Abstract Algebra

Feb 19, 2024
algebra math

Section 4 - Groups

A group \( G \):

Has an identity \( e \) where \( ex = xe = x \space \forall x\in G \)
Satisfies \( \forall x \in G \), there exists an inverse \(x^{-1} \) where \( x^{-1}x = xx^{-1} = e \)
Satisfies associative property \( a(bc) = (ab)c \space \forall a, b, c \in G \)

A group \( \langle G, * \rangle \) is isomorphic to another group \( \langle H, *^\prime \rangle \) if:

There is a well defined function, \( f : G \rightarrow H \)
\( f(x * y) = f(x) *^\prime f(y) \), the homomorphism property, is satisfied \( \forall x, y \in G \)
\( f \) is onto, which means if \( y\in H \), then there exists an \( x\in G \) such that \( f(x) = y \). It feels like finding the \( f(y^\prime) \) that yields \( x \)
\( f \) is one-to-one, which means if \( f(x) = f(y) \), then \( x = y \)

Section 5 - Subgroups

A subset \( H \) is a subgroup of \( \langle G, * \rangle \) if:

\( H \) is also closed under \( * \)
\( H \) has the same identity
\( \forall a\in H \), \( a^{-1}\in H \) too

A cyclic subgroup \( \langle a \rangle = \{ a^{n} \mid n \in \mathbb{Z} \} \) is the smallest subgroup containing \( a \).

Section 6 - Cyclic Groups

A subgroup, \( H \), of a cyclic group, \( G \), is cyclic. Sketch of proof: The empty case works, and in the non-empty case, the subgroup is going to be generated by \( a^m \) where \( m \) is the smallest integer that satisfies \( a^m \in H \).

If \( G \) is a cyclic group, if its order is infinite, it’s isomorphic to \( \langle \mathbb{Z}, + \rangle \). The isomorphism is \( \phi(a^i) = i \). If its order is \( n \), it’s isomorphic to \( \langle \mathbb{Z}_n, +_n \rangle \). The isomorphism is \( \phi(a^i) = i \bmod{n} \).

A cyclic subgroup, \( H = \langle a^m \rangle \), of a cyclic group, \( G = \langle a \rangle \), is of order \( x = n / \gcd(n, m) \) where \( \vert G \vert = n \). This is because it satisfies the smallest \( x \) in \( (a^m)^x = e = a^n \).

Section 8 - Permutation Groups

An example of a permutation \( \sigma \) is \( \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix} \), where the first element is mapped to the second, the second element is mapped to the first, and third is mapped to itself. Permutations can be composed like \( (\sigma \sigma) (1) = \sigma (\sigma (1)) = \sigma (2) = 1 \). The set of all permutations of any set \( A \) forms the symmetric group \( S_A \) with the composition operation. A subgroup of \( S_A \) is the permutation group.

The nth dihedral group \( D_n \) is the group of symmetries of the regular n-gon. Rotations, mirror images, and diagonal flips characterize its elements, which form a subgroup of \( S_n \).

Cayley’s Theorem states every group is isomorphic to a permutation group. This is not the same as saying every group is isomorphic to a symmetric group, as symmetric groups contain \( n! \) elements and there are groups that don’t contain \( n! \) elements, like \( D_4 \). The isomorphism is \( \phi(x) : G \rightarrow S_G = \lambda_x \), where \( \lambda_x : G \rightarrow G = xg \space \forall g\in G \). It means that an element in a set can be used to shuffle the set into a new permutation. Because \( \phi(x) \) is one-to-one and \( \phi(xy) = \phi(x)\phi(y) \space \forall x, y \in G \), the image of \( G \) under \( \phi \), or \( \phi[G] = \{ \phi(g) \mid g\in G \} \), is a subgroup of \( S_G \) and \( \phi \) is an isomorphism relating \( G \) with \( \phi[G] \).

Section 9 - Orbits and Alternating Groups

The orbits are the partitions/equivalence classes of a set by a permutation. \( a, b \in A \) are in the same orbit if \( a = \sigma^n (b) \) where \( \sigma \) is the permutation. That is an equivalence relation (reflexive, symmetric, and transitive). An orbit of length two is a transposition and all orbits of length greater than two can be expressed as a product of transpositions.

A cycle \( (1, 2) \) in \( S_3 \) can be expressed as a permutation like \( \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix} \). Basically, interchange 1 and 2 and fix 3. Multiplying two cycles would be the composition of two permutations.

A permutation in \( S_n \) can either be expressed as an odd or even number of transpositions. That is, a transposition \( (i, j) \) multiplied with an orbit will either split the orbit if it was already part of it or join two orbits together if the transposition and the orbit were disjoint. A subgroup of \( S_n \) of even permutations is the alternating group \( A_n \).

Section 10 - Cosets and Lagrange’s Theorem

If \( H \) is a subgroup of \( G \), \( G \) will be divided into \( \lvert G \rvert / \lvert H \rvert \) disjoint partitions/subgroups/equivalence classes called cosets. The number of partitions is also known as the index \( ( G : H ) \). The left coset of \( H \) containing \( a \) is \( \{ ah \mid h \in H \} \).

If \( H = 3\mathbb{Z} \) and \( G = \mathbb{Z} \), the cosets would be \( 0 + 3\mathbb{Z} = \{0, 3, 6, … \} \), \( 1 + 3\mathbb{Z} = \{ 1, 4, 7… \} \), …

Reordering a Cayley table, grouping together elements in cosets and compressing those elements to new ones, gives rise to coset groups.

Lagrange’s Theorem states the order of a subgroup \( H \) must be a divisor of \( G \). This is because every coset has the same order as \( H \), because \( \phi(a) = ah \mid h \in H \), which converts one coset into another, is one-to-one and onto.

Section 11 - Direct Products and Finite Generated Abelian Groups

The product of groups is denoted \( \prod_{i=1}^n G_i \) and is the Cartesian product of groups. Elements look like \( (g_1, g_2, … g_n), … \). It also looks like \( \oplus_{i=1}^n G_i \) for sums. An element of a product of groups has an order which is the LCM of the individual orders.

The group \( \mathbb{Z}_{m} \times \mathbb{Z}_n \) is isomorphic to \( \mathbb{Z}_{mn} \) when \( m \) and \( n \) are relatively prime. Every element of \( \mathbb{Z}_3 \) has order \( 3 \). Therefore, every element of \( \mathbb{Z}_3 \times \mathbb{Z}_3 \) has order \( 3 \) too, easily seen with addition. So, \( \mathbb{Z}_3 \times \mathbb{Z}_3 \) cannot be cyclic because it cannot generate all \( 9 \) elements of the group.

The fundamental theorem of finitely generated abelian groups is every finitely generated abelian group \( G \) is isomorphic to \( \mathbb{Z}_{{p_1}^{r_1}} \times \mathbb{Z}_{{p_2}^{r_2}} \times … \times \mathbb{Z}_{{p_1}^{g_1}} \times … \), where \( p_1^{r_1 + g_1} \mid \lvert G \rvert \).

Section 13 - Homomorphisms

Let \( \phi : X \rightarrow Y \), \( A \subseteq X \), and \( B \subseteq Y \). The image of \( A \) under \( Y \) is \( \phi[A] = \{ \phi(a) \mid a \in A \} \). \( \phi[X] \) is the range. The inverse image of \( B \) in \( X \) is \( \phi^{-1}[B] = \{ x\in X \mid \phi(x)\in B \} \).

If \( \phi \) were a homomorphism from group \( G \) to \( G^\prime \),

If \( e \) is \( G \)’s identity, \( \phi(e) = e^\prime \) is \( G^\prime \)’s identity.
If \( a\in G \), then \( \phi(a)^{-1} = \phi(a^{-1}) \).
If \( H \) is a subgroup of \( G \), \( \phi[H] \) is a subgroup of \( G^\prime \).
If \( K \) is a subgroup of \( G^\prime \), \( \phi^{-1}[K] \) is a subgroup of \( G \).

If \( \phi : G \rightarrow G^\prime \) were a homomorphism, \( \phi^{-1}[\{e^\prime\}] = \{x\in G \mid \phi(x) = e^\prime \} \) is the kernel of \( \phi \). If \( H \) were the kernel of \( \phi \), \( \phi^{-1}[\{\phi(g)\}] = \{ x\in G \mid \phi(x) = \phi(g) \} = gH = Hg \). With homomorphisms, opposed to isomorphisms whose kernel is singular, we need to worry about the left and right cosets of the kernel whenever thinking about inverse images. If the kernel of \( \phi \) were \( \{e\} \), then \( \phi \) is one-to-one.

Another way to show \( \phi : G \rightarrow G^\prime \) is an isomorphism is:

Show \( \phi \) is a homomorphism.
Show the kernel of \( \phi \) is \( \{e\} \).
Show \( \phi \) maps \(G\) onto \(G^\prime\).

A subgroup \( H \) of a group \( G \) is normal if \( gH = Hg \space \forall g\in G \).

Section 14 - Factor Groups

If \( \phi : G \rightarrow G^\prime \) is a group homomorphism with kernel \( H \), then the cosets of \( H \) form a factor/quotient group \( G/H \). \( (aH)(bH) = (ab)H \space \forall a, b \in G\) can be derived by knowing that inverse images are related to cosets and sets with the same cardinality as a group are isomorphic. \( \mu: G/H \rightarrow \phi[G] = \mu(aH) = \phi(a) \) is an isomorphism and relates images to cosets.

However, \( (aH)(bH) = (ab)H \) does not hold if \( H \) is a subgroup of \( G \) instead of the kernel of a group homomorphism. It only holds if \( H \) is a normal subgroup.

If \( H \) is a normal subgroup of \( G \), then \( \gamma(x): G \rightarrow G/H = xH \) is a homomorphism with kernel \( H \). It follows that \( \phi = \mu\gamma \). The LHS is the image of \( G \) and the RHS converts \( G \) into a factor group and turns the factor group into the image of \( G \).

To classify \( (\mathbb{Z}_4 \times \mathbb{Z}_2) / (\{0\} \times \mathbb{Z}_2) \), we must find a homomorphism with kernel \( H = (\{0\} \times \mathbb{Z}_2) \). This must be the map \( \phi: (G = \mathbb{Z}_4 \times \mathbb{Z}_2) \rightarrow ? \) where \( \phi(H) = \phi(e) = e^\prime \) does not affect the identity of the new group. Well, the identity is \( e^\prime = (0, 0) \), so the only way \( he^\prime \forall h \in H \) doesn’t affect its status as the identity is if we ignore the second element in the pair. Hence the map is a projective map \( \pi_1: G \rightarrow \mathbb{Z}_4 \).

If \( H \) is a normal subgroup of \( G \), either of these conditions apply:

\( ghg^{-1} \in H \forall g\in G, h\in H \)
\( gHg^{-1} = H \forall g\in G \)
\( gH = Hg \forall g\in G \)

An automorphism is an isomorphism of a group onto itself. The inner automorphism is \( i_g : G \rightarrow G = i_g(x) = gxg^{-1} \forall x\in G \).

Section 15 - Factor Group Computations and Simple Groups

\( \mathbb{Z} / \{ 0 \} \cong \mathbb{Z} \) since there is no collapsing. \( \mathbb{R} / n\mathbb{R} \cong \{1\} \) since it is the reals, not the integers, and every real is of form \( n\mathbb{R} \).

Every subgroup that is half the order of the parent group is normal. The normal relation \( gH = Hg \) would be apparent when \( g \) is every element not in the subgroup.

Computing a factor group should involve:

Expanding the kernel. This is the first coset.
Finding an element not in the kernel and applying the operation to it. This is the second coset. Repeat for the third, etc.
Finding an isomorphism would involve trying to find a single element/representation for the coset

A simple group is nontrivial and has no proper nontrivial normal subgroups. A maximal normal subgroup is a normal subgroup where no other normal subgroups contain it. If \( G / M \) is simple, then \( M \) is a maximal normal subgroup. The proof involves thinking about \( \gamma \).

The center \( Z(G) \) of a nonabelian group \( G \) is defined as \( \{ z\in G \mid zg = gz \forall g \in G \} \).

If \( G \) is a group, the set of all commutators \( aba^{-1}b^{-1} \forall a, b \in G \) generates the commutator subgroup \( C \) of \( G \), attempting to transform \( G \) into an abelian group by forcing abelian elements to be the identity.

\( C \) is a normal subgroup of \( G \). This is because \( g^{-1}xg\in C \forall g\in G, c\in C \), since \( g^{-1}(cdc^{-1}d^{-1})g = g^{-1}cdc^{-1} \cdot gd^{-1}dg^{-1} \cdot d^{-1}g = efe^{-1}f^{-1}jkj^{-1}k^{-1} \), where \( e = g^{-1}c, f = d, j = d, k = g^{-1} \).

If \( N \) is a normal subgroup of \( G \), then \( G / N \) is abelian iff \( C \) is a subgroup of \( N \). Saying \( G / N \) is abelian implies \( (aN)(bN) = (bN)(aN) \). This works because \( (aN)(bN) = (ab)N = ab(b^{-1}a^{-1}ba)N = baN = (bN)(aN) \) where \( b^{-1}a^{-1}ba\in C \).

Section 16 - Group Action on a Set

If \( X \) is a set and \(G\) is a group, an action of G on X is a map \( * : G \times X \rightarrow X = * (g, x) = g * x = gx \) such that:

\( ex = x \forall x \in X \)
\( (g_1g_2)(x) = g_1(g_2x) \forall x\in X, g_1, g_2\in G \)

I only care about the definition for now.

Section 18 - Rings and Fields

A ring \( \langle R, +, \cdot \rangle \) satisfies:

\( \langle R, + \rangle \) is an abelian group
Multiplication is associative
\( a(b + c) = ab + ac, (a + b)c = ac + bc \forall a, b, c\in R \)

\( n\mathbb{Z} \), or the cyclic group of \(\mathbb{Z}\) under addition, is a ring. So is \(\mathbb{Z}_n \), which holds the set of integers generated by the product of them modulo \( n \).

If a ring \( R \) has an additive identity \( 0 \), it seems ridiculously obvious that:

\( 0a = a0 = 0 \)
\( a(-b) = (-a)b = -(ab) \)
\( (-a)(-b) = ab \)

but this requires applications of the ring axioms. 2 requires the distributive property by \( a(-b) + (ab) = 0 \implies a(-b + b) = 0 \), 1 requires it also by \( 0a + 0a = (0 + 0)a = 0a \implies 0a = 0 \), and 3 requires the associative property by \( (-a)(-b) = -(a(-b)) = -(-a(b)) = ab \).

\( \phi: R \rightarrow R^\prime \) is a homomorphism from rings \(R, R^\prime \) if:

\( \phi(a + b) = \phi(a) + \phi(b) \)
\( \phi(ab) = \phi(a)\phi(b) \)

An evaluation homomorphism is \( \phi_a : F \rightarrow \mathbb{R} \), where \( F \) is the functions over the reals and \( a\in\mathbb{R} \). \( \phi_a(f) = 0 \) is used to find roots. The isomorphism is a homomorphism if it is one-to-one and onto.

A unity is the multiplicative identity. A unit of a ring has a multiplicative inverse. If every element of the ring has a unit, it’s a division ring. If the division ring is commutative, it’s a field, otherwise it’s a strictly skew field. Subrings and subfields are subsets of rings and fields with the same algebraic properties.

Section 19 - Integral Domains

Solving \( x^2 - 5x + 6 = 0 \) in \( \mathbb{Z}_{12} \) would require solving \( x^2 - 5x + 6 = 12, x^2 - 5x + 6 = -12, … \).

The zero divisors of a ring are nonzero elements \( a \) and \( b \) such that \( ab = 0 \). In \( \mathbb{Z}_n \) they are the numbers that aren’t relatively prime to \( n \), since \( a(n/d) = 0 = n(a/d) \) where \( d \) is the non-zero gcd of \( a \) and \( n \), so \( n/d = b \).

An integral domain \( D \) is a commutative ring with unity \( 1 \neq 0 \) which contains no zero divisors. A direct product cannot be an integral domain because it always contains zero divisors: \( (r, 0)(0, s) \).

Every field is an integral domain. Since every element is a unit/has a multiplicative inverse, if \( ab = 0 \) and \( a \neq 0 \), then \( (a^{-1}a)b = 0 \implies b = 0 \).

Every finite integral domain is a field. If you list out all the elements of the integral domain and multiply it by \( a \), the two sets will be the same size and none of the elements will be zero. At least one of the elements in the new set will be \( 1 \) so \( a \) has an inverse. Matrices are always in rings and not integral domains or fields since they have zero divisors.

If for a ring \( a + a … \) is done \( n \) times, or \( na = 0 \forall a \in R \), then \( n \) is the characteristic of the ring. If there is no \( n \), the characteristic is \( 0 \).

Section 21 - The Field of Quotients of an Integral Domain

Every integral domain can be contained in a field of quotients. If \( D \) is an integral domain, we can enlarge it to a field of quotients \( F \) via:

Defining what \( F \) looks like
Defining binary operations of addition and multiplication in \( F \)
Checking field axioms
Showing that \( F \) contains \( D \) as an integral subdomain

In 1, \( F = \{ (a, b) \mid a, b \in D, b \neq 0, (a, b) \sim (c, d) \iff ad = bc \} \).

In 2, \( [(a, b)] + [(c, d)] = [(ad+bc, bd)] \) and \( [(a, b)][(c, d)] = [(ac, bd)] \). Ensure that different elements in the same equivalence class do not change the result.

In 3, check ring axioms, commutativity for operations, and no zero divisors (every element has inverse).

In 4, check \( i : D \rightarrow F = i(a) = [(a, 1)] \) is an isomorphism on \( i[D] \).

\( F \) is the minimal field containing \( D \), and any other fields (\( L \)) containing \( D \) contain \( F \) by \( \psi : F \rightarrow L = \psi(a) = a, a \in D \). Show that with \( \psi \) the conditions in 1’s set apply to \( L \), the homomorphism laws apply, and it’s one-to-one.

Section 22 - Rings of Polynomials

Let \( R \) be a ring. A polynomial \( f(x) \) with coefficients in \( R \) is an infinite formal sum \( \sum_{i=0}^{\infty} a_ix^i \) where \( a_i \in R \) and \( a_i \neq 0 \) for a finite number of values of \( i \). The ring \( R \) is a subring of the polynomial ring \( R[x] \). If there are two indeterminates, we denote it as \( R[x, y] \). The field of quotients of \( F[x_1,…,x_n] \) is \( F(x_1,…,x_n) \).

To prove \( R[x] \) is a ring, we have to show the homomorphism property, which requires us to derive polynomial multiplication, which can be found algorithmically through guessing patterns from examples: \( f(x)g(x) = d_0 + d_1x + … + d_nx^n, d_n = \sum_{i=0}^n a_ib_{n-i} \). This assumes \( f(x) \) and \( g(x) \) have coefficients already ordered.

To prove \( R[x] \) is associative, we’ll make sense of the same polynomial multiplication rule twice by applying it to \( (ab)c = (abc) \) on each side for \( a, b, c \in R[x] \) and collapsing the summations to a sum condition, then re-ordering the summations around again to show \( (ab)c = a(bc) \).

The evaluation homomorphism \( \phi_2 : \mathbb{Q}[x] \rightarrow \mathbb{R} = \phi_2(a_0 + a_1x + … + a_nx^n) = a_0 + a_12 + … \). Evaluating \( \phi_2(x^2 + x - 6) = 2^2+2-6 = 0 \), so the kernel contains \(x^2+x-6\).

Section 23 - Factorization of Polynomials over a Field

If \( E \) and \( F \) are fields, we can find zeroes of \( f(\alpha) = \phi_\alpha(f(x)) = g(\alpha)h(\alpha) \) if \( f \) can be factored into \( g \) or \( h \). In fact, every polynomial \( f(x) \in F[x] \) satisfies \( f(x) = g(x)q(x) + r(x) \). If the degree of \( r(x) \) is just greater than that of \( g(x) \), just apply the division algorithm to it and keep moving it to the quotient. If \( \alpha \in F \) is a zero, then \( x - \alpha \) is a factor of \( f(x) \in F[x] \). An irreducible polynomial \( f(x) \) cannot be factored into \( g(x)h(x) \). Finally, the Eisenstein Criterion says that if \( p\in Z \) is a prime, and \( f(x) = {a_n}^{x}+…+a_0 \in \mathbb{Z}[x] \) with \( a_n \neq 0 \bmod{q} \) but \( a_i = 0 \bmod{q} \forall i < n \) and \( a_0 \neq 0 \bmod{q^2} \), then \( f(x) \) is irreducible over \( \mathbb{Q} \). This is clear after you try to factor \( f(x) \) into \( g(x)h(x) \) and you realize one of the factors cannot have factor \( p \), because none of the leading coefficients have \(p\) as a factor, only one of the constants has \(p\) as a factor, and the rest of the coefficients only have \(p\) as factors. This can be used to prove that \(f(x)=(x^{p}-1)/(x-1)\) is irreducible over \(\mathbb{Q}\) by applying \(g(x)_{x+1}\) to it, and the numerator after applying the binomial theorem satisfies the Eisenstein Criterion.

Section 26 - Homomorphisms and Factor Rings

Ring homomorphisms deal with the additive side of the ring more so than the multiplicative side. The kernel also relates to the additive identity.

The specific properties are for a ring homomorphism \( \phi : R \rightarrow R^\prime \):

\( \phi(0) = 0^\prime \)
\( \phi(-a) = -\phi(a) \forall a \in R \)
If \( S \) is a subring of \( R \), \( \phi[S] \) is a subring of \( R^\prime \)
If \( S^\prime \) is a subring of \( R^\prime \), \( \phi^{-1}[S^\prime] \) is a subring of \( R \)
If \( R \) has unity \( 1 \), \( \phi(1) \) is the unity for \( \phi[R] \) (not \( R^\prime \))

The kernel of \( \phi \) is \( \phi^{-1}[0^\prime] = \{ r\in R \mid \phi(r) = 0^\prime \} \). The cosets are \( \phi^{-1}[\phi(a)] = a + H = H +a \).

Some coset operations on a ring homomorphism \( \phi: R \rightarrow R^\prime \) with kernel \( H \) are:

\( (a+H)+(b+H) = (a+b)+H \)
\( (a+H)(b+H) = (ab)+H \)
\( \mu: R/H \rightarrow \phi[R] = \mu(a+H) = \phi(a) \)

2 is proven by finding \( h_1, h_2 \in H \) such that \( \phi((a+h_1)(b+h_2)) = \phi(ab) \) because \( \phi(ah_1) = \phi(h_2b) = 0^\prime \). So similarly, if we change \( H \) to be a subring instead of the kernel, 2 still holds iff \( ab \in H, hb\in H \forall a,b \in R, h\in H \).

An ideal is an additive subgroup \( N \) of a ring \( R \) such that \( aN \subseteq N, Nb \subseteq N \forall a, b \in R \). The operation to generate members with respect to \( R \) is still multiplication, but its members are related to each other with addition. The ideals form a quotient ring \( R/N \) with the coset operations 1 and 2 replacing \( H \) with \( N \). My conceptual understanding is that ideals and normal subgroups are like the quotients which summarize information about the parent group/ring, cosets are like the divisors which combine each isomorphic quotient to become the dividend, and the parent group/ring is like the dividend which is the set needing to be compressed.

The fundamental homomorphism of rings can be summarized as \( \phi(x) = \mu\gamma(x) \), where \( \phi(x) : R \rightarrow R^\prime \) is the ring homomorphism with kernel \( N \), \(\mu : R/N \rightarrow \phi[R] = \mu(x + N) = \phi(x) \), and \( \gamma : R \rightarrow R/N = x + N \).

Section 27 - Prime and Maximal Ideals

The factor ring of a ring with zero divisors may be an integral domain or a field because it compresses the ring down to not have any zero divisors. It may also be worse (have more zero divisors) with more zero divisors.

Proper ideals cannot contain units, otherwise the unity would be a part of the ideal since at least one element \( r \in R \) is the multiplicative inverse of the unit. Then, every element of \( R \) would be in that ideal.

A maximal ideal of a ring \( R \) is a proper ideal \( M \) such that there is no proper ideal \( N \) of \( R \) containing \( M \).

An ideal is prime if \( ab\in N \) implies either \( a \in N \) or \( b \in N \) for \( a, b \) in a commutative ring \( R \).

In a commutative ring \( R \) with unity:

An ideal \( M \) of \( R \) is maximal iff \( R/M \) is a field.
An ideal \( N \) of \( R \) is a prime iff \( R/N \) is an integral domain.
Every maximal ideal of \( R \) is a prime ideal.

A principal ideal is an ideal, \( \langle a \rangle = \{ra\mid r\in R\} \) generated by \( a \in R \). A maximal ideal \( \langle p(x) \rangle \) of \( F[x] \) is irreducible.

Section 28 - Grobner Bases for Ideals

An algebraic variety \( V(S) \in F^n \) is the set of all common zeroes of \( S \subset F[x] \). It is represented as an ideal with \( I = \langle f_1, f_2, … f_i \rangle = \{ c_1f_1 + c_2f_2 \mid c_i\in R \} \), where \( c_i,f_i\in F[x] \). The basis \( \langle f_1, f_2, … f_i \rangle \) can be reduced and transformed.

If we want to find all solutions of

\[ x + y - 3z = 8 \] \[ 2x + y + z = -5 \]

we can use a basis that is \( \{ x + y - 3z - 8, 2x + y + z + 5 \} \), or replace the second element with the remainder with the second divided by the first.

We also want to transform the polynomials so that their power products, or modified expanded terms in the form of \(\prod_i {x_i}^{m_i} \forall m_i \ge 0\), are ordered, so that any simplifications can terminate. Polynomials can then be mulitplied and added in the basis to reduce the size of power products. A Grobner basis ensures that picking any polynomial from the ideal, there will be a power product of a term before the polynomial in the ideal that divides it.

Sections 29 - Extension Fields

A field \( E \) is an extension field of \( F \) if \( F \le E \). If \( f(x) \in F[x] \), then there exists an extension field \( E = F[x] / \langle p(x) \rangle, \alpha\in E \) where \( f(\alpha) = 0 \) and \( p(x) \) is an irreducible factor of \( f(x) \).

An element \( \alpha \) of an extension field \( E \) of a field \( F \) is algebraic over \( F \) if \( f(\alpha) = 0 \) for some \( f(x)\in F[x] \). Otherwise, it is transcendental. Algebraic numbers are elements of \( \mathbb{C} \) that are algebraic over \( \mathbb{Q} \); transcendental numbers the same with transcendentals. Every element \( \beta \in E = F(\alpha) = b_0 + b_1\alpha + … + b_{n-1}\alpha^{n-1}, b_i \in F \).

Section 30 - Vector Spaces

\( F \)-vector spaces consist of an abelian group of vectors \( V \) (vectors) interacting with a field \( F \) (scalars). \( F[x] \) is like a vector space, where elements are like vectors and multiplication by elements in \( F \) is like multiplying by a scalar. If \( E \) is an extension field of \( F(\alpha) \), \( E \) is a vector space over \( F \). The vector would be \( \alpha \) and the scalars would be the elements in \( F \). \( F(\alpha) \) would be a vector space over \( \{1, \alpha, …, \alpha^{n-1}\} \) (the basis which spans the vector space) where \( n = [E : F] \) is the degree of \( \alpha \) in \( F \).

Section 31 - Algebraic Extensions

An extension field \( E \) of \( F \) is an algebraic extension of \( F \) if every element in \( E \) is algebraic over \( F \). If \( K \) is an extension field of \( E \), then \(K\) is an extension field of \(F\) as well, and \([K:F] = [K:E][E:F]\). By thinking about each of these in terms of bases (\([K:E]\implies\{\alpha_i\}, [E:F]\implies\{\beta_j\}\)), the product of these bases \(\{\alpha_i\beta_j\}\) produces the new basis. So the basis of \(\mathbb{Q}(\sqrt{2}, \sqrt{3})\) is \(\{1, \sqrt{2}, \sqrt{3}, \sqrt{6}\}\). The basis may be reduced. Check if an element has a different degree than it seems.

The algebraic closure of \( F \) in \( E \) is the set of all \( \alpha\in E \) that are algebraic over \( F \). It’s a subfield of \( E \). Every field has an algebraic closure.

The field \( \mathbb{C} \) is algebraically closed, or that is, every element of \(\mathbb{C}[z]\) has zeroes. Suppose a complex function does not have any zeroes. That is, \(\lim_{\lvert z\rvert \to \infty} = \infty\). Then \( 1/f \) has a zero as it approaches infinity.

You can also prove that every field has an algebraic closure by using Zorn’s Lemma, which states that if a chain, or a sequence of sets where each next set is a superset of the previous one, in a partially ordered set (most elements are comparable with each other) has upper bounds, then the ordered set has a maximal element. To prove, think of \(K, E, F\) as chains and show that the algebraic closure is the maximal algebraic extension.

Section 33 - Finite Fields

A Galois field is a finite field of order \(p^n\) where \(p\) is prime and \(n\) is a positive integer. If \(E\) is a finite extension of degree \(n\) over a field with \(p\) elements, \(E\) has \(p^n\) elements. The elements of \(E\) are the zeroes of \(x^{p^{n}}-x \in \mathbb{Z}_p[x]\) because you translate the order calculated from the previous sentence to the degree of the extension field and retrieve the polynomial from there.

An element \(\alpha\) in a field is an nth root of unity if \(\alpha^n=1\). It’s primitive if it’s the smallest exponent that yields \(1\).