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math/abstract-algebra-charles-pinter-notes.tex
abstract-algebra-charles-pinter-notes.texRaw
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 23\title{Notes on "A book of Abstract Algebra", Charles C. Pinter}
 24\author{arnaucube}
 25\date{February 2022}
 26
 27\begin{document}
 28
 29\maketitle
 30
 31\begin{abstract}
 32    Notes on \emph{"A book of Abstract Algebra - by Charles C. Pinter"}, is a $LaTeX$ version of handmade notes taken while reading the book. It contains only some definitions and theorems (without proofs), so it is highly recommended to read the actual book instead of the current notes.
 33    \\
 34    \emph{This is an unfinished and 'work in progress' document.}
 35\end{abstract}
 36
 37\tableofcontents
 38% \newpage
 39
 40% TODO maybe this 'Introduction' section should be removed
 41% \section{Introduction}
 42% 
 43% $\mathbb{R} \supseteq \mathbb{Q} \supseteq \mathbb{Z} \supseteq \mathbb{N}$
 44% \begin{itemize}
 45%     \item $\mathbb{R}$ (reals): $\{ \ldots, -3, 0, \sqrt{2}, 2, e, \pi, 4, \ldots \}$
 46%     \item $\mathbb{Q}$ (rationals): $\{ \frac{a}{b} \mid a, b \in \mathbb{Z} \wedge b \neq 0 \}$
 47%     \item $\mathbb{Z}$ (integers): $\{ \ldots, -3, -1, 0, 1, 3, \ldots \}$
 48%     \item $\mathbb{N}$ (naturals): non negative integers, $\{1, 2, 3, ...\}$
 49% \end{itemize}
 50
 51\section{Groups}
 52
 53\begin{definition}[Group]
 54    A set $G$ with an operation $*$ which satisfies the axioms:
 55    \begin{enumerate}[i.]
 56	\item $*$ is \emph{associative}
 57	\item \emph{(identity element)} there is an element $e \in G$ s.t. $a * e = a$ and $e * a = a$ $\forall a \in G$
 58	\item \emph{(inverse)} $\forall a \in G$, there is an element $a^{-1} \in G$ s.t. $a*a^{-1} = e$ and $a^{-1} * a =e$
 59    \end{enumerate}
 60\end{definition}
 61
 62\begin{definition}[Abelian Group]
 63    A group $G$ is said to be \emph{commutative} if $\forall a, b \in G$, $ab = ba$. A commutative group is also called \emph{Abelian}.
 64\end{definition}
 65
 66\begin{definition}[Order of an element]
 67    In a group $G$, the order of an element $a \in G$ is the least positive integer $n$ such that $a \cdot a \cdots a = a^n = e$. It is represented by $ord(a)$.
 68\end{definition}
 69
 70\begin{definition}[Order of a group]
 71    Order of a group $G$, is the number of elements in $G$. It is represented by $|G|$.
 72\end{definition}
 73
 74\begin{definition}[Cyclic group]
 75    Let $G$ be a group, and $a \in G$. If $G$ consists of all the powers of $a$ and nothing else:
 76    $$G = \{a^n : n \in \mathbb{Z}\}$$
 77    then, $G$ is called a \emph{cyclic group}, and $a$ is called its \emph{generator}.
 78    \\
 79    The group $G$ generated by $a$ is defined by $G=\langle a \rangle$.
 80\end{definition}
 81
 82\begin{theorem}
 83    The \emph{order of a cyclic group} is the same as the \emph{order of it's generator}. In other words, for a cyclic group, $|\langle a \rangle | = ord(a)$.
 84    \begin{itemize}
 85	\item[] $\langle a \rangle$ defines a cyclic group generated by $a$. $\langle a \rangle = \{e, a, a^2, ..., a^{n-1}\}$
 86	\item[] $| \langle a \rangle |$ defines the order of the cyclic group generated by $a$.
 87    \end{itemize}
 88\end{theorem}
 89
 90\begin{theorem}
 91Every subgroup of a cyclic group is cyclic.
 92\end{theorem}
 93
 94
 95
 96\section{Subgroups}
 97\begin{definition}[Subgroup]
 98    Let $G$ be a group, and $H$ a non-empty subset of $G$. If
 99    \begin{enumerate}[i.]
100	\item the identity $e$ of $G$ is in $H$.
101	\item $H$ is closed with respect to the operation. Which is for $a, b \in H$, $ab \in H$.
102	\item $H$ is closed with respect to inverses. Which is for $a \in H$, $a^{-1} \in H$.
103    \end{enumerate}
104    we call $H$ a \emph{subgroup} of $G$. The operation of $H$ is the same as the operation of $G$.
105\end{definition}
106
107\section{Functions}
108\begin{definition}[Function]
109    If $A$ and $B$ are sets, then a function from $A$ to $B$ is a rule which to every element $x$ in $A$ assigns a unique element $y$ in $B$.
110    \\
111    Functions are represented by $f: A \rightarrow B$, where $\forall a \in A \Rightarrow f(a) \in B$.
112\end{definition}
113
114\begin{definition}[Injective (monomorphism)]
115    A function $f:A \rightarrow B$ is called \emph{injective} if each element of $B$ is the image of no more than one element of $A$.
116\end{definition}
117
118\begin{definition}[Surjective (epimorphism)]
119    A function $f:A \rightarrow B$ is called \emph{surjective} if each element of $B$ is the image of at least one element of $A$.
120    \\ In other words, does not repeat outputs.
121\end{definition}
122
123\begin{definition}[Bijective (isomorphism)]
124    A function $f:A \rightarrow B$ is called \emph{bijective} if it is both \emph{injective} and \emph{surjective}.
125    \\
126    A function $f: A \rightarrow B$ has an inverse iff it is \emph{bijective}. In that case, the inverse $f^{-1}$ is a bijective function from $B$ to $A$.
127\end{definition}
128
129In finite sets, if $f: A \rightarrow B$ is injective then $|A| \leq |B|$, and if $f$ is surjective then $|B| \leq |A|$. And if $f$ is bijective, then $|A| = |B|$.
130
131\begin{definition}[Composite function]
132    A function $f:A \rightarrow B$ and $g: B \rightarrow C$ be functions. The \emph{composite function} denoted by $g \circ f$ is a function from $A$ to $C$ defined as follows:
133    $$[g \circ f](x) = g(f(x)),  \forall x \in A$$
134\end{definition}
135
136
137\begin{definition}[Permutation]
138    By a \emph{permutation} of a set $A$ we mean a \emph{bijective function from $A$ to $A$}, that is, a one-to-one correspondence between $A$ and itself.
139    \\
140    The set of all the permutations of $A$, with the operation $\circ$ of composition, is a group.
141    \\
142    For any positive integer $n$, the symmetric group on the set ${1,2, 3,..., n}$ is called the \emph{symmetric group on $n$ elements}, and is denoted by $S_n$.
143\end{definition}
144
145% TODO define 'group of permutations' & 'cycle of permutations'
146
147\section{Isomorphism}
148
149\begin{definition}[Isomorphism]
150    Let $G_1$ and $G_2$ be groups. A bijective function $f: G_1 \rightarrow G_2$ with the property that for any two elements $a, b \in G_1$,
151    $$f(ab) = f(a)f(b)$$
152    is called an \emph{isomorphism} from $G_1$ to $G_2$.
153    \\
154    If there exists an isomorphism from $G_1$ to $G_2$, we say that $G_1$ is \emph{isomorphic} to $G_2$, symbolized by $G_1 \cong G_2$.
155\end{definition}
156
157
158\begin{theorem}[Cayley's Theorem]
159    Every group is isomorphic to a group of permutations.
160\end{theorem}
161
162\begin{theorem}
163    (Isomorphism of cyclic groups)
164    \begin{enumerate}[i.]
165	\item For every positive integer $n$, every cyclic group of order $n$ is isomorphic to $\mathbb{Z}_n$. Thus, any two cyclic groups of order $n$ are isomorphic to each other.
166	\item Every cyclic group of order infinity is isomorphic to $\mathbb{Z}$, and therefore any two cyclic groups of order infinity are isomorphic to each other.
167    \end{enumerate}
168\end{theorem}
169
170\section{Cosets}
171
172\begin{definition}[Coset]
173    Let $G$ be a group, and $H$ a subgroup of $G$. For any element $a$ in $G$, the symbol $aH$ denotes the set of all products $ah$, as $a$ remains fixed and $h$ ranges over $H$. $aH$ is called a \emph{left coset} of $H$ in $G$.
174    \\
175    In similar fashion, $Ha$ denotes the set of all products $ha$, as $a$ remains fixed an $h$ ranges over $H$. $Ha$ is called a \emph{right coset} of $H$ in $G$.
176\end{definition}
177
178\begin{theorem}
179    If $Ha$ is any coset of $H$, there is a one-to-one correspondence from $H$ to $Ha$ (there is a bijection between $H$ and $Ha$).\\
180    If $a \in G$, then $|H| = |Ha|$.
181\end{theorem}
182
183\begin{theorem}[Lagrange's theorem]
184    Let $G$ be a finite group, and $H$ any subgroup of $G$. The order of $G$ is a multiple of the order of $H$. $|H|$ divides $|G|$.
185\end{theorem}
186    Lagrange's theorem can be easily seen by the facts that:
187    \begin{enumerate}[i.]
188	\item cosets partition the group G
189	\item $|Ha| = |H|$ (each coset has the same order as H). 
190    \end{enumerate}
191
192By consequence,
193\begin{theorem}
194    If $G$ is a group with a prime number $p$ of elements, then $G$ is a cyclic group. Furthermore, any element $a \neq e$ in $G$ is a generator of $G$.
195\end{theorem}
196
197Thus,
198\begin{theorem}
199    The order of any element of a finite group divides the order of the group.
200\end{theorem}
201
202\begin{definition}[Index of H in G]
203    Number of cosets of H in G. Represented by $(G:H)$.\\
204    Combined with \emph{Lagrange Theorem}, we know that $|G| = |H| \cdot |G:H|$, so,
205    $$(G:H) = \frac{|G|}{|H|}$$
206\end{definition}
207
208
209\section{Homomorphisms}
210
211\begin{definition}[Homomorhism]
212    If $G$ and $G$ are groups, a \emph{homomorphism} from $G$ to $H$ is a function $f: G \rightarrow H$ s.t. for any two elements $a, b \in G$,
213    $$f(ab) = f(a)f(b)$$
214    If there exists a homomorphism from $G$ \emph{onto} $H$, we say that $H$ is a \emph{homomorphic image} of $G$.
215\end{definition}
216Note: an \emph{isomorphism} is a \emph{bijective} \emph{homomorphism}.
217\\
218Example of an \emph{homomorphism}: $f: \mathbb{Z}_6 \rightarrow \mathbb{Z}_3$.
219
220
221\begin{theorem}
222    Let $G$ and $G$ be groups, and $f: G \rightarrow H$ a homomorphism. Then
223    \begin{enumerate}[i.]
224	\item $f(e) = e$
225	\item $f(a^{-1}) = [f(a)]^{-1}, \quad \forall a \in G$
226    \end{enumerate}
227\end{theorem}
228
229\begin{definition}[Conjugate]
230    A \emph{conjugate} of $a$ is any element of the form $xax^{-1}$, where $x \in G$.
231\end{definition}
232
233\begin{definition}[Normal subgroup]
234    Let $H$ be a subgroup of a group $G$. $H$ is called a \emph{normal} subgroup of $G$ if it is closed with respect to conjugates, that is, if\\
235    for any $a \in H$ and $x \in G$, $xax^{-1} \in H$.
236    \\
237    Alternatively, we can see that $H$ is a \emph{normal} subgroup iff $\forall a \in G, aH = Ha$.
238    \\
239    In an abelian group, every subgroup is normal.
240\end{definition}
241
242\begin{definition}[Kernel]\label{def:homomorphismkernel}
243    Let $f: G \rightarrow H$ be a homomorphism. The \emph{kernel} of $f$ is the set $K$ of all the elements of $G$ which are carried by $f$ onto the neutral element of $H$. That is,
244    $$K = {x \in G : f(x) = e}$$
245    \emph{Kernel in the context of Extension fields: \ref{def:extensionkernel}}
246\end{definition}
247For every homomorphism, the $e \in G$ maps to $e \in H$, so the \emph{kernel} is never empty, it always contains the identity $e_G$, and if the kernel only contains the identity, then $f$ is one-to-one (injective).
248
249
250\section{Quotient Groups}
251
252Quotient group construction is useful as a way of actually manufacturing all the homomorphic images of any group G. Additionally, we can often choose $H$ so as to "factor out" unwanted properties of $G$, and prserve in $G/H$ only "desirable" traits.
253
254\begin{definition}[Coset multiplication]
255    The coset of $a$, multiplied by the coset of $b$, is defined to be the coset of $ab$. In symbols, $Ha \cdot Hb = H(ab)$.
256\end{definition}
257
258\begin{theorem}
259    Let $H$ be a normal subgroup of $G$. If $Ha = Hc$ and $Hb = Hd$, then $H(ab) = H(cd)$.
260\end{theorem}
261
262\begin{definition}
263    $G/H$ denotes the set which consists of \emph{all the cosets of $H$}.
264    \\
265    Thus, if $Ha, Hb, Hc, \ldots$ are cosets of $H$, then $G/H = \{ Ha, Hb, Hc, ... \}$.
266\end{definition}
267
268\begin{theorem}[Quotient group]
269    $G/H$ with coset multiplication is a group.
270\end{theorem}
271
272\begin{theorem}
273    $G/H$ is a homomorphic image of G.
274    \\
275    Conversely, every homomorphic image of $G$ is a quotient group of $G$.
276\end{theorem}
277
278\begin{theorem}
279    Let $G$ be a group and $H$ a subgroup of $G$. Then
280    \begin{enumerate}[i.]
281	\item $Ha = Hb$ iff ${ab}^{-1} \in H$
282	\item $Ha = H$ iff $a \in H$
283    \end{enumerate}
284\end{theorem}
285
286
287\section{Rings and Fields}
288
289\begin{definition}[Ring]
290    A set $A$ with operations called \emph{addition} and \emph{multiplication} which satisfy the following axioms:
291    \begin{enumerate}[i.]
292	\item $A$ with addition alone is an abelian group.
293	\item Multiplication is associative.
294	\item Multiplication is distributive over addition. That is, $\forall a,b,c \in A$,
295	    $$a(b+c) = ab + ac$$
296	    $$(b+c)a = ba + ca$$
297    \end{enumerate}
298\end{definition}
299
300\begin{definition}[Commutative ring]
301    By definition, addition is commutative in every ring but multiplication is not. When multiplication also is commutative in a ring, we call that ring a \emph{commutative} ring.
302\end{definition}
303
304\begin{definition}[Unity]
305    A ring does not necessarily have a neutral element for multiplication. If there is in $A$ a neutral element for multiplication, it is called the \emph{unity} of $A$, and is denoted by the symbol $1$.
306    \\
307    If $A$ has a unity, we call $A$ a \emph{ring with unity}.
308\end{definition}
309
310\begin{definition}[Field]
311    If $A$ is a commutative ring with unity in which every nonzero element is invertible, $A$ is called a \emph{field}.
312\end{definition}
313
314\begin{theorem}[Finite Field must be over p prime ($\mathbb{F}_p$)]
315    Proof from \href{https://github.com/aragonzkresearch/blog/blob/main/pdf/Aragon_Math_Seminar.pdf}{Matan Prasma seminars}:\\
316    One of the axioms of a field is $\exists$ multiplicative inverse.\\
317    If $\mathbb{Z}_n$ with $n$ no prime, then $n= k \cdot l$ for some $1 \leq k,~l \leq n-1$.\\
318    Then in $\mathbb{Z}_n$, $k \cdot l = 0$, but if $k \cdot l=0$ means that either $k=0$ or $l=0$ (otherwise, we could multiply by (eg) $k^{-1}$ and get $k^{-1} \cdot k \cdot l = k^{-1} \cdot 0$, which leads to $1 \cdot l = 0$).\\
319    which is a contradiction here (since $1 \leq k,~l \leq n-1$).\\
320    Thus $\mathbb{Z}_n$ with $n$ not prime can not be a field.\\
321    Conversely, if $n = p$ prime,\\
322    for $0 \neq x \in \mathbb{Z}_p$, $gcd(x, p)=1$, so Extended Euclidean Algorithm gives $u, v \in \mathbb{Z}$ such that $u x + v p = 1$.\\
323    Then, $ux=1 \pmod p$, so $u=x^{-1}$, so inverses exist.\\
324    Thus $\mathbb{Z}_p$ is a field.
325\end{theorem}
326
327\begin{definition}[Divisor of zero]
328    In any ring, a nonzero element a is called a \emph{divisor of zero} if there is a
329    nonzero element b in the ring such that the product ab or ba is equal to
330    zero.
331\end{definition}
332
333\begin{definition}[Cancellation property]
334A ring is said to have the cancellation property if $ab = ac$ or $ba = ca$ implies $b = c$
335for any elements a, b, and c in the ring if $a \neq 0$.
336\end{definition}
337
338\begin{theorem}
339    A ring has the \emph{cancellation property} iff it has no \emph{divisors of zero}.
340\end{theorem}
341
342\begin{definition}[Ideal]
343    A nonempty subset $B$ of a ring $A$ is called an \emph{ideal} of $A$ if $B$ is closed with respect to addition and negatives, and $B$ absorbs products in $A$.
344    \\
345    (\emph{Absorbs product}: $\forall b \in B$ and $x \in A$, then $xb, bx \in B$).
346\end{definition}
347
348\begin{definition}[Principal ideal]
349    A \emph{principal ideal} is an ideal $I$ in a ring $R$ that is generated by a single element $a \in R$ through multiplication by every element of $R$. In other words $I = aR = \{a r : r \in R \}$.
350    \\
351    (eg. Every ideal of $\mathbb{Z}$ is principal).
352\end{definition}
353
354\begin{definition}[Integral domain]
355    An \emph{integral domain} is defined to be a commutative ring with unity having the cancellation property.
356\end{definition}
357
358Every field is an integral domain, but the converse is not true (eg. $\mathbb{Z}$ is an integral domain but not a field).
359
360\begin{definition}[Characteristic n]
361    Let $A$ be a ring with unity, the \emph{characteristic} of $A$ is the least positive integer $n$ such that
362    $$\underbrace{1 + 1 + \cdots + 1}_{n-times} = 0$$
363    If there is no such positive integer $n$, $A$ has characteristic $0$.
364\end{definition}
365
366\section{Elements of number theory}
367
368\begin{definition}[Euclid's lemma]
369    Let $m$ and $n$ be integers, and let $p$ be a prime. If $p|(mn)$, then either $p|m$ or $p|n$.
370\end{definition}
371
372\begin{theorem}[Factorization into primes]
373    Ever integer $n>1$ can be expressed as a product of positive primes. That is, there are one or more primes $p_1, \ldots, p_r$ such that $n=p_1 p_2 \cdots p_r$.
374\end{theorem}
375
376\begin{theorem}[Unique factorization]
377    Suppose $n$ can be factored into positive primes in two ways, namely,
378    $$n= p_1 \cdots p_r = q_1 \cdots q_t$$
379    Then $r=t$, and the $p_i$ are the same numbers as the $q_j$ except, possibly, for the order in which they appear.
380\end{theorem}
381
382From the last two theorems: every integer $m$ can be factored into primes, and the prime factors of $m$ are unique (except for the order).
383
384\begin{theorem}[Little theorem of Fermat]
385    Let $p$ be a prime. Then,
386    $$a^{p-1} \equiv 1 \pmod p, \forall a \not\equiv 0 \pmod p$$
387    \\
388    So, by taking $a^{p-2} \cdot a \equiv 1 \pmod p$, where $a^{p-2} \equiv a^{-1} \pmod p$ (the inverse modulo p), we see that $a^p \equiv a \pmod p, \forall a \in \mathbb{Z}$, so $a^p - a$ is a multiple of $p$.
389
390    ~\\\emph{Relation to Lagrange's theorem:}\\
391    Let $G = \mathbb{Z}_p$, and let $H$ be the multiplicative subgroup of $G$ generated by $a$ (ie. $H = \{ 1, a, a^2, \ldots \}$). The order of $H$ ($h = |H|$), is also the order of $a$ (ie. smallest $n>1$ s.t. $a^n=1~mod~p$).
392
393    By Lagrange's theorem, $h~|~|G| = p - 1$, so $p-1 = h \cdot m$, thus
394    $$
395    a^{p-1} = (a^h)^m \equiv 1^m \equiv 1~mod~p
396    $$
397
398    ~\\\emph{Another perspective:}\\
399    We have $a^p \equiv a \pmod{p}$, by dividing by $a$ on both sides, we obtain $a^{p-1} \equiv 1 \pmod{p}$.
400
401\end{theorem}
402
403\begin{theorem}[Euler's $\phi$ function]
404    \emph{Euler's $\phi$ function} describes the number of integers in $\mathbb{Z} / n \mathbb{Z}$ which are relatively prime (coprime) to $n$.
405\end{theorem}
406
407\begin{theorem}[Euler's theorem]
408    If $a$ and $n$ are relatively prime, $$a^{\phi(n)} \equiv 1 \pmod n$$
409\end{theorem}
410
411
412\section{Polynomials}
413
414\begin{definition}
415    Let $A$ be a commutative ring with unity, and $x$ an arbitrary symbol. Every expression of the form
416    $$a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$$
417    is called a \emph{polynomial in $x$ with coefficients in $A$}, or more simply, a \emph{polynomial in $x$ over $A$}.
418\end{definition}
419 The expressions $a_k x^k$, for $k \in \{ 1, \ldots, n \}$, are called the \emph{terms} of the polynomial, being $a_n x^n$ the \emph{leading term}, and $a_0$ the \emph{constant term}.
420 The $a_k$ are called the \emph{coefficients} of $x^k$, being $a_n$ the \emph{leading coefficient}. And the \emph{degree} of a polynomial $a(x)$ is the greatest $n$ such that the coefficient of $x^n$ is not zero.
421 The polynomial whose leading coefficient is equal to $1$ is called \emph{monic}.
422
423\begin{theorem}[Division algorithm for polynomials]
424    If $a(x)$ and $b(x)$ are polynomials over a field $F$, and $b(x) \neq 0$, there exist polynomials $q(x)$ and $r(x)$ over $F$ such that
425    $a(x) = b(x) q(x) + r(x)$ and
426    [$r(x) = 0$ or $\deg r(x) < \deg b(x)$].
427\end{theorem}
428
429\begin{theorem}
430    Any two nonzero polynomials $a(x), b(x) \in F[x]$ have a $\gcd d(x)$. Furthermore, $d(x)$ can be expressed as a \emph{linear combination}
431    $$d(x)= r(x) a(x) + s(x) b(x)$$
432    where $r(x), s(x) \in F[x]$.
433\end{theorem}
434
435\begin{theorem}[Factorization into irreducible polynomials]
436    Every polynomial $a(x)$ of positive degree in $F[x]$ can be written as a product
437    $$a(x) = k p_1(x) p_2(x) \cdots p_r(x)$$
438    where $k$ is a constant in $F$ and $p_1(x), \ldots, p_r(x)$ are monic irreducible polynomials of $F[x]$.
439\end{theorem}
440
441\begin{theorem}[Unique factorization]
442    If $a(x)$ can be written in two ways as a product of monic irreducibles, say
443    $$a(x) = k p_1(x) \cdots p_r(x) = l q_1(x) \cdots q_s(x)$$
444    then $k=l$, $r=s$, and $p_i(x) = q_j(x)$.
445\end{theorem}
446
447\begin{theorem}
448    $c$ is a root of $a(x)$ iff $x - c$ is a factor of $a(x)$.
449\end{theorem}
450\begin{theorem}
451    If $a(x)$ has distinct roots $c_1, \ldots, c_m$ in $F$, then $(x-c_1)(x-c_2)\cdots(x-c_m)$ is a factor of $a(x)$.
452\end{theorem}
453
454\begin{theorem}
455    If $a(x)$ has degree $n$, it has at most $n$ roots.
456\end{theorem}
457
458In finite $F$, polynomial $\neq$ polynomial function. If $F$ is infinite, polynomial $=$ polynomial function.
459
460For every polynomial with rational coefficients, there is a polynomial with integer coefficients having the same roots. See:
461$$a(x) = \frac{k_0}{l_0} + \frac{k_1}{l_1} x + \cdots + \frac{k_n}{l_n} x^n$$
462$$=\frac{1}{l_0 \cdots l_n} \cdot \underbrace{(k_0 l_1 \cdots l_n + k_1 l_0 l_2 \cdots l_n x + \cdots + k_n l_0 \cdots l_{n-1} x^n)}_{b(x)}$$
463$a(x)$ has rational coefficients, $b(x)$ has integer coefficients. $b(x)$ differs from $a(x)$ only by a constant factor ($\frac{1}{l_0\cdots l_n}$), so $a(x)$ and $b(x)$ have the same roots.
464
465$\Longrightarrow~~\forall~p(x) \in \mathbb{Q}[x]$, there is a $f(x) \in \mathbb{R}$ with the same roots (for every polynomial with rational coefficients, there is a polynomial with integer coefficients having the same roots).
466
467\begin{theorem}
468    If $s/t$ is a root of $a(x)$, then $s|a_0$ and $t|a_n$.
469\end{theorem}
470
471\begin{theorem}
472    Suppose $a(x)$ can be factored as $a(x) = b(x)c(x)$, where $b(x), c(x)$ have rational coefficients. Then there are polynomials $B(x), C(x)$ with integer coefficients, which are constant multiples of $b(x)$ and $c(x)$ respectively, such that $a(x) = B(x)C(x)$.
473\end{theorem}
474
475\begin{theorem}[Eisenstein's irreducibility criterion]
476    Let $a(x) = a_0 + a_1 x + \cdots + a_n x^n$ be a polynomial with integer coefficients.
477
478    If there is a prime $p$ such that $p | a_i, ~\forall i\in\{0, n-1\}$, and $p \ndiv a_n$ and $p^2 \ndiv a_0$, then $a(x)$ is irreducible over $\mathbb{Q}$.
479    % Suppose there is a prime number $p$ which divides every coefficient of $a(x)$ except the leading coefficient $a_n$; suppose $p$ does not divide $a_n$ and $p^2$ does not divide $a_0$. Then $a(x)$ is irreducible over $\mathbb{Q}$.
480\end{theorem}
481
482\section{Extensions of fields}
483
484\begin{definition}[Kernel]\label{def:extensionkernel}
485    The \emph{kernel} of $\sigma_c$ consists of all polynomials $a(x) \in F[x]$ such that $c$ is a root of $a(x)$.
486
487    \emph{Kernel in the context of Homomorphisms: \ref{def:homomorphismkernel}}
488\end{definition}
489
490\begin{definition}[Algebraic]
491    $c \in E$ is called \emph{algebraic over} $F$ if it is the root of some nonzero polynomial $a(x) \in F[x]$.
492
493    Otherwise, $c$ is called \emph{transcendental over} $F$.
494\end{definition}
495
496$E/K$ denotes the (field) extension of $E$ over $K$.
497
498\begin{theorem}[Basic theorem of field extensions]
499    Let $F$ be a field and $a(x) \in F[x]$ a nonconstant polynomial. There exists an extension field $E/F$ and an element $c \in E$ such that $c$ is a root of $a(x)$.
500\end{theorem}
501
502Let $a(x) \in F[x]$ be a polynomial of degree $n$. There is an extension field $E/F$ which contains all $n$ roots of $a(x)$.
503
504\section{Vector spaces}
505
506\begin{definition}[Vector space]
507    A \emph{vector space} over a field $F$ is a set $V$, with two operations $+, \cdot$, called \emph{vector addition} and \emph{scalar multiplication}, such that
508
509    \begin{itemize}
510	\item $V$ with vector addition is an abelian group
511	\item $\forall k \in F$ and $\overrightarrow{a} \in V$, the scalar product $k \overrightarrow{a}$ is an element of $V$, subject to the following conditions:
512	    $\forall k, l \in F,~\overrightarrow{a},\overrightarrow{b} \in V$
513	    \begin{enumerate}[i.]
514		\item $k(\overrightarrow{a} + \overrightarrow{b}) = k\overrightarrow{a} + k\overrightarrow{b}$
515		\item $(k + l)\overrightarrow{a} = k\overrightarrow{a} + k\overrightarrow{b}$
516		\item $k(l\overrightarrow{a}) = (kl)\overrightarrow{a}$
517		\item $1 \overrightarrow{a} = \overrightarrow{a}$
518	    \end{enumerate}
519    \end{itemize}
520\end{definition}
521
522\begin{definition}[Linear combination]
523    If $\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n} \in V$, and $k_1, k_2, \ldots, k_n$ are scalars, then the vector
524    $$k_1 \overrightarrow{a_1} + k_2 \overrightarrow{a_2} + \cdots + k_n \overrightarrow{a_n}$$
525    is called a \emph{linear combination} of $\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n}$.
526
527    The set of all the linear combinations of $\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n}$ is a \emph{subspace of} $V$.
528\end{definition}
529
530\begin{definition}[Linear dependency]
531    Let $S = \{$\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n}$\}$ be a set of distinct vectors in a vector space $V$. $S$ is said to be \emph{linearly dependent} if there are scalars $k_1, \ldots, k_n$, not all zero, such that $k_1 \overrightarrow{a_1} + k_2 \overrightarrow{a_2} + \cdots + k_n \overrightarrow{a_n} = 0$.
532    Which is equivalent to saying that at least one of the vectors in $S$ is a linear combination of the others.
533
534    If $S$ is not linearly dependent, then it is \emph{linearly independent}. $S$ is linearly independent iff $k_1 \overrightarrow{a_1} + k_2 \overrightarrow{a_2} + \cdots + k_n \overrightarrow{a_n}=0$ implies $k_1 = k_2 = \cdots =  k_n =0$.
535    Which is equivalent to saying thatno vector in $S$ is equal to a linear combination of the other vectors in $S$.
536\end{definition}
537
538If $\{ \overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n} \}$ is linearly dependent, then some $a_i$ is a linear combination of the preceding ones.
539
540If $\{ \overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n} \}$ spans $V$, and $a_i$ is a linear combination of the preceding vectors, then $\{ \overrightarrow{a_1}, \ldots, \crossover{\overrightarrow{a_i}}, \ldots, \overrightarrow{a_n} \}$ still spans $V$.
541
542
543\begin{theorem}
544    Any two bases of a vector space $V$ have the same number of elements.
545
546    (This comes from the fact that all bases of $\mathbb{R}^n$ contain exactly $n$ vectors)
547\end{theorem}
548
549If the  set $(\overrightarrow{a_1}, \overrightarrow{a_2}, \ldots, \overrightarrow{a_n})$ spans $V$, it contains a basis of $V$.
550
551\framebox{WIP: covered until chapter 28, work in progress.}
552
553\end{document}