983: For Field, Polynomials Ring over Field, and Larger-Than-1-Degree Irreducible Polynomial, Field Can Be Extended to Have Root of Polynomial in Polynomials Ring over Extended Field, Which Is 'Fields - Homomorphisms' Isomorphic to Any Smallest Such

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description/proof of that for field, polynomials ring over field, and larger-than-1-degree irreducible polynomial, field can be extended to have root of polynomial in polynomials ring over extended field, which is 'fields - homomorphisms' isomorphic to any smallest such

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Topics

About: field

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note 1
• 3: Proof
• 4: Note 2

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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 irreducible element of commutative ring.
• The reader knows a definition of root of polynomial in polynomials ring over commutative ring.
• The reader knows a definition of principal ideal of ring.
• The reader knows a definition of quotient ring of ring by ideal.
• The reader knows a definition of field generated by subset of elements of field over subfield.
• The reader knows a definition of polynomial extended over extended field.
• The reader admits the proposition that for the polynomials ring over any field, the units are the nonzero constants.
• The reader admits the proposition that for the polynomials ring over any field, the principal ideal by any irreducible polynomial is a maximal ideal.
• The reader admits the proposition that the quotient of any commutative ring by any maximal ideal is a field.
• The reader admits the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
• The reader admits the proposition that any ring homomorphism between any fields is a field homomorphism.
• The reader admits the proposition that the range of any field homomorphism is a subfield of the codomain.
• The reader admits the proposition that any bijective field homomorphism is a 'fields - homomorphisms' isomorphism.
• The reader admits the proposition that the polynomials ring over any field is a Euclidean domain.
• The reader admits the proposition that the kernel of any ring homomorphism is an ideal of the domain.
• The reader admits the proposition that for any commutative ring, the ring is a field if and only if the ideals of the ring are the 0 ideal and the whole ring.
• The reader admits the proposition that any group homomorphism is injective if and only if the kernel is the 1 subgroup.

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Target Context

• The reader will have a description and a proof of the proposition that for any field, the polynomials ring over the field, and any larger-than-1-degree irreducible polynomial, the field can be extended to have a root of the polynomial in the polynomials ring over the extended field, which (the extended field) is 'fields - homomorphisms' isomorphic to any smallest such.

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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:
$F$: $\in \{\text{ the fields }\}$
$F[x]$: $=\text{ the polynomials ring over }F$
$p(x)$: $\in \{\text{ the n-degree irreducibles of }F[x]\}$ where $2\le n$
$I_b(p(x))$: $=\text{ the principal ideal of }F[x]\text{ by }p(x)$
$F[x]/I_b(p(x))$: $=\text{ the quotient ring }$, $\in \{\text{ the fields }\}$
$f$: $:F\to F[x]/I_b(p(x)),r\mapsto [r{x}^0]$
$\overline{F[x]/I_b(p(x))}$: $=F[x]/I_b(p(x))\text{ with }f(F)\text{ replaced with }F$
$\overline{p(x)}$: $=p(x)\text{ extended in }\overline{F[x]/I_b(p(x))}[x]$
//

Statements:
$[x]\in \{\text{ the roots of }\overline{p(x)}\}$
$\wedge$
$\mathrm{\forall }{F}^{\prime }\in \{\text{ the extended fields of }F\}\text{ such that }\mathrm{\exists }\alpha \in F^{\prime }\text{ such that }\widetilde{p(x)}(\alpha )=0(F_{F^{\prime }}(\alpha )\cong \overline{F[x]/I_b(p(x))})$, where $\widetilde{p(x)}$ is the extended polynomial in $F^{\prime }$, $F_{F^{\prime }}(\alpha )$ is the field generated by $\{\alpha \}$ in $F^{\prime }$ over $F$, and $\cong$ denotes 'fields - homomorphisms' isomorphic, by the isomorphism, $\varphi :\overline{F[x]/I_b(p(x))}\to F_{F^{\prime }}(\alpha ),[p^{\prime }(x)]\mapsto p^{\prime }(\alpha )$
//

Note that we sometimes use notations like "$r{x}^0$", which is usually denoted as just "$r$", whose ("$r{x}^0$"'s) intention is to distinguish $r=r{x}^0\in F[x]$ from $r\in F$.

As an immediate corollary, $\overline{F[x]/I_b(p(x))}$ is minimal such that $[x]$ is a root of $\overline{p(x)}$, which means that no element can be eliminated from $\overline{F[x]/I_b(p(x))}$ to keep it a field: supposing that there is an extended field of $\overline{F[x]/I_b(p(x))}$, $F^{\prime }$, $F_{F^{\prime }}([x])=\overline{F[x]/I_b(p(x))}$: while there is the 'fields - homomorphisms' isomorphism, $\varphi :\overline{F[x]/I_b(p(x))}\to F_{F^{\prime }}([x]),[p^{\prime }(x)]\mapsto p^{\prime }([x])$, in fact, $[p^{\prime }(x)]=p^{\prime }([x])$, so, it is indeed the identity map.

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2: Note 1

When $p(x)$ is 1-degree, $p(x)$ has the root in $F$: $p(x)=p_{1}x+p_0=(x+p_0/p_1)p_1$ with the root, $-p_0/p_1$, and $F$ does not need to be extended at all.

$F[x]/I_b(p(x))$ is not really any extension of $F$, because $F$ is not really contained in $F[x]/I_b(p(x))$, although rather prevalently, $f(F)$ is sloppily identified with $F$. We do not favor such sloppy-ness and go an extra mile to construct a real extension.

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3: Proof

Whole Strategy: Step 1: see that $F$ has no root of $p(x)$; Step 2: see that $F[x]/I_b(p(x))$ is a field; Step 3: see that $f$ is an injective field homomorphism; Step 4: see that $\overline{F[x]/I_b(p(x))}$ is an extended field of $F$; Step 5: see that $[x]$ is a root of $\overline{p(x)}$; Step 6: see that $F_{F^{\prime }}(\alpha )\cong \overline{F[x]/I_b(p(x))}$.

Step 1:

Let us see that $F$ has no root of $p(x)$.

If $p(x)$ had a root, $p(x)=(x-r)q(x)$ where $q(x)\in F[x]$. $q(x)$ would not be any constant, because otherwise, $p(x)$ would be of 0-or-1-degree, a contradiction. Then, $x-r$ nor $q(x)$ would be any unit, by the proposition that for the polynomials ring over any field, the units are the nonzero constants. So, $p(x)$ would not be any irreducible, by the definition of irreducible element of commutative ring.

Step 2:

$F[x]/I_b(p(x))$ is a field, because $I_b(p(x))$ is a maximal ideal, by the proposition that for the polynomials ring over any field, the principal ideal by any irreducible polynomial is a maximal ideal, and the quotient is a field, by the proposition that the quotient of any commutative ring by any maximal ideal is a field.

Step 3:

Let us see that $f$ is an injective field homomorphism.

For each $r_1,r_2\in F$ such that $r_1\ne r_2$, $[r_1{x}^0]\ne [r_2{x}^0]$, because $[r_1{x}^0]=[r_2{x}^0]$ would mean that $r_1{x}^0-r_2{x}^0\in I_b(p(x))$, which would mean that $(r_1-r_2)x^0=p(x)q(x)$, but as $p(x)$ is of larger-than-1-degree, the only possibility is that $q(x)=0$ and $(r_1-r_2)x^0=0$, which would mean that $r_1=r_2$, a contradiction.

$f$ is an additive group homomorphism: $f(r_1+r_2)=[(r_1+r_2)x^0]=[r_1{x}^0+r_2{x}^0]=[r_1{x}^0]+[r_2{x}^0]=f(r_1)+f(r_2)$: see the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.

$f(1)=[1x^0]$, which is $1$ in $F[x]/I_b(p(x))$.

$f(r_1{r}_2)=[r_1{r}_2{x}^0]=[r_1{x}^0{r}_2{x}^0]=[r_1{x}^0][r_1{x}^0]=f(r_1)f(r_2)$.

So, $f$ is a ring homomorphism, and is a field homomorphism, by the proposition that any ring homomorphism between any fields is a field homomorphism.

Step 4:

$f(F)\in F[x]/I_b(p(x))$ is a subfield of $F[x]/I_b(p(x))$, by the proposition that the range of any field homomorphism is a subfield of the codomain.

$F$ is not exactly any subfield of $F[x]/I_b(p(x))$, because any $r\in F$ is not any element of $F[x]/I_b(p(x))$, while $f(r)=[r{x}^0]$ is an element of $F[x]/I_b(p(x))$.

So, in order to construct a real extension of $F$, we define $\overline{F[x]/I_b(p(x))}$ as $F[x]/I_b(p(x))$ with $f(F)$ replaced by $F$. The operations on $\overline{F[x]/I_b(p(x))}$ are canonically defined: for each $r_1,r_2\in \overline{F[x]/I_b(p(x))}$, when $r_1$ or $r_2$ is in $F$, it is mapped to $f(r_j)$, the operations are done in $F[x]/I_b(p(x))$, and when the result is in $f(F)$, it is mapped back into $F$. $\overline{F[x]/I_b(p(x))}$ is indeed a field, because $F$ and $f(F)$ are 'fields - homomorphisms' isomorphic, by the proposition that any bijective field homomorphism is a 'fields - homomorphisms' isomorphism.

So, $\overline{F[x]/I_b(p(x))}$ is a field extension of $F$.

Step 5:

Let us see that $[x]\in \overline{F[x]/I_b(p(x))}$ is a root of $\overline{p(x)}$.

Let $p(x)=p_n{x}^n+...+p_0$, which means that $\overline{p(x)}=p_n{x}^n+...+p_0$.

$\overline{p(x)}([x])=p_n[x{]}^n+...+p_0=p_n[x^n]+...+p_0=[p_n{x}^0][x^n]+...+[p_0{x}^0]$, which means that $p_j\in F$ has been mapped to $f(p_j)$ and now we are doing the operations in $F[x]/I_b(p(x))$, $[p_n{x}^n]+...+[p_0{x}^0]=[p_n{x}^n+...+p_0{x}^0]=[0]$, which is mapped back to $0\in F$, so, $=0$.

Step 6:

Let us see that $F_{F^{\prime }}(\alpha )\cong \overline{F[x]/I_b(p(x))}$.

In fact, let us see that $F_{F^{\prime }}(\alpha )\cong F[x]/I_b(p(x))$ instead, which implies the above, because $F[x]/I_b(p(x))\cong \overline{F[x]/I_b(p(x))}$.

Let us define $\varphi :F[x]/I_b(p(x))\to F_{F^{\prime }}(\alpha ),[p^{\prime }(x)]\to p^{\prime }(\alpha )$.

Let us see that that is well-defined. For each $[p^{\prime }(x)]=[p^{″}(x)]$, $p^{\prime }(x)-p^{″}(x)=p(x)q(x)$. $p^{\prime }(\alpha )-p^{″}(\alpha )=p(\alpha )q(\alpha )=0q(\alpha )=0$. So, $p^{\prime }(\alpha )=p^{″}(\alpha )$, which means that $\varphi ([p^{\prime }(x)])$ does not depend on the representation.

Let us see that $\varphi$ is field homomorphic.

$\varphi ([p^{\prime }(x)]+[p^{″}(x)])=\varphi ([p^{\prime }(x)+p^{″}(x)])=p^{\prime }(\alpha )+p^{″}(\alpha )=\varphi ([p^{\prime }(x)])+\varphi ([p^{″}(x)])$. So, $\varphi$ is an additive group homomorphism, by the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.

$\varphi ([1x^0])=1$.

$\varphi ([p^{\prime }(x)][p^{″}(x)])=\varphi ([p^{\prime }(x)p^{″}(x)])=p^{\prime }(\alpha )p^{″}(\alpha )=\varphi ([p^{\prime }(x)])\varphi ([p^{″}(x)])$.

So, $\varphi$ is a ring homomorphism, and is a field homomorphism, by the proposition that any ring homomorphism between any fields is a field homomorphism.

Let us see that $\varphi$ is bijective. $Ker(\varphi )$ is an ideal of $F[x]/I_b(p(x))$, by the proposition that the kernel of any ring homomorphism is an ideal of the domain, and the ideal is $\{0\}$ or $F[x]/I_b(p(x))$, by the proposition that for any commutative ring, the ring is a field if and only if the ideals of the ring are the 0 ideal and the whole ring. But the ideal is not $F[x]/I_b(p(x))$, because $[1x^0]$ is mapped to $1$, so, the ideal is $\{0\}$. $\varphi$ is injective, by the proposition that any group homomorphism is injective if and only if the kernel is the 1 subgroup. $\varphi$ is surjective, because while $\varphi (F[x]/I_b(p(x)))$ is a field, by the proposition that the range of any field homomorphism is a subfield of the codomain, $F$ is contained in the range and $\alpha$ is contained in the range, which means that $\varphi (F[x]/I_b(p(x)))$ is a field that contains $F$ and $\{\alpha \}$, but as $F_{F^{\prime }}(\alpha )$ is the smallest such, $F_{F^{\prime }}(\alpha )\subseteq \varphi (F[x]/I_b(p(x)))$.

So, $\varphi$ is a 'fields - homomorphisms' isomorphism, by the proposition that any bijective field homomorphism is a 'fields - homomorphisms' isomorphism.

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4: Note 2

Now, $\overline{p(x)}$ has the root, ${\alpha }_1:=[x]\in \overline{F[x]/I_b(p(x))}$, which means that $\overline{p(x)}=(x-{\alpha }_1)q_{n-1}(x)$, where $q_{n-1}(x)\in \overline{F[x]/I_b(p(x))}[x]$ is of '$n-1$'-degree.

$q_{n-1}(x)$ may be or may not be irreducible (at least, we have not proved that it is not irreducible). If $q_{n-1}(x)$ is not irreducible, it can be factorized into some irreducibles. Anyway, if a larger-than-1-degree irreducible factor, $q(x)$, remains, we can apply this proposition for $q(x)$ and have the extended field of $\overline{F[x]/I_b(p(x))}$ in which $q(x)$ has the root, ${\alpha }_2:=[x]$, and so on. Note that ${\alpha }_2:=[x]$ is of course different from the previous ${\alpha }_1:=[x]$: ${\alpha }_1:=[x]$ is an element of $\overline{F[x]/I_b(p(x))}$, while ${\alpha }_2:=[x]$ is an element of $\overline{\stackrel{―}{F[x]/I_b(p(x))}[x]/I_b(q(x))}$, not of $\overline{F[x]/I_b(p(x))}$.

After all, any not-necessarily-irreducible polynomial, $p(x)\in F[x]$, can be factorized as $r(x-{\alpha }_1)...(x-{\alpha }_n)\in F^{\prime }[x]$ where $F^{\prime }$ is an extended field of $F$. $\{{\alpha }_1,...,{\alpha }_n\}$ may not be distinct.

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References

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