description/proof of that for polynomials ring over field, units are nonzero constants
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of polynomials ring over commutative ring.
- The reader knows a definition of field.
- The reader knows a definition of units of ring.
- The reader admits the proposition that for the polynomials ring over any integral domain, any unit is a nonzero constant.
Target Context
- The reader will have a description and a proof of the proposition that for the polynomials ring over any field, the units are the nonzero constants.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(F [x]\): \(= \text{ the polynomials ring over } F\)
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Statements:
\(\{\text{ the units of } F [x]\} = \{\text{ the nonzero constants in } F [x]\}\)
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2: Note
When \(F\) is just an integral domain, a nonzero constant may not be any unit: for a \(p (x) = p_0 \in F [x]\), \(p_0 \in F\) may not have any inverse, then, \(p (x) = p_0 \in F [x]\) will not have any inverse: compare with the proposition that for the polynomials ring over any integral domain, any unit is a nonzero constant.
3: Proof
Whole Strategy: Step 1: see that each unit is a nonzero constant; Step 2: see that each nonzero constant is a unit.
Step 1:
As any field is an integral domain, each unit is a nonzero constant, by the proposition that for the polynomials ring over any integral domain, any unit is a nonzero constant.
Step 2:
Let \(p (x) = p_0 x^0 \in F [x]\) be any nonzero constant: as we use notations like "\(p_0 x^0\)", it is usually written as \(p_0\): we use such notations in order to distinguish \(p_0 \in F\) and \(p_0 \in F [x]\).
As \(p_0 \in F\) and \(F\) is a field, there is the inverse, \(p_0^{-1} \in F\). \(p_0^{-1} x^0 \in F [x]\) and \(p_0 x^0 p_0^{-1} x^0 = p_0 p_0^{-1} x^0 = 1 x^0\) and \(p_0^{-1} x^0 p_0 x^0 = p_0^{-1} p_0 x^0 = 1 x^0\).
As \(1 x^0\) is the identity element of \(F [x]\), \(p_0^{-1} x^0\) is an inverse of \(p_0 x^0\).
