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
Topics
About: field
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
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.
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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(F [x]\): \(= \text{ the polynomials ring over } F\)
\(p (x)\): \(\in \{\text{ the n-degree irreducibles of } F [x]\}\) 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:
\([x] \in \{\text{ the roots of } \overline{p (x)}\}\)
\(\land\)
\(\forall F' \in \{\text{ the extended fields of } F\} \text{ such that } \exists \alpha \in F' \text{ such that } \widetilde{p (x)} (\alpha) = 0 (F_{F'} (\alpha) \cong \overline{F [x] / I_b (p (x))})\), where \(\widetilde{p (x)}\) is the extended polynomial in \(F'\), \(F_{F'} (\alpha)\) is the field generated by \(\{\alpha\}\) in \(F'\) over \(F\), and \(\cong\) denotes 'fields - homomorphisms' isomorphic, by the isomorphism, \(\phi: \overline{F [x] / I_b (p (x))} \to F_{F'} (\alpha), [p' (x)] \mapsto p' (\alpha)\)
//
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 \(\overline{F [x] / I_b (p (x))}\), \(F'\), \(F_{F'} ([x]) = \overline{F [x] / I_b (p (x))}\): while there is the 'fields - homomorphisms' isomorphism, \(\phi: \overline{F [x] / I_b (p (x))} \to F_{F'} ([x]), [p' (x)] \mapsto p' ([x])\), in fact, \([p' (x)] = p' ([x])\), so, it is indeed the identity map.
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.
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 \(F_{F'} (\alpha) \cong \overline{F [x] / I_b (p (x))}\).
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 \(r_1 \neq r_2\), \([r_1 x^0] \neq [r_2 x^0]\), 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.
\(f (1) = [1 x^0]\), 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 \(F_{F'} (\alpha) \cong \overline{F [x] / I_b (p (x))}\).
In fact, let us see that \(F_{F'} (\alpha) \cong F [x] / I_b (p (x))\) instead, which implies the above, because \(F [x] / I_b (p (x)) \cong \overline{F [x] / I_b (p (x))}\).
Let us define \(\phi: F [x] / I_b (p (x)) \to F_{F'} (\alpha), [p' (x)] \to p' (\alpha)\).
Let us see that that is well-defined. For each \([p' (x)] = [p'' (x)]\), \(p' (x) - p'' (x) = p (x) q (x)\). \(p' (\alpha) - p'' (\alpha) = p (\alpha) q (\alpha) = 0 q (\alpha) = 0\). So, \(p' (\alpha) = p'' (\alpha)\), which means that \(\phi ([p' (x)])\) does not depend on the representation.
Let us see that \(\phi\) is field homomorphic.
\(\phi ([p' (x)] + [p'' (x)]) = \phi ([p' (x) + p'' (x)]) = p' (\alpha) + p'' (\alpha) = \phi ([p' (x)]) + \phi ([p'' (x)])\). So, \(\phi\) 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.
\(\phi ([1 x^0]) = 1\).
\(\phi ([p' (x)] [p'' (x)]) = \phi ([p' (x) p'' (x)]) = p' (\alpha) p'' (\alpha) = \phi ([p' (x)]) \phi ([p'' (x)])\).
So, \(\phi\) 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 \(\phi\) is bijective. \(Ker (\phi)\) 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 \([1 x^0]\) is mapped to \(1\), so, the ideal is \(\{0\}\). \(\phi\) is injective, by the proposition that any group homomorphism is injective if and only if the kernel is the 1 subgroup. \(\phi\) is surjective, because while \(\phi (F [x] / I_b (p (x)))\) 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 \(\phi (F [x] / I_b (p (x)))\) is a field that contains \(F\) and \(\{\alpha\}\), but as \(F_{F'} (\alpha)\) is the smallest such, \(F_{F'} (\alpha) \subseteq \phi (F [x] / I_b (p (x)))\).
So, \(\phi\) is a 'fields - homomorphisms' isomorphism, by the proposition that any bijective field homomorphism is a 'fields - homomorphisms' isomorphism.
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 \(\overline{\overline{F [x] / I_b (p (x))} [x] / I_b (q (x))}\), 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 \(r (x - \alpha_1) ... (x - \alpha_n) \in F' [x]\) where \(F'\) is an extended field of \(F\). \(\{\alpha_1, ..., \alpha_n\}\) may not be distinct.
