description/proof of that polynomials ring over integral domain is integral domain
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: Proof
Starting Context
- The reader knows a definition of integral domain.
- The reader knows a definition of polynomials ring over commutative ring.
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.
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:
\(R\): \(\in \{\text{ the integral domains }\}\)
\(R [x]\): \(= \text{ the polynomials ring over } R\)
//
Statements:
\(R [x]\): \(\in \{\text{ the integral domains }\}\)
//
2: Natural Language Description
For any integral domain, \(R\), the polynomials ring over \(R\), \(R [x]\), is an integral domain.
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 \(p (x), p' (x) \in R [x]\) such that \(p (x) p' (x) = 0\), \(p (x) = 0\) or \(p' (x) = 0\).
Step 2 Strategy: Step 2-1: suppose that \(p (x)\) is with certain coefficients with n-degree with \(0 \le n\) and \(p (x) \neq 0\) and \(p' (x)\) is with certain coefficients with m-degree, and see that \(p' (x) = 0\); Step 2-2: suppose that \(p' (x)\) is with certain coefficients with m-degree with \(0 \le m\) and \(p' (x) \neq 0\) 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 \(p_n \neq 0\) and \(p' (x) = p'_m x^m + ... + p'_0\). \(p_n\) may be \(p_0\) when \(n = 0\). The supposition for \(p' (x)\) means that \(p' (x)\) is equal to or smaller than \(m\) degree, which does not lose any generality.
\(p_n p'_m\) has to be \(0\), because that is the coefficient of the \(n + m\) degree, which implies that \(p'_m = 0\), because \(p_n \neq 0\) and \(R\) is an integral domain.
Then, \(p_n p'_{m - 1}\) has to be \(0\), because that is the coefficient of the \(n + m - 1\) degree, which implies that \(p'_{m - 1} = 0\), because \(p_n \neq 0\) and \(R\) is an integral domain.
And so on, after all, \(p_n p'_0\) has to be \(0\), because that is the coefficient of the \(n\) degree, which implies that \(p'_0 = 0\), because \(p_n \neq 0\) and \(R\) is an integral domain.
So, all the coefficients of \(p' (x)\) are \(0\), so, \(p' (x) = 0\).
Step 2-2:
Let us suppose that \(p' (x) = p'_m x^m + ...+ p'_0\) with \(p'_m \neq 0\) and \(p (x) = p_n x^n + ... + p_0\). \(p'_m\) may be \(p'_0\) 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.
\(p_n p'_m\) has to be \(0\), because that is the coefficient of the \(n + m\) degree, which implies that \(p_n = 0\), because \(p'_m \neq 0\) and \(R\) is an integral domain.
Then, \(p_{n - 1} p'_m\) has to be \(0\), because that is the coefficient of the \(n - 1 + m\) degree, which implies that \(p_{n - 1} = 0\), because \(p'_m \neq 0\) and \(R\) is an integral domain.
And so on, after all, \(p_0 p'_m\) has to be \(0\), because that is the coefficient of the \(m\) degree, which implies that \(p_0 = 0\), because \(p'_m \neq 0\) and \(R\) is an integral domain.
So, all the coefficients of \(p (x)\) are \(0\), so, \(p (x) = 0\).
