956: For Polynomials Ring over Field, Principal Ideal by Irreducible Polynomial Is Maximal Ideal

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description/proof of that for polynomials ring over field, principal ideal by irreducible polynomial is maximal ideal

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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: Proof

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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.

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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.

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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:
$F$: $\in \{\text{ the fields }\}$
$F[x]$: $=\text{ the polynomials ring over }F$
$p(x)$: $\in F[x]$, $\in \{\text{ the irreducible polynomials }\}$
$I_b(p(x))$: $=\text{ the principal ideal of }F[x]\text{ by }p(x)$
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Statements:
$I_b(p(x))\in \{\text{ the maximal ideals }\}$
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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 $p(x)\in I_b(p(x))\subseteq I_b(q(x))=I_l(q(x))$, $p(x)=p^{\prime }(x)q(x)$ for a $p^{\prime }(x)\in F[x]$.

As $p(x)$ is irreducible, $p^{\prime }(x)$ is with $0$-degree or $q(x)$ is with $0$-degree.

When $p^{\prime }(x)$ 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, $I=I_b(q(x))=I_l(q(x))=F[x]$.

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References

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