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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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About:
field
The table of contents of this article
Starting Context
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.
References
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definition of kernel of ring homomorphism
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About:
ring
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of kernel of ring homomorphism.
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:
\( R_1\): \(\in \{\text{ the rings }\}\)
\( R_2\): \(\in \{\text{ the rings }\}\)
\( f\): \(: R_1 \to R_2\), \(\text{ the ring homomorphisms }\)
\(*Ker (f)\): \(= f^{-1} (0) \subseteq R_1\)
//
Conditions:
//
2: Note
Note that it is not "\(f^{-1} (1)\)".
References
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description/proof of that group homomorphism is injective iff kernel is 1 subgroup
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group
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any group homomorphism is injective if and only if the kernel is the 1 subgroup.
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:
\(G_1\): \(\in \{\text{ the groups }\}\)
\(G_2\): \(\in \{\text{ the groups }\}\)
\(f\): \(: G_1 \to G_2\), \(\in \{\text{ the group homomorphisms }\}\)
//
Statements:
\(Ker (f) = \{1\}\)
\(\iff\)
\(f \in \{\text{ the injections }\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(Ker (f) = \{1\}\), take any \(g, g' \in G_1\) such that \(g \neq g'\), suppose that \(f (g) = f (g')\), and find a contradiction; Step 2: suppose that \(f\) is injective, and see that \(Ker (f) = \{1\}\).
Step 1:
Let us suppose that \(Ker (f) = \{1\}\).
Let \(g, g' \in G_1\) be any such that \(g \neq g'\).
Let us suppose that \(f (g) = f (g')\) and find a contradiction.
\(1 = f (g) f (g')^{-1} = f (g) f (g'^{-1}) = f (g g'^{-1})\), which implies that \(g g'^{-1} = 1\), which implies that \(g = g'\), a contradiction.
Step 2:
Let us suppose that \(f\) is injective.
\(f (1) = 1\).
Let us suppose that there is a \(g \in G_1\) such that \(f (g) = 1\). As \(f (1) = 1\) and \(f\) is injective, there is no other \(g\) than \(1\), which means that \(Ker (f) = \{1\}\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
definition of polynomial extended over extended field
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ring
The table of contents of this article
Starting Context
Target Context
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The reader will have a definition of polynomial extended over extended field.
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:
\( F\): \(\in \{\text{ the fields }\}\)
\( F'\): \(\in \{\text{ the extended fields of } F\}\)
\( F [x]\): \(= \text{ the polynomials ring over } F\)
\( F' [x]\): \(= \text{ the polynomials ring over } F'\)
\( p (x)\): \(\in F [x]\)
\(*\overline{p (x)}\): \(\in F' [x]\), \(= p (x) \text{ regarded to be in } F' [x]\)
//
Conditions:
//
2: Note
The coefficients of \(p (x)\) are in \(F\) and are in \(F'\), and so, \(p (x)\) can be regarded to be in \(F' [x]\), which this definition is saying.
\(\overline{p (x)}\) is sometimes (or rather usually) denoted as \(p (x)\), but sometimes we need to distinguish \(\overline{p (x)}\) from \(p (x)\): for example, it may be that \(p (x)\) is irreducible while \(\overline{p (x)}\) is not irreducible.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
definition of extended field of field
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field
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of extended field of field.
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:
\( F\): \(\in \{\text{ the fields }\}\)
\(*F'\): \(\in \{\text{ the fields }\}\)
//
Conditions:
\(F \subseteq F'\)
//
2: Note
There can be some multiple extended fields of \(F\).
For example, for \(F = \mathbb{Q}\), \(\mathbb{R}\) and \(\mathbb{C}\) are some extended fields of \(\mathbb{Q}\) among possible others.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for ring and set of subfields, intersection of set is subfield
Topics
About:
ring
About:
field
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any ring and any set of subfields, the intersection of the set is a subfield.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(B\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{F_\beta \vert \beta \in B\}\): \(F_\beta \in \{\text{ the subfields of } R\}\)
//
Statements:
\(\cap \{F_\beta \vert \beta \in B\} \in \{\text{ the subfields of } R\}\)
//
2: Note
A point is that \(R\) does not need to be any field.
But \(R\) is allowed to be a field for this proposition.
3: Proof
Whole Strategy: Step 1: see that \(\cap \{F_\beta \vert \beta \in B\}\) is a subring of \(R\); Step 2: see that \(\cap \{F_\beta \vert \beta \in B\}\) is a commutative ring; Step 3: see that each element of \(\cap \{F_\beta \vert \beta \in B\}\) has an inverse.
Step 1:
\(\cap \{F_\beta \vert \beta \in B\}\) is a subring of \(R\), by the proposition that for any ring and any set of subrings, the intersection of the set is a subring: each \(F_\beta\) is a subring of \(R\).
Step 2:
Let us see that \(\cap \{F_\beta \vert \beta \in B\}\) is a commutative ring.
For each \(r_1, r_2 \in \cap \{F_\beta \vert \beta \in B\}\), \(r_1, r_2 \in F_\beta\) for each \(\beta\), so, \(r_1 r_2 = r_2 r_1\), because \(F_\beta\) is a field.
Step 3:
Let us see that each element of \(\cap \{F_\beta \vert \beta \in B\}\) has an inverse.
For each \(r \in \cap \{F_\beta \vert \beta \in B\}\), \(r \in F_\beta\) for each \(\beta\), so, the inverse, \(r_\beta\) (we cannot rule out the possibility that it depends on \(\beta\), at this point), is on \(F_\beta\) for each \(\beta\), but the inverse is an inverse also on \(R\), because \(r r_\beta = r_\beta r = 1\) holds on \(R\), and it is unique on \(R\), by the proposition that for any ring, if an element has an inverse, the inverse is unique; so, the \(r_\beta\) s are in fact the same, which can be denoted as \(r^{-1}\), so, \(r^{-1} \in F_\beta\) for each \(\beta\) and \(r^{-1} \in \cap \{F_\beta \vert \beta \in B\}\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for ring, if element has inverse, inverse is unique
Topics
About:
ring
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any ring, if an element has an inverse, the inverse is unique.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(r\): \(\in R\)
//
Statements:
\(\exists r' \in R (r' r = r r' = 1) \land \exists r'' \in R (r'' r = r r'' = 1)\)
\(\implies\)
\(r' = r''\)
//
2: Proof
Whole Strategy: Step 1: multiply \(r r'' = 1\) by \(r'\) from left, and see that \(r'' = r'\).
Step 1:
From \(r r'' = 1\), \(r' r r'' = r' 1 = r'\).
The left hand side is \(r' r r'' = (r' r) r'' = 1 r'' = r''\).
So, \(r'' = r'\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for ring and set of subrings, intersection of set is subring
Topics
About:
ring
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any ring and any set of subrings, the intersection of the set is a subring.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(B\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{R_\beta \vert \beta \in B\}\): \(R_\beta \in \{\text{ the subrings of } R\}\)
//
Statements:
\(\cap \{R_\beta \vert \beta \in B\} \in \{\text{ the subrings of } R\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(\cap \{R_\beta \vert \beta \in B\}\) is an Abelian group under the addition; Step 2: see that \(\cap \{R_\beta \vert \beta \in B\}\) is a monoid under the multiplication; Step 3: see that the multiplication is distributive with respect to the addition.
Step 1:
Let us see that \(\cap \{R_\beta \vert \beta \in B\}\) is an Abelian group under the addition.
\(\cap \{R_\beta \vert \beta \in B\}\) is a subgroup of \(R\) under the addition, by the proposition that for any group, the intersection of any possibly uncountable number of subgroups of the group is a subgroup of the group.
\(\cap \{R_\beta \vert \beta \in B\}\) is an Abelian group, because for each \(r_1, r_2 \in \cap \{R_\beta \vert \beta \in B\}\), \(r_1 + r_2 = r_2 + r_1\), because \(r_1, r_2 \in R\) and \(r_1 + r_2 = r_2 + r_1\) on \(R\).
Step 2:
Let us see that \(\cap \{R_\beta \vert \beta \in B\}\) is a monoid under the multiplication.
\(\cap \{R_\beta \vert \beta \in B\}\) is closed under the multiplication, because for each \(r_1, r_2 \in \cap \{R_\beta \vert \beta \in B\}\), \(r_1, r_2 \in R_\beta\) for each \(\beta\), so, \(r_1 r_2 \in R_\beta\) for each \(\beta\), and so, \(r_1 r_2 \in \cap \{R_\beta \vert \beta \in B\}\).
For each \(r_1, r_2, r_3 \in \cap \{R_\beta \vert \beta \in B\}\), \((r_1 r_2) r_3 = r_1 (r_2 r_3)\), because it is so on the ambient \(R\).
\(1 \in \cap \{R_\beta \vert \beta \in B\}\), because \(1 \in R_\beta\) for each \(\beta\), and for each \(r \in \cap \{R_\beta \vert \beta \in B\}\), \(1 r = r 1 = r\), because it is so on the ambient \(R\).
Step 3:
Let us see that the multiplication is distributive with respect to the addition.
For each \(r_1, r_2, r_3 \in \cap \{R_\beta \vert \beta \in B\}\), \(r_1 (r_2 + r_3) = (r_1 r_2) + (r_1 r_3)\) and \((r_1 + r_2) r_3 = (r_1 r_3) + (r_2 r_3)\), because it is so on the ambient \(R\).
References
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description/proof of that intersection of subgroups of group is subgroup of group
Topics
About:
group
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any group, the intersection of any possibly uncountable number of subgroups of the group is a subgroup of the group.
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:
\(G'\): \(\in \{\text{ the groups }\}\)
\(S\): \(\subseteq \{\text{ the subgroups of } G'\}\)
//
Statements:
\(\cap S \in \{\text{ the subgroups of } G'\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(\cap S\) satisfies the requirements to be a group.
Step 1:
For the identity element, \(1 \in G'\), \(1 \in \cap S\), because for each \(G \in S\), \(1 \in G\).
For any \(g_1, g_2 \in \cap S\), \(g_1 g_2 \in \cap S\), because for each \(G \in S\), \(g_1, g_2 \in G\), and so, for each \(G \in S\), \(g_1 g_2 \in G\).
For any \(g \in \cap S\), \(g^{-1} \in \cap S\), because for each \(G \in S\), \(g \in G\), and so, for each \(G \in S\), \(g^{-1} \in G\).
The associativity of multiplications holds, because it holds in the ambient \(G'\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
definition of field generated by subset of superfield over subfield
Topics
About:
field
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of field generated by subset of superfield over subfield.
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:
\( F'\): \(\in \{\text{ the fields }\}\)
\( F\): \(\in \{\text{ the fields }\}\) such that \(F \subseteq F'\)
\( S\): \(\subseteq F'\), just a subset
\(*F_{F'} (S)\): \(= \text{ the smallest field }\) such that \(F \cup S \subseteq F_{F'} (S) \subseteq F'\)
//
Conditions:
//
2: Note
\(F_{F'} (S)\) is uniquely determined, because it is the intersection of the fields such that \(F \cup S \subseteq F_{F'} (S) \subseteq F'\), while there is at least 1 such, \(F'\): refer to the proposition that for any ring and any set of subfields, the intersection of the set is a subfield.
\(F_{F'} (S)\) seems to be widely denoted as \(F (S)\), but \(F_{F'} (S)\) is clearer because it depends on \(F'\).
\(S \subseteq F\) is possible, and then, \(F_{F'} (S) = F\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that ring homomorphism between fields is field homomorphism
Topics
About:
field
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any ring homomorphism between any fields is a field homomorphism.
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:
\(F_1\): \(\in \{\text{ the fields }\}\)
\(F_2\): \(\in \{\text{ the fields }\}\)
\(f\): \(F_1 \to F_2\), \(\in \{\text{ the ring homomorphisms }\}\)
//
Statements:
\(f \in \{\text{ the field homomorphisms }\}\)
//
2: Proof
Whole Strategy: Step 1: see that all what needs to be checked is that for each \(r_1 \in F_1\), \(f (r_1^{-1}) = f (r_1)^{-1}\); Step 2: see that \(f (r_1^{-1}) = f (r_1)^{-1}\).
Step 1:
As any field is a ring, calling \(f\) "ring homomorphism" makes sense.
While only the differences of 'field' from 'ring' are being multiplicative commutative and each nonzero element's having the inverse, the commutativity is just about inside \(F_1\) or \(F_2\), not about \(f\), and the only concern about \(f\) is whether for each \(r_1 \in F_1\), \(f (r_1^{-1}) = f (r_1)^{-1}\).
So, let us check that \(f (r_1^{-1}) = f (r_1)^{-1}\).
Step 2:
\(r_1 r_1^{-1} = r_1^{-1} r_1 = 1\). \(f (r_1 r_1^{-1}) = f (r_1^{-1} r_1) = f (1) = 1\). But \(f (r_1 r_1^{-1}) = f (r_1) f (r_1^{-1})\) and \(f (r_1^{-1} r_1) = f (r_1^{-1}) f (r_1)\). So, \(f (r_1) f (r_1^{-1}) = f (r_1^{-1}) f (r_1) = 1\), which means that \(f (r_1)^{-1} = f (r_1^{-1})\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of root of polynomial in polynomials ring over commutative ring
Topics
About:
ring
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of root of polynomial in polynomials ring over commutative ring.
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:
\( R\): \(\in \{\text{ the commutative rings }\}\)
\( R [x]\): \(= \text{ the polynomials ring over } R\)
\( p (x)\): \(\in R [x]\)
\(*r\): \(\in R\)
//
Conditions:
\(\exists q (x) \in R [x] (p (x) = (x - r) q (x))\)
//
2: Note
We have not defined 'root' as \(p (r) = 0\).
We need to distinguish what are implied when \(R\) is a field and when \(R\) is not necessarily a field.
When \(R\) is a field, this definition equals \(p (r) = 0\): if \(p (x) = (x - r) q (x)\), \(p (r) = (r - r) q (r) = 0 q (r) = 0\); if \(p (r) = 0\), \(p (x) = (p - r) q (x)\), by the proposition that for the polynomials ring over any field and any nonconstant polynomial, if and only if the evaluation of the polynomial at a field element is 0, the polynomial can be factorized with x - the element: when \(p (x)\) is any constant, \(p (x) = 0\) and \(p (x) = (x - r) 0\) anyway.
But when \(R\) is not necessarily a field, if \(p (x) = (x - r) q (x)\), \(p (r) = (r - r) q (r) = 0 q (r) = 0\); but the proposition that for the polynomials ring over any field and any nonconstant polynomial, if and only if the evaluation of the polynomial at a field element is 0, the polynomial can be factorized with x - the element cannot be applied because it uses the fact that \(R [x]\) is a Euclidean domain, which (the fact) has been proved based on the requirement that \(R\) is a field.
So, in general, this definition guarantees that \(p (r) = 0\), but we have not proved that \(p (r) = 0\) guarantees this definition.
For a \(p (x)\), there may be no root and there may be some multiple roots.
It is crucial to be aware that what \(R [x]\) we are thinking in: \(p (x) = x^2 - 2\) has no root with \(p (x)\) regarded in \(\mathbb{Q} [x]\), but it has the roots, \(\sqrt{2}, - \sqrt{2}\), with \(p (x)\) regarded in \(\mathbb{R} [x]\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for field, positive natural number, and nonzero element of field, if field has primitive natural-number-th root of 1 and natural-number-th root of element, roots of element are products of root and roots of 1
Topics
About:
field
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any field, any positive natural number, and any nonzero element of the field, if the field has a primitive the-number-th root of 1 and a the-number-th root of the element, the the-number-th roots of the element are the products of the root and the the-number-th roots of 1.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(r\): \(\in F \setminus \{0\}\)
\(R_{r, n}\): \(= \{\alpha \in F \vert \alpha^n = r\}\)
//
Statements:
(
\(\exists \omega \in F \setminus \{1\} (\omega^n = 1 \land \forall j \in \{1, ..., n - 1\} (\alpha^j \neq 1))\)
\(\land\)
\(\exists \alpha \in F (\alpha^n = r)\)
)
\(\implies\)
\(R_{r, n} = \{\alpha \omega, ..., \alpha \omega^n\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(\{\omega, ..., \omega^n\}\) are the \(n\)-th roots of \(1\); Step 2: see that \(\{\alpha \omega, ..., \alpha \omega^n\}\) is distinct; Step 3: see that each element of \(\{\alpha \omega, ..., \alpha \omega^n\}\) is in \(R_{r, n}\); Step 4: see that there is no other element in \(R_{r, n}\).
Step 1:
\(\{\omega, ..., \omega^n\}\) are the \(n\)-th roots of \(1\), by the proposition that for any field, if the field has a primitive positive-natural-number-th root of 1, the 1 to the-natural-number powers of the primitive root are the the-natural-number-th roots of 1.
Step 2:
Let us see that \(\{\alpha \omega, ..., \alpha \omega^n\}\) is distinct.
Let us suppose that \(\alpha \omega^j = \alpha \omega^k\) where \(1 \le j, k \le n\) with \(j \le k\) without loss of generality.
\(\alpha \neq 0\), because if \(\alpha = 0\), \(\alpha^n = 0 \neq r\): the proposition that for any field, any positive-natural-number-th root of 0 is 0, a contradiction.
So, \(\alpha\) has a inverse, \(\alpha^{-1}\).
\(\omega^j = \alpha^{-1} \alpha \omega^j = \alpha^{-1} \alpha \omega^k = \omega^k\), which means that \(j = k\), because we already know that \(\{\omega, ..., \omega^n\}\) is distinct.
Step 3:
Let us see that each element of \(\{\alpha \omega, ..., \alpha \omega^n\}\) is in \(R_{r, n}\).
\((\alpha \omega^j)^n = \alpha^n (\omega^j)^n = r \omega^{j n} = r (\omega^n)^j = r 1^j = r 1 = r\).
Step 4:
\(\{\alpha \omega, ..., \alpha \omega^n\}\) has \(n\) elements, and \(R_{r, n}\) cannot have any other element, by the proposition that over any field, any n-degree polynomial has at most n roots.
References
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