description/proof of that for polynomials ring over field, principal ideal by irreducible polynomial is maximal ideal
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of field.
- The reader knows a definition of polynomials ring over commutative ring.
- The reader knows a definition of principal ideal of ring.
- The reader knows a definition of maximal ideal of ring.
- The reader admits the proposition that the polynomials ring over any field is a Euclidean domain with the size function as taking the degree of the polynomial.
- The reader admits the proposition that any Euclidean domain is a principal integral domain.
Target Context
- The reader will have a description and a proof of the proposition that for the polynomials ring over any field, the principal ideal by any irreducible polynomial is a maximal ideal.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(F [x]\): \(= \text{ the polynomials ring over } F\)
\(p (x)\): \(\in F [x]\), \(\in \{\text{ the irreducible polynomials }\}\)
\(I_b (p (x))\): \(= \text{ the principal ideal of } F [x] \text{ by } p (x)\)
//
Statements:
\(I_b (p (x)) \in \{\text{ the maximal ideals }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(F [x]\) is a principal integral domain; Step 2: see that \(I_b (p (x)) \subset F [x]\); Step 3: suppose that there is an ideal, \(I\), such that \(I_b (p (x)) \subseteq I\) and see that \(I = I_b (p (x))\) or \(I = F [x]\).
Step 1:
Let us see that \(F [x]\) is a principal integral domain.
\(F [x]\) is a Euclidean domain, by the proposition that the polynomials ring over any field is a Euclidean domain with the size function as taking the degree of the polynomial.
\(F [x]\) is a principal integral domain, by the proposition that any Euclidean domain is a principal integral domain.
Step 2:
Let us see that \(I_b (p (x)) \subset F [x]\).
As \(F [x]\) is commutative, \(I_b (p (x)) = I_l (p (x)) = F [x] p (x)\).
\(p (x)\) is with positive-degree, because each 0-degree polynomial is \(0\) or a unit, which is not any irreducible element by the definition of irreducible element of commutative ring.
So, each element of \(F [x] p (x)\) is \(0\) or with positive-degree, which means that no nonzero 0-degree polynomial is contained in it, which means that \(I_b (p (x)) \subset F [x]\).
Step 3:
Let us suppose that there is an ideal, \(I\), such that \(I_b (p (x)) \subseteq I\).
As \(F [x]\) is a principal integral domain, \(I\) is a principal ideal, \(I_b (q (x))\).
\(I_b (q (x)) = I_l (q (x))\) as before.
As \(p (x) \in I_b (p (x)) \subseteq I_b (q (x)) = I_l (q (x))\), \(p (x) = p' (x) q (x)\) for a \(p' (x) \in F [x]\).
As \(p (x)\) is irreducible, \(p' (x)\) is with \(0\)-degree or \(q (x)\) is with \(0\)-degree.
When \(p' (x)\) is with \(0\)-degree, \(q (x)\) is a constant multiple of \(p (x)\), which implies that \(I_b (p (x)) = I_b (q (x)) = I\).
When \(q (x)\) is with \(0\)-degree, \(I = I_b (q (x)) = I_l (q (x)) = F [x]\).
