definition of polynomials ring over commutative ring
Topics
About: ring
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Note
Starting Context
- The reader knows a definition of ring.
Target Context
- The reader will have a definition of polynomials ring over commutative ring.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( R\): \(\in \{\text{ the commutative rings }\}\)
\(*R [x]\): \(= \{\sum_{j \in \mathbb{N}} p_j x^j \vert p_j \in R \text{ such that only a finite number of } p_j \text{ s are not zero }\}\), with the addition and the multiplication specified below, \(\in \{\text{ the commutative rings }\}\)
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Conditions:
\(\forall \sum_{j \in \mathbb{N}} p_j x^j, \sum_{j \in \mathbb{N}} p'_j x^j \in R [x] (\sum_{j \in \mathbb{N}} p_j x^j + \sum_{j \in \mathbb{N}} p'_j x^j = \sum_{j \in \mathbb{N}} (p_j + p'_j) x^j)\)
\(\land\)
\(\forall \sum_{j \in \mathbb{N}} p_j x^j, \sum_{j \in \mathbb{N}} p'_j x^j \in R [x] (\sum_{j \in \mathbb{N}} p_j x^j \sum_{j \in \mathbb{N}} p'_j x^j = \sum_{j \in \mathbb{N}} \sum_{l \in \{0, ..., j\}} p_l p'_{j - l} x^j)\)
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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 \(p (x) + p' (x)\) 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.
\(\sum_{j \in \mathbb{N}} p_j x^j \sum_{k \in \mathbb{N}} p'_k x^k = \sum_{j \in \mathbb{N}} \sum_{k \in \mathbb{N}} p_j p'_k x^{j + k}\) 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.
2: Natural Language Description
For any commutative ring, \(R\), \(R [x] := \{\sum_{j \in \mathbb{N}} p_j x^j \vert p_j \in R \text{ such that only a finite number of } p_j \text{ s are not zero }\}\), with the addition and the multiplication such that \(\forall \sum_{j \in \mathbb{N}} p_j x^j, \sum_{j \in \mathbb{N}} p'_j x^j \in R [x] (\sum_{j \in \mathbb{N}} p_j x^j + \sum_{j \in \mathbb{N}} p'_j x^j = \sum_{j \in \mathbb{N}} (p_j + p'_j) x^j)\) and \(\forall \sum_{j \in \mathbb{N}} p_j x^j, \sum_{j \in \mathbb{N}} p'_j x^j \in R [x] (\sum_{j \in \mathbb{N}} p_j x^j \sum_{j \in \mathbb{N}} p'_j x^j = \sum_{j \in \mathbb{N}} \sum_{l \in \{0, ..., j\}} p_l p'_{j - l} x^j)\).
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: \(\sum_{j} p_j x^j (\sum_{k} p'_k x^k + \sum_{k} p''_k x^k) = \sum_{j} p_j x^j (\sum_{k} (p'_k + p''_k) x^k) = \sum_{j} \sum_{k} p_j (p'_k + p''_k) x^{j + k} = \sum_{j} \sum_{k} (p_j p'_k + p_j p''_k) x^{j + k} = \sum_{j} \sum_{k} p_j p'_k x^{j + k} + \sum_{j} \sum_{k} p_j p''_k x^{j + k} = \sum_j p_j x^j \sum_k p'_k x^k + \sum_j p_j x^j \sum_k p''_k x^k\), \((\sum_{k} p'_k x^k + \sum_{k} p''_k x^k) \sum_{j} p_j x^j = \sum_{k} p'_k x^k \sum_{j} p_j x^j + \sum_{k} p''_k x^k p_j x^j\) is likewise.
Let us see that \(R [x]\) is indeed a commutative ring.
\(\sum_j p_j x^j \sum_k p'_k x^k = \sum_j \sum_k p_j p'_k x^{j + k} = \sum_k \sum_j p'_k p_j x^{j + k} = \sum_k p'_k x^k \sum_j p_j x^j\).
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 \(p' \in R\), \(f_{p'}: R [x] \to R, p (x) \mapsto p (p')\). When \(R\) is not commutative, \(f_{p'} ((p_1 x) (p'_1 x)) = f_{p'} (p_1 p'_1 x^2) = p_1 p'_1 {p'}^2 \neq p_1 p' p'_1 p' = f_{p'} (p_1 x) f_{p'} (p'_1 x)\), 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 \(R [x]\).
