finite number of coefficients in the addition or the multiplication are nonzero, so, the addition and the multiplication are closed.

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2: Natural Language Description

For any commutative ring, $R$, $R[x]:=\{\sum _{j\in \mathbb{N}}{p}_j{x}^j|p_j\in R\text{ such that only a finite number of }{p}_j\text{ s are not zero }\}$, with the addition and the multiplication such that $\mathrm{\forall }\sum _{j\in \mathbb{N}}{p}_j{x}^j,\sum _{j\in \mathbb{N}}{p}_j^{\prime }{x}^j\in R[x](\sum _{j\in \mathbb{N}}{p}_j{x}^j+\sum _{j\in \mathbb{N}}{p}_j^{\prime }{x}^j=\sum _{j\in \mathbb{N}}(p_j+p_j^{\prime })x^j)$ and $\mathrm{\forall }\sum _{j\in \mathbb{N}}{p}_j{x}^j,\sum _{j\in \mathbb{N}}{p}_j^{\prime }{x}^j\in R[x](\sum _{j\in \mathbb{N}}{p}_j{x}^j\sum _{j\in \mathbb{N}}{p}_j^{\prime }{x}^j=\sum _{j\in \mathbb{N}}\sum _{l\in \{0,...,j\}}{p}_l{p}_{j-l}^{\prime }{x}^j)$.

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3: Note

Let us see that $R[x]$ is indeed a ring.

1) it is an Abelian group under addition: the addition is associative, because it comes down to the associativity of $R$, $0$ is the additive identity, each $p(x)=p_n{x}^n+...+p_0$ has the additive inverse, $(-p_n)x^n+...+(-p_0)$, the addition is commutative, because it comes down to the commutativity of $R$; 2) it is a monoid under multiplication: the multiplication is associative, because for the multiplication of any 3 elements, while the coefficient of $x^j$ is the sum of the multiplications of the appropriate coefficients of the 3 elements, the selections of such trios are the same regardless of the associations, and the issue comes down to the associativity of each term, which is guaranteed by the associativity of $R$, $1$ is the multiplicative identity; 3) multiplication is distributive with respect to addition: $\sum _j{p}_j{x}^j(\sum _k{p}_k^{\prime }{x}^k+\sum _k{p}_k^{″}{x}^k)=\sum _j{p}_j{x}^j(\sum _k(p_k^{\prime }+p_k^{″})x^k)=\sum _j\sum _k{p}_j(p_k^{\prime }+p_k^{″})x^{j+k}=\sum _j\sum _k(p_j{p}_k^{\prime }+p_j{p}_k^{″})x^{j+k}=\sum _j\sum _k{p}_j{p}_k^{\prime }{x}^{j+k}+\sum _j\sum _k{p}_j{p}_k^{″}{x}^{j+k}=\sum _j{p}_j{x}^j\sum _k{p}_k^{\prime }{x}^k+\sum _j{p}_j{x}^j\sum _k{p}_k^{″}{x}^k$, $(\sum _k{p}_k^{\prime }{x}^k+\sum _k{p}_k^{″}{x}^k)\sum _j{p}_j{x}^j=\sum _k{p}_k^{\prime }{x}^k\sum _j{p}_j{x}^j+\sum _k{p}_k^{″}{x}^k{p}_j{x}^j$ is likewise.

Let us see that $R[x]$ is indeed a commutative ring.

$\sum _j{p}_j{x}^j\sum _k{p}_k^{\prime }{x}^k=\sum _j\sum _k{p}_j{p}_k^{\prime }{x}^{j+k}=\sum _k\sum _j{p}_k^{\prime }{p}_j{x}^{j+k}=\sum _k{p}_k^{\prime }{x}^k\sum _j{p}_j{x}^j$.

Why is $R$ required to be commutative? That question arises because the above argument to confirm that $R[x]$ is a ring does not use the commutativity of $R$. In fact, it should be possible to define $R[x]$ with a non-commutative $R$. $R$ is usually required to be commutative probably in order to make each evaluation ring homomorphism a ring homomorphism. The evaluation ring homomorphism is for each $p^{\prime }\in R$, $f_{p^{\prime }}:R[x]\to R,p(x)\mapsto p(p^{\prime })$. When $R$ is not commutative, $f_{p^{\prime }}((p_{1}x)(p_1^{\prime }x))=f_{p^{\prime }}(p_1{p}_1^{\prime }{x}^2)=p_1{p}_1^{\prime }{p^{\prime }}^2\ne p_1{p}^{\prime }{p}_1^{\prime }{p}^{\prime }=f_{p^{\prime }}(p_{1}x)f_{p^{\prime }}(p_1^{\prime }x)$, so, not any ring homomorphism. Why does it have to be a ring homomorphism? Well, it does not have to be so, but it is usually desired to be so. That is because evaluation is almost an expected accessory to $R[x]$.

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References

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