Rings and Modules MATH326

Chapter 01 - Basics

Skipped the following out of triviality:

• Operations and their properties (definitions of identity, inverse, associativity, commutativity)
• Review of groups (see MATH 328 notes for more)

Rings and Fields

• In this course, we assume rings are commutative and have identity unless otherwise stated
• We often assume that \(1_R \ne 0_R\) if \(1_R\) exists; if \(1_R=0_R\), we get the trivial ring with one element
• E.g. \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\), \(\mathbb{C}\) are rings under "regular" addition and multiplication
• E.g. The set \(C(I)\) of continuous functions \(f : I \to \mathbb{R}\) for closed interval \(I\) is a ring under the addition and multiplication of functions
• E.g. The set \(\text{End}_{\mathbb{R}}(V)\) of \(\mathbb{R}\)-linear maps from vector space \(V\) to itself is a ring under addition and composition of linear maps (another example of the common multiplication \(\iff\) composition correspondence)

We can define the direct product \(R \times S\) of rings \(R,S\) in the same way they are defined for groups, i.e. the operations are inherited and applied to the corresponding "type" of element in the tuple.

We can trivially find the following properties of arithmetic in rings (proofs in text):

• The additive identity \(0_R\) is unique
• Additive cancellation law: \(a+c=b+c\implies a=b\)
• For \(a,b\in R\), the equation \(a+x=b\) has a unique solution for \(x\), namely \(-a+b\)
• Zero-absorption/Zero-annihilation: for all \(a\in R\), \(0_R\cdot a=a\cdot 0_R=0_R\)
• If a ring has multiplicative identity, it is unique
• \((-a)b=a(-b)=-(ab)\)
• \(-(-a)=a\) (involution of \(a \mapsto -a\))
• \((-a)(-b)=ab\)
• \(-(a+b)=(-a)+(-b)\)
• \(-(a-b)=b-a\)
• If a ring has multiplicative identity \(1_R\), then \(-(1_R)\cdot a=-a\)

• E.g. \(\mathbb{Z}\) under addition and multiplication is a subring of \(\mathbb{Q}\) under addition and multiplication, which is a subring of (etc.)
• E.g. the ring of integers divisible by \(n\in \mathbb{N}\), denoted \(n\mathbb{Z}\), is a subring of \(\mathbb{Z}\)
• E.g. the set of differentiable functions is a subring of the ring of continuous functions on a given closed interval (mentioned above)

Note: it is not generally true that a subring of a ring with identity also has identity, i.e. \(1_R\in R\) does not imply \(1_R\in S\). However, if \(S\) does (happen to) have identity, it will be the same element as the identity of \(1_R\), i.e. \(1_S=1_R\) if \(1_S\) exists.

• E.g. \(2\mathbb{Z}\) is a subring of \(\mathbb{Z}\) that does not contain the identity \(1_\mathbb{Z}:=1\).

• This is the ring equivalent of the subgroup criterion; the proof is similar and fairly trivial

Units, Zero-divisors, Fields

• E.g. the units of \(\mathbb{Z}\) are \(\set{\pm 1}\)
• E.g. the units of \(M_{n}(\mathbb{R})\) are the matrices with nonzero determinants
• E.g. if \(R\) has identity, then \(1_R\) is trivially a unit
• \(a\in R\) is a unit iff \(aR =R\)
• We must check both \(ab=1\) and \(ba=1\) (unless \(R\) is commutative, in which case they imply each)

The set \(R^{\times}:=\set{r\in R\mid r\text{ is a unit }}\) is set the of units of \(R\). If \(R\) is commutative and has identity, \(R^\times\) is a group under multiplication, called the multiplicative group of \(R\).

The units of the ring are the elements that we can divide by, i.e. if \(a\) is a unit in \(R\), for any \(b\in R\), the equation \(ax=b\) has a unique solution, namely \(b^{-1}\). So, units obey a cancellation law: if \(a\) is a unit of \(R\), then \(ab=ac\implies b=c\) and \(ba=ca\implies b=c\) for \(b,c\in R\).

• E.g. In \(\mathbb{Z}_n\), the zero divisors are the elements \([a]\) such that \(\text{GCD}(a,n)\ne 1\).

Regular elements also have the cancellation property. In fact, since all units are regular, this generalizes that set of elements that have the cancellation property to its characteristic form.

• We have \(ab=ac\iff ab-ac=0 \iff a(b-c)=0\); if \(a\) is not a zero-divisor, we must have \(b-c=0\iff b=c\).
• So, zero-divisors can't have the cancellation property, namely because the product \(ab\) could equal \(0\) if \(a\) is a zero-divisor.

• Proof: Let \(a\) be regular and \(f : R\to R, x \mapsto xa\). We have \(f(x_1)=f(x_2)\implies ax_1=ax_2\implies a(x_1-x_2)=0\). \(a\) is regular and thus not a zero-divisor, so \(x_1=x_2\), and thus \(f\) is injective. We can pick \(x\) such that \(x:=ba^{-1}\) so that \(f(x)=b\) for arbitrary \(b\in R\). So, since \(R\) is finite, \(f\) is surjective too, and thus \(x\) exists such that \(f(x)=1\); \(a\) is thus a unit.

• E.g. \(\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}\)
• E.g. \(\mathbb{Z}_p\) for prime \(p\).

• E.g. \(\mathbb{Q}, \mathbb{R}, \mathbb{C}\) are fields but \(\mathbb{Z}\) is not.
• E.g. \(\mathbb{Z}_p\) is a field if and only if \(p\) is prime (if \(p\) is not prime, then \(p=ab\) for \(a,b > 1\), so \([a][b]=[0]\), meaning \([a]\) is a zero-divisor, a contradiction).
• E.g. \(F := \set{a + b\sqrt{2} \mid a,b\in \mathbb{Q}}\) is a field (routine check, inverses given by \((a+b\sqrt{2})^{-1}=\dfrac{1}{a^2-b^2}(a-b\sqrt{2})\))
  • Aside: this mirrors the structure of complex inverses because these structures are isomorphic (to that of the rational complex numbers) up to the substitution \(\sqrt{2} \leftrightarrow i\)

Polynomial Rings

There are two natural ways to generate examples of rings

• Defining a set of matrices over an existing ring \(R\) by forcing their entries to be members of \(R\)
• Using polynomials and ideals

We denote a polynomial over ring \(R\) as \(R[x]\). \(R[x]\ni p(x)=a_0 + a_1x + a_2x^{2}+ \dots + a_nx^{n}=\displaystyle\sum\limits_{i=0}^{n} a_i x^i\) where coefficients \(a_k\) are in \(R\). Such a polynomial has degree \(n\) (assuming that \(a_n\) is nonzero).

• We define two polynomials as equal when they have the same degree and all of their corresponding coefficients are equal (thus, equality is inherited from the set to which the coefficients belong)
• A monic polynomial has a leading coefficient \(a_n\) of \(1_R\).

Polynomials are added pair-wise and multiplied according to the distributive law.

Fraction Fields

We can generalize the construction of \(\mathbb{Q}\) from \(\mathbb{Z}\) (as seen in MATH 217) to the construction of a fraction field from any integral domain.

Define \(R \times (R \setminus \set{0})\) as the set of "fractions", where \((a,b)\in R \times (R \setminus \set{0})\) is conventionally denoted \(\dfrac{a}{b}\) or \(b^{-1}a\). We define over \(R \times (R \setminus \set{0})\) the binary relation \((a,b)\sim(c,d)\iff ad=bc\) Without proof, this is an equivalence relation.

\(R \times (R \setminus \set{0})\) is a field when equipped with addition given by \(\dfrac{a}{b} + \dfrac{c}{d} := \dfrac{ad+bc}{bd}\) and multiplication given by \(\dfrac{a}{b}\cdot\dfrac{c}{d}:=\dfrac{ac}{bd}\), with zero \(\dfrac{0}{a}, a\ne 0\) and multiplicative identity \(\dfrac{a}{a}, a\ne 0\). This is straightforwardly proven by showing \(\sim\) is well-defined and checking against the ring axioms.

• Clearly, \(\mathbb{Q} := \text{Frac}({\mathbb{Z}})\)

If \(f : R \to F\) is a homomorphism from ID \(R\) to field \(F\), then there exists a unique ring homomorphism \(g : \text{Frac}(R)\to F\) such that \(g \circ j_R=f\). \(g\) is defined by \(g(b^{-1}a)=f(b)^{-1}\cdot f(a)\).

Polynomial Modular Arithmetic

We construct \(\mathbb{Z}_n\) by partitioning \(\mathbb{Z}\) into equivalence classes of the relation \(a\sim b \iff a \equiv b \mod n\), i.e. if \(n \mid (a-b)\). So, we can convincingly use the notation \(\mathbb{Z}/n \mathbb{Z}\), or simply \(\mathbb{Z}/n\). We extend the ideal of modular arithmetic to any structure that can define divisibility. We will do this in general later, but we define "polynomial modular arithmetic" now:

Chapter 02 - Homomorphisms, Ideals, Quotients

Homomorphisms

A ring homomorphism is a map \(f : R \to S\) between rings \(R\) and \(S\) such that \(f(a +_R b) = f(a) +_S + f(b)\) and \(f(a \cdot_R b) = f(a) \cdot_S f(b)\), i.e. \(f\) preserves addition and multiplication.

• E.g. For subring \(S\) of \(R\), the inclusion map \(f : S \to R\) given by \(x \mapsto x\) is a homomorphism
• E.g. The mapping \(f : \mathbb{R}\to M_2(\mathbb{R})\) given by \(r \mapsto \begin{bmatrix}r & 0 \\ 0 & r\end{bmatrix}\) is a homomorphism.
• E.g. The evaluation map \(C(\mathbb{R}) \to \mathbb{R}\) from the set of continuous functions on \(\mathbb{R}\) to \(\mathbb{R}\) taken by evaluating some function \(f \in C(\mathbb{R})\) at \(0\) is a homomorphism
• E.g. The map \(f : \mathbb{Z} \to R\) given by \(n \mapsto \displaystyle\sum\limits_{i=0}^{n} 1_R\) (sometimes denoted \(n\cdot 1_r\)) is a homomorphism. Any homomorphism \(f: \mathbb{Z}\to R\) where \(1 \mapsto 1_R\) and \(f(a+b)\mapsto f(a)+f(b)\) follows the formula above
• E.g. The map \(f: R[x]\to R\) given by \(\displaystyle\sum\limits_{i=0}^{n}a_i x^{i}\mapsto \sum\limits_{i=0}^{n}a_i a^i\) is a homomorphism.
• Aside: just like kinds of operators have "names" in universal algebra, maps (or at least homomorphisms) seem to as well

For any ring \(R\), the map \(f : R \to \set{0_R}\) given by \(x \mapsto 0_R\) is a homomorphism. A trivial homomorphism is a ring homomorphism \(f : R \to S\) given by \(x \mapsto 0_S\). A homomorphism is trivial if and only if \(f(1_R)=0_S\).

If \(f_1 : R \to S_1\) and \(f_2 : R \to S_2\) are homomorphisms, then \(f : R \to S_1 \times S_2\) given by \(f(x):=(f_1(x), f_2(x))\) is also a homomorphism. So, the direct product preserves homomorphism.

A unital homomorphism is a homomorphism \(f : R \to S\) between unital rings \(R\) and \(S\) such that \(1_R\) is mapped to \(1_S\), i.e. the multiplicative identity is preserved.

• Proof: exercise

Any homomorphism \(f : R[x]\to R\) is determined uniquely by the value \(f(x)\). So, the map \(\gamma : \text{Hom}(R[x], R)\to R\) given by \(f \mapsto f(x)\) is a bijection.

• If \(\gamma: R\to S\) and \(s\in S\), \(\Gamma_s : R[x] \to S\) given by \(\displaystyle\sum_{i=0}^{m}a_i x^{i}\mapsto \displaystyle\sum_{i=0}^{m} \gamma(a_i)s^i\) (i.e. that replaces coefficients with their evaluation by \(\gamma\) and evaluates the polynomial at \(s\)) is a ring homomorphism.
• If we fix \(\gamma = \text{id} : R\to R\), then \(\Gamma\) is the evaluation map at \(s\), i.e. \(\Gamma_s = f(s)\) where \(f(x)\) is the polynomial defined by \(\displaystyle\sum_{i=0}^{m}a_i x^{i}\).
• ==is the map \(R \to S\) a homomorphism?

An isomorphism \(f : R\to S\) is a bijective homomorphism between rings \(R\) and \(S\) (i.e. \(f\) and \(f^{-1}\) are both ring homomorphisms \(R \to S\) and \(S \to R\), respectively). \(R\) and \(S\) are isomorphic, denoted \(R \cong S\), if a isomorphism between them exists.

• An automorphism is an isomorphism from a ring to itself, i.e. an isomorphism \(f : R\to R\)

Isomorphism implies that two rings have the same structure (literally, "equal shape"). So, if rings \(R\) and \(S\) are isomorphic, \(R\) is commutative \(\implies\) \(S\) is commutative, \(R\) is unital \(\implies\) \(S\) is unital,

Properties of Homomorphisms

• Proof (1): \(0_S+f(0_R)=f(0_R)=f(0_R+0_R)=f(0_R)+f(0_R) \implies 0_S=f(0_R)\)
• Proof (2): \(f(a)+f(-a)=f(a + (-a))=f(0_R)=0_S \implies f(-a)=-f(a)\)
• Proof (3): \(f\) is injective and \(f(a)=0_S\) \(\implies\) \(a=0_R\). Since \(f\) is injective, for \(x,y\in R\) where \(f(x)=f(y)\), we have \(f(x)-f(y)=0_S\implies f(x+(-y))=0_S\implies x+(-y)=0_R\implies x=y\)

The kernel \(\ker{f}\) of homomorphism \(f : R\to S\) is the set \(\ker{f}:=\set{r\in R\mid f(r)=0_S}\)

• So, units, inverses, and zero-divisors are preserved under homomorphism between unital rings

Ideals

It turns out that the equivalence class construction used to define \(\mathbb{Z}_n\) and \(F[x]/(p(x))\) can be extended to any commutative ring with multiplicative identity; this is sufficient to define divisibility, and thus construct congruence classes.

• In this section, we assume all rings are commutative and have identity

• Trivially, all ideals are subrings, but it turns out not all subrings are ideals
• Ideals are somewhat analogous to normal subgroups in group theory: both are substructures that are closed under multiplication by elements of the entire structure.
• They are more closely analogous to subspaces of vector spaces as well (since vector spaces are defined over similar structures to rings. But we'll get to that later)
• We aren't defining anything with \(rIr^{-1}\) (yet) since \(rIr^{-1}\subseteq I\) is trivial by commutativity

Examples of ideals:

• For \(n\in \mathbb{N}\), \(n \mathbb{Z}\) is an ideal of the ring \(\mathbb{Z}\)

• Proof: Let \(a\) be a unit in \(I\). Then, since \(a^{-1}\in R\), we have \(a^{-1}a = 1_R \in I\). Thus, arbitrary \(r\in R\) must also be in \(I\) since \(r1_R=r\in I\) by definition.
• \(R\) and \(\set{0_R}\) are the trivial ideals of \(R\).

• \(a\) is a unit if and only if \((a)=R\)
• E.g. every ideal in \(\mathbb{Z}\) is principal (proof uses divisibility rule, much like the group-theoretic proof of \(n \mathbb{Z}\leq \mathbb{Z}\) for all \(n\in \mathbb{N}\))

A ring \(R\) is also a field if and only if all of its ideals are trivial, i.e. iff all of its nonzero elements are units, which is what classically defines a field (as opposed to a ring or ID).

• An ideal \(I\) of \(R\) is finitely generated if there exists \(a_1, \dots, a_n\in R\) such that \(I=(a_1, \dots, a_n)\). So, any member of \(I\) can be written as a linear combination of its system of generators.

• Proof: exercise

Operations on Ideals

• Note that the union of ideals isn't generally an ideal. The intuition is similar for why the union of subgroups is not generally a subgroup.
• Proof: exercise

• If \(\mathcal{I}\) is set the set of ideals in \(R\) that contain \(S\), then \((S)=\displaystyle\bigcap_{I\in \mathcal{I}} I\).
  • Aside: this is also how an open set is defined in point-set topology. Can we express ideals generated by sets in terms of an interior operator and/or topological space?
• We have \((S)=\displaystyle\sum\limits_{i=1}^{m} r_{j_i}s_{j_i}\) where \(r_{j_1}, \dots, r_{j_i}\in R\) and \(s_{j_1}, \dots, s_{j_i}\in S\)

• We have \((r_1, \dots, r_n)=(r_1)+\dots + (r_n)\)

• We have \(\displaystyle\prod_{i=1}^{n}I_i = \set{\sum\limits_{j=1}^{m} {r_1}_j \times {r_2}_j \times \dots \times {r_n}_j \mid \text{ for all } r_{a,b}\in I_1, \dots, I_n}\)

We have the following algebraic properties of operations on ideals

• Associativity: \(I \times (J \times K) = (I \times J) \times K\)
• Commutativity: \(I \times J = J \times I\)
• Distributivity: \(I \times (J + K) = I\times J + I \times K\)
• \((a)\times (b)=(ab)\)
• \(I \times J \subseteq I \cap J\)

• If \(a\in R\) is nilpotent and \(R\) has identity, \(a + 1_R\) is a unit (proof: A1 question)

• Proof: let \(a, b \in R\) be nilpotent → \(a^m=0_R\) and \(b^n=0_R\) for some \(m,n\in \mathbb{N}\). Clearly \(ab\) is nilpotent since \((ab)^{\max{m,n}}=a^{\max{m,n}}b^{\max{m,n}}=0_R 0_R = 0_R\). We have \((a+b)^{m+n}=\displaystyle\sum\limits_{i=0}^{m+n}{m+n \choose i} \times r^{i} \times s^{m+n-i}\) by the binomial theorem. Note that for \(0 \leq i \leq m-1\) we must have \(m+n-1 \geq n\), so \(s^{m+n-i}=0_R\) by nilpotence, and if \(i \geq m\) we have \(r^{i}=0_R\) again by nilpotence. Thus, every term in the sum is \(0_R\), and thus \((a+b)^{m+n}=0_R\). So, the set of nilpotent elements satisfies the requirements to be an ideal.

• So, the nil radical is simply the radical of the ideal \((0)\)

• Proof is a trivial application of exponent laws
• Aside: is there a lattice here? Will the the radical of an idea always be smaller (or maybe always be bigger) than the ideal itself? What structure does repeatedly applying ideals yield? A DAG?

Prime and Maximal Ideals

A prime ideal is a proper ideal \(I \subsetneq R\) of ring \(R\) such that for all \(a,b\in R\), \(ab\in I\) implies that either \(a\in I\) or \(b\in I\).

• So, prime ideals have a "cancellation law" with respect to inclusion (wording?)
• This is clearly connected to the idea of zero divisors and "prime" elements
• E.g. \(p \mathbb{Z}\) is an ideal of \(\mathbb{Z}\) if and only if \(p\) is prime
• E.g. for field \(F\) and polynomial \(p(x)\in F[x]\), \(F[x]/p(x)\) is prime if and only if \(p(x)\) is irreducible in \(F[x]\) (!)
• E.g. The set \(\set{0_D}\) for \(0_D\in D\) is a prime ideal if and only if \(D\) is an integral domain

• Proof: clearly \(\sqrt{I}\subseteq I\). Let \(x\in \sqrt{I}\); let \(n\) be the minimal where \(x^n\in I\). If \(n=1\), then we are done. Otherwise, we have \(x\) and \(x^{n-1}\) both not in \(I\). But, \(x^{n-1}x=x^{n}\in I\); both of these factors are not in \(I\), which contradicts that \(I\) is a prime ideal. Thus, \(n=1\) and \(I\subseteq \sqrt{I}\).

A maximal ideal is a proper ideal \(I \subsetneq R\) of ring \(R\) if there are no proper ideals \(J\) of \(R\) that contain \(I\).

• Proof: Let \(M\) be maximal and let \(a,b\in R\) such that \(ab\in M\). If \(a\not\in M\) (WLOG), then \(M+(a)\) is an ideal containing \(M\) that is bigger than \(M\); since \(M\) is maximal, this implies \(R=M+(a)\). So, there exists \(r\in R\) and \(x\in M\) where \(1_R=x+ra\) (since \(1_R \in R\) by implicit assumption that all rings are unital in this chapter). So, we have \(b\in M\) because \(b=b1_R =bx+r(ab)\in M\). Thus, \(a\not\in M\implies b\in M\); at least one of \(a\) or \(b\) must be in \(M\), so \(M\) is prime.

Existence of Maximal Ideals

The existence of maximal (and thus prime) ideals is derived from Zorn's Lemma.

We define a partial order relation \(\prec\) as a relation over some (maybe all) of set \(K\) that obeys reflexivity, transitivity, and antisymmetry (\(x\prec y \land y \prec x \implies x=y\)). We call the structure \((K, \prec)\) a partially ordered set.

• A set is totally ordered when \(\prec\) is defined over all of its elements
• E.g. For a set \(S\), the subset relation \(\subseteq\) defines a partial order over its power set \(\mathcal{P}(S)\), i.e. \((\mathcal{P}(S), \subseteq)\) is a partially ordered set. It is not a total order for \(|S|\geq 2\) since there exist pairs of subsets where neither is a subset of the other

An upper bound of \(K\) is some \(a\) (not in \(K\)) such that we have \(k \prec a\) for all \(k\in K\). A maximal element of \(K\) is some \(m\in K\) (this time, must be in \(K\)) that is an upper bound of \(K\), i.e. such that \(k\prec m\implies k=m\) for all \(k\in K\), i.e. an element such that no bigger element exists in \(K\).

A partially ordered set \((K, \prec)\) is inductive if every totally ordered subset of \(K\) has an upper bound.

• This is equivalent to the axiom of choice (!!)

Establishing Zorn's Lemma lets us prove more facts about rings in general.

• Proof: let \(K\) be the set of "sub-ideals" of \(I_0\), not including \(R\) if \(I_0\) happens to be \(R\) itself. \(K\) is partially ordered by \(\subseteq\). For any totally ordered subset \(T\subseteq K\), the union \(J\) (intersection?? typo?) of ideals in \(T\) is itself an ideal in \(R\) (huh??). \(1_R\) isn't in every ideal in \(T\), so it isn't in \(J\). So, \(J\) is in \(K\) and is an upper bound for \(T\). Since \(K\) is inductive and partially ordered, it has a maximal element by Zorn's Lemma.
• Gotta review this proof its a bit fuzzy and I don't understand it well enough to know if some of the things rae typos

• Proof in text

-- get to : R/I is a field if I is a maximal ideal --

Quotient Rings and Fields

We can extend the idea of divisibility/quotient classes/modular arithmetic to ideals; these might be the minimal structure that we can define modular arithmetic over.

For ring \(R\) with ideal \(I\subseteq R\), we define \(a, b\in R\) as being congruent modulo \(I\) if \(a-b\in I\). We denote this \(a\equiv b\mod I\) . This congruence is an equivalence relation. We define \([a]:=a+I\) as the coset of representative \(a\) mod \(I\).

We define \(R/I\) as the set of all cosets of \(R\) induced by \(I\), i.e. all the equivalence classes of the congruence relation. Treating \(R/I\) like a set of elements, we can define addition as \([a]+[b]=[a+b]\) and multiplication as \([a][b]=[ab]\) (i.e. that addition and multiplication are well-defined with respect to the equivalence classes). \(R/I\) is itself a ring under these operations; this is a quotient ring.

• Proof \(\implies\): Let \(R/I\) be an integral domain and let \(a,b\in R\) such that \(ab\in I\). We have \([a][b]=[ab]\), which must equal \([0]\) since we are taking the quotient by \(I\ni [0]\). By assumption \(R/I\) is an integral domain and thus has no zero divisors, so either \([a]=[0]\) or \([b]=[0]\), implying \(a\in I\) or \(b\in I\). So, \(I\) is a prime ideal
• Proof \(\impliedby\): Let \(I\) be a prime ideal. Let \([a],[b]\in R/I\) such that \([a][b]=[ab]=[0]\). Thus, \(ab\in I\); since \(I\) is a prime ideal by assumption, either \(a\in I\) or \(b\in I\), and thus either \([a]=[0]\) or \([b]=[0]\). So, \(R/I\) cannot have zero-divisors, and is thus an integral domain.

• Proof \(\implies\): Let \(R/I\) be a field. \(I\) is maximal if for any \(a\not\in I\), \((a)+I=R\), i.e. we can write \(1=ar+i\) for some \(r\in R\) and \(i\in I\). Note that \(a\not\in I\implies [a]\ne [0]\), so since \(R/I\) is a field, there exists \([a^{-1}]\in R/I\) such that \([a][a^{-1}]=[1]\). Thus, \(1-ar\in I\), so define \(i:=1-ar\) to conform to the equation above.
• Proof \(\impliedby\): Assume \(I\) is maximal. Consider \([a]\ne [0]\); we must have \(a\not\in I\). So, by the previous proof, we can write \(1=ar+i\) for some \(r\in R\) and \(i\in I\). So, we have \([1]=[a][r]+[i]=[a][r]+[0]=[a][r]\), i.e. \([a]\) has inverse \([r]\). So, since an inverse for arbitrary nonzero \([a]\) can be constructed, \(R/I\) must be a field.

Note that since we know all fields are also integral domains, we can deduce from the last two theorems that all maximal ideals are prime ideals, a result we proved earlier. Everything lines up.

Quotient Rings and Homomorphisms

For ring \(R\) and ideal \(I\) of \(R\), we define the canonical homomorphism \(\gamma : R\to R/I\) as the one that maps an element to the equivalence class it represents, i.e. \(\gamma : a\mapsto [a]\), in other notation \(\gamma : a \mapsto a+I\). This homomorphism is surjective, with kernel \(\ker{\gamma}=I\).

• It turns out all surjective homomorphisms resemble this one

• Proof: Let \(a,b\in \ker{f}\). We have \(f(a+b)=f(a)+f(b)=0+0=0\), so \(f(a+b)\in \ker{f}\). We have, for \(r\in R\), \(f(ra)=f(r)f(a)=f(r)\cdot 0=0\), so \(ra\in\ker{f}\).
• These proofs inherit from the fact that \(0+0=0\) and \(0\cdot 0=0\)

Since kernels of homomorphisms are ideals, we can take quotients by them.

• Proof (1): Let \(a+I\in \ker{\varphi}\). By construction of \(\varphi\), we have \(f(a)=0\), so \(a\in \ker{f}\). Sim \(a + \ker{f}=\ker{f}\), implying \(\varphi\) is injective.
• Proof (2): Let \(f\) be surjective and let \(s\in S\). Pick \(r\in R\) such that \(f(r)=s\). By construction, we have \(\varphi(r+\ker{f})=f(r)=s\), so \(h\) is surjective.
• Aside: is this the factorization theorem / first iso theorem for rings? Sure looks like it

\usepackage{tikz-cd}
\begin{document}

\begin{tikzcd}[scale=50]
    R \arrow[r, "f"] \arrow[d, "\gamma"'] & S \\
    R/\ker{f} \arrow[ru, "{\varphi}"']
\end{tikzcd}

\end{document}

By corollary, if \(f:R\to S\) is surjective, then \(R/\ker{f}\cong S\).

Informally, show that (generally) every surjective homomorphism \(f\) is of the form \(R\to R/I\), \(a\mapsto [a]\) for a proper choice of ideal \(I\) of \(R\).

We also showed that \(\text{im }f\) is a subring of \(S\); it turns out specifically, we have \(\text{im }{f}\cong R/\ker{f}\).

Chapter 03 - Field Extensions

As standard notation, we use \(k\) to refer to a field, and \(E\) to refer to an extension of that field.

Characteristic and \(k\)-algebras

For arbitrary ring \(R\), recall the canonical homomorphism \(\pi_R : \mathbb{Z} \to R\) given by \(n\mapsto n\cdot 1_R\) (in the sense of repeated addition, i.e. \(n\cdot 1_R:=\displaystyle\sum\limits_{i=1}^{n} 1_R\)). Since \(\ker{\pi_R}\) must be an ideal of \(\mathbb{Z}\), it must be equal to \((n)\) for some \(n\in \mathbb{N}\) (every ideal of \(\mathbb{Z}\) is principal).

• So, the characteristic of \(R\) is the smallest \(n\in \mathbb{N}\) such that adding \(1_R\) to itself \(n\) times yields \(0_R\).
• Adding \(1_R\) to itself lets us create a copy of \(\mathbb{N}\) (characteristic \(0\)) or \(\mathbb{Z}_n\) (non-zero characteristic);
• Rings/fields where no such \(n\) exists (i.e. whose copy of \(\mathbb{N}\) is unbounded) have the Archimedean property, and are defined to have characteristic \(0\).

Examples:

• \(\mathbb{Z}\) has characteristic \(0\) since it is archimedean
• \(\mathbb{Z}/n\mathbb{Z}=:\mathbb{Z}_n\) has characteristic \(n\) since clearly \([1]+\dots+[1]=[n]=[0]\)

• Proof: let \(\text{char}R =: n\ne 0\), so \(n=ab\) for \(a,b\in \mathbb{N}\) where \(1\leq a,b\leq n\). We have \(0_R=n\cdot 1_R=(ab)\cdot 1_R=(a\cdot 1_R)(b\cdot 1_R)\). Since \(R\) is an ID, either \(a\cdot 1_R=0_R\) or \(b\cdot 1_R=0_R\). By definition of characteristic, \(n\) is minimal, we must either have \(a=n\) or \(b=n\), implying the other is \(1_R\). So, \(n\) has no divisors except for itself and \(1_R\); \(n\) must be prime.
  • ==it seems like we're mixing ring multiplication and integer scaling here; it's not clear if we're talking about primality in \(\mathbb{Z}\) or in the ring itself, and whether \(1\) is in \(\mathbb{N}\) or in \(R\). Disambiguate this==.

• Note that any homomorphism of rings \(f:K\to R\) must be injective since \(K\) is a field. \(\ker{f}\) must be an ideal in \(K\), but again since \(K\) is a field, it only has the trivial ideals \(\set{0_K}\) and \(K\). We must have \(f : 1_K \mapsto 1_R\ne 0_R\), so we have \(\ker{f}=K\).
• We (abusively) use \(K\) to denote the subring \(\iota(K)\) of \(A\)
• A \(k\)-subalgebra is a subring \(B\subseteq R\) containing \(K\) (i.e. containing \(\iota(K)\))
• Aside: what is the point?? who cares? needs more motivation
• Aside: what characterizes "algebras" as a class? Is an operation with some sort of properties like a homomorphism needed?

==inner product on a vector space is a \(r\) algebra?

vector space axioms, k algebra can be viewed as a vector space??==

A \(K\)-algebra homomorphism \(f\) is a ring homomorphism \(f:R_A\to R_B\) between \(k\)-algebras \(A=(R_A, \iota_A)\) and \(B=(R_B, \iota_B)\) such that for all \(x\in K\), we have \(f(x)=x\), i.e. \(f(\iota_A(x))=\iota_B(x)\).

\usepackage{tikz-cd}
\begin{document}

\begin{tikzcd}[scale=50]
    A \arrow[rr, "f"] & {} & B \\
    {} & K \arrow[lu, "{\iota_A}"'] \arrow[ru, "{\iota_B}"']
\end{tikzcd}

\end{document}

Generally, \(f\) is a \(K\)-morphism (and a \(K\)-isomorphism if it is a ring isomorphism)

• what is a morphism? define properly

Field Extensions and Prime Subfields

• If \(E\) extends \(K\), \(K\) has the natural structure of a \(K\)-algebra (??? intuition?)
• If \(K\) is a subring of \(E_K\), it is also a subfield of \(E_K\) if every \(x\in K\) has an inverse \(x^{-1}\in K\) as well.
• E.g. if \(K[T]\) is the fraction field of the polynomial ring with coefficients in \(K\) over variable \(T\), then the fraction field of \(K[T]\) is a field extension of \(K\).
  • This also for polynomials over \(K\) of multiple variables, i.e. \(K[\vec{T}]\)

• The extension from \(\pi\) to \(\pi_\mathbb{Q}\) is suggested by the universal property of fraction fields.
• As alluded to before, fraction fields (and the idea of inverses) are the "bones" of \(\mathbb{Q}\) in the sense that the Peano axioms are the "bones" of \(\mathbb{N}\); they describe the minimal conditions for a field to have a subfield isomorphic to \(\mathbb{Q}\). Note that any ring with substructure isomorphic to \(\mathbb{Q}\) must be a field (I think) because fields require multiplicative inverses, while rings do not.
• \(\pi_\mathbb{Q}(\mathbb{Q})\) is the smallest subfield of \(F\) because any subfield would contain \(1_F\) and \(0_F\), which guarantee the fraction field construction (i.e. \(\pi_\mathbb{Q}(\mathbb{Q})\)) is a subfield of it.
• ==Aside: connection to 217: peano axioms and every comoplex enough thing has a subset isomorphic to \(\mathbb{N}\). Did we make any claims about what is needed to have a subset isomorphic to \(\mathbb{Q}\)?
• note: I may very well have \(E\) and \(F\) mixed up==

• Explanation: by definition, \(\ker{\pi_F}\) must be generated by \(p\), so by properties of homomorphisms (??), we have \(\text{Im}(\pi_F)\cong \mathbb{Z}/(p)=\mathbb{Z}_p\), which is also denoted \(\mathbb{F}_p\) when \(\mathbb{Z}_p\) is a field (i.e. when \(p\) is prime). We know that the characteristic of a field must be prime.
• \(F\) is a field extension of \(\mathbb{F}_p\).

Algebraic and Finite Extensions

Let \(K\) be a field with field extension \(E_K\). We have the following definitions:

• Extension \(E_K\) is finite if the vector space over \(K\) it suggests has finite dimension. The degree \([E_K : K]\) of field extension \(E_K\) of \(K\) is the dimension of this vector space. \(E_K\) is infinite if it is not finite.
• Element \(a\in E_K\) is algebraic over \(K\) if a polynomial \(f(T)\in K[T]\) exists such that \(f(a)=0_F\). \(E_K\) itself is algebraic if every \(a\in E_K\) is algebraic
  • All elements in \(K\) are trivially algebraic over \(K\). So, \(K\) is algebraic
• The subfield \(K(S)\) generated by subset \(S\subseteq E_K\) over \(K\) is the smallest subfield of \(E_K\) containing both \(K\) and \(S\).
  • If \(S=\set{a_1, \dots, a_n}\) is finite, then we can write \(K(S)\) as \(K(a_1, \dots, a_n)\)
• Extension \(E_K\) of \(K\) is finitely generated if \(E=K(a_1, \dots, a_n)\) for some \(a_1, \dots, a_n\in E_K\).
  • We say that \(a_1, \dots, a_n\) generate \(E_K\)
• If \(E_K=K(a)\) for single \(a\in E_K\), then \(E_K\) is a simple extension of \(K\) and \(a\) is a primitive element of \(E_K\).

• Proof: By definition, \(a_1, \dots, a_k\) form a \(K\)-basis for \(E\) (??), so every element in \(E\) can be written as a linear combination of \(a_1, \dots, a_k\).

Note: Finitely generated field extensions don't need to be finite; infinite field extensions can be generated by some \(a_1, \dots, a_k\) as well.

• E.g. The function field \(K[T]\) is a field extension of \(K\) with single generator \(T\) (the variable). \(K[T]\) has infinite dimension over \(K\): The elements \(T^{0}:=1,T, T^{2}, T^{3}, \dots\) are all linearly independent over \(K\).

==\(\mathbb{Q}(\sqrt{2})\) is a field extension? example? integrate this somewhere, thanks pianzola for this example

function fields? used as an example but we didn't define them yet

• Proof: If we define \(n:= \dim_K E_K\), then for arbitrary \(x\in E_K\), the elements \(1, x, x^{2}, \dots, x^{n}\) cannot be linearly independent by the pigeonhole principle. Thus, there exist \(b_0, \dots, b_n\in K\) (not all \(0_K\)) (==why in \(K\)?==) such that \(\displaystyle\sum\limits_{i=0}^{\ell}b_i x^{i}=0_K\). Thus, by definition, \(x\) is algebraic over \(K\).
• E.g. \(\mathbb{C}\) is a finite, and thus algebraic extension of \(\mathbb{R}\). However, \(\mathbb{C}\) is not a finite extension of \(\mathbb{Q}\). It is not even an algebraic extension of \(\mathbb{Q}\).

Minimal Polynomials

• proof

capital k?

If \(a\) is algebraic, we have minimal polynomial \(k(a) = \set{g(a)\mid g(T)\in k[T]}\), which is given by \(\set{\displaystyle\sum\limits_{i=0}^{n-1}a_i a^{i}\mid a_i\in k \text{ for all } 0\leq i \leq n-1}\), where \(n\) is the degree of \(f(T)\). These elements \(1, a, \dots, a^{n-1}\) form a basis of the \(k\)-vector space \(k(a)\).

• So, \(\deg{f(T)}=[k(a) : k]=\deg_k{ k(a)}\).

The minimal polynomial of algebraic \(a\in E\supset k\) is the unique monic irreducible polynomial \(f(T)\in k[T]\) such that \(f(a)=0\).

• E.g. \(T^{2}-2\) is the minimal polynomial of \(\sqrt{2}\) over \(\mathbb{Q}\). So, \(\mathbb{Q}[T]/(T^{2}-2)\cong \mathbb{Q}(\sqrt{2})\), which is a vector space of dimension \(2\) over \(\mathbb{Q}\) with basis \(\set{1, \sqrt{2}}\)
• E.g. the general pattern is true for non-square \(a\): \(\mathbb{Q}[T]/(T^{2}-a)\cong \mathbb{Q}(\sqrt{a})\).
• E.g. \(\mathbb{C}\cong \mathbb{R}[T]/(T^2 +1)\)

Properties of Field Extensions

• Proof

By corollary:

• If \(a_1, \dots, a_n\) are algebraic over \(K\), then \(K(a_1, \dots, a_n)\) is a finite extension of \(K\)
• The set of elements in \(E\) that are algebraic over \(K\) is a subfield of \(E\) called the algebraic closure of \(K\) in \(E\).
• If \(S\subseteq E\) are all algebraic, then \(k(S)\) is an algebraic extension of \(K\)
• Algebraic field extensions are also transitive (what's the difference??)

Note: showing something is algebraic often involves using the evaluation map

Adjoining Roots of Polynomials

Let \(k\) be a field, \(f(T)\in k[T]\) be a non-constant polynomial, and let irreducible \(g(T)\) divide \(f(T)\). Since \(g(T)\) is irreducible, \(E:=k[T]/(g(T))\) is a field extension of \(k\). A member of \(E\) (i.e. a congruence class) given by \(\theta:=T+(g(T))\) is a root of \(g(T)\), and thus \(f(T)\), since \(g(T)(\theta)=g(T)+(g(T))=0\).

• This root generates \(E\): \(E=k(\theta)\)
• We can keep iterating this construction to generate more and more roots of \(f(T)\)

• If \(f[T]\) is already linear or "splittable" in \(k\), then \(k\) is the only splitting field of \(f[T]\)
• Splitting fields are unique up to \(k\)-isomorphism (??)

look at the remark about finite/infinite here

Uniqueness of Splitting Fields

For a morphism of fields \(\sigma : k\to k_1\), a homomorphism of polynomial rings \(k[T]\to k_1[T]\) given by \(\displaystyle\sum\limits_{i=0}^{n}a_i T^{i}\mapsto \sum\limits_{i=0}^{n}\sigma(a_i)T^{i}\) is implied; we says that polynomial \(f\) gets sent to \(\sigma(f)\).

• If \(\sigma\) is an isomorphism of fields, then the implied polynomial homomorphism is also an isomorphism, and thus reducibility is preserved.

![[Pasted image 20250225205231.png]]

![[Pasted image 20250225210316.png]]

• Proof

WAYYYYYY more here

Normal Extensions

An extension \(E\) of field \(k\) is a normal extension if every irreducible polynomial \(g(T)\in k[T]\) that has a root in \(E\) is completely decomposed into a product of linear factors in \(E[T]\).

• Proof. Wayyyy more proof

relate to the minimal polynomial here

E.g. For field \(k\) such that \(\text{char}{K}\ne 2\) and non-square \(a\in F\), the polynomial \(T^{2}-a\in F[T]\) is irreducible and \(F(\sqrt{a}):=F[T]/(T^{2}-a)\) is a normal extension of \(F\).

• \(T^{2}-a\) is irreducible because it would need to reduce into \((T-b)(T+b)\)
• Normal because the roots of \(T^2-a\) are \(\pm \sqrt{a}\), which exist in \(F(\sqrt{a})\) by construction. So, \(T^{2}-a\) factors into \((T-\sqrt{a})(T+\sqrt{a})\) in \(F(\sqrt{a})\), and is thus normal

Separable Polynomials

A polynomial \(f(T)\in k[T]\) is separable if \(f(T)\) has no root that occurs more than once in splitting field \(E\) of \(f(T)\).

• This is equivalent to saying that \(f(T)\) has no multiple roots in any field extension of \(k\), since splitting fields of \(f(T)\) are isomorphic and don't depend on the choice of \(E\).

• Proof: relies on being able to show that in \(k[T]\), we have \(d(T)=r(T)\cdot f(T)+s(T)\cdot g(T)\) (bezout's lemma)

The formal derivation of polynomial \(f(T)=\displaystyle\sum\limits_{i=0}^{n}a_i T^{i}\) in \(k[T]\) is defined \(f'(T):=\displaystyle\sum\limits_{i=1}^{n}i\cdot a_i\cdot T^{i-1}\), i.e. is defined like the derivative in calculus.

• The formal derivation is linear: \((f(T)+g(T))'=f'(T)+f'(T)\)
• Product rule: \((f(T)g(T))'=f'(T)g(T)+f(T)g'(T)\)

If \(f(T)\) is not constant and \(\text{char}{k} = 0\), then \(f'(T)\) will be nonzero.

• Proof: Assume towards a contradiction that \(f(T)\) has some multiple roots, so \(f(T)=(T-a)^{2} g(T)\) in \(E\) for some \(g(T)\). We find \(f(T)\) and \(f'(T)\) both have common root \((T-a)\). So, the GCD of \(f(T)\) and \(f'(T)\) in \(k[T]\) is not \(1\). Since \(f(T)\) is irreducible, the degree of this GCD must be the same as \(f(T)\), which is impossible since this CGD must also divide \(f'(T)\), which has lower degree. Thus, \(f(T)\) has no multiple roots.

not sure what that example is

Theorem on Primitive Elements

An element \(x\in E\) is separable over \(k\subseteq E\) if \(x\) is algebraic over \(k\) and its minimal polynomial is separable.

The extension \(E\) of \(k\) is separable if all elements in \(E\) are separable over \(k\).

• So, if \(\text{char}{k}=0\), then all algebraic field extensions of \(k\) are separable

• Proof: this proof is the important one

A Galois finite extension \(E\) of field \(k\) is an extension \(E\) of \(k\) that is normal and separable.

• We define the Galois group \(\text{Gal}(E/k)\) of the Galois extension \(E\) as the group of isomorphisms \(E\to E\) (automorphisms) under composition of maps.

Finite Fields

Finite fields must have a nonzero characteristic, and thus a prime characteristic \(p\). So, it will contain a subfield \(\mathbb{F}_p:=\mathbb{Z}/(p)\). Since the whole field is an extension of this subfield, any finite field is a finite-dimensional vector space over its \(\mathbb{F}_p\).

• If the dimension of this vector space is \(n\), then \(K\) has \(p^{n}\) elements

E.g. The polynomial \(p(T)=T^{2}+T+1\) in \(\mathbb{F}_2[T]\) clearly has no roots in \(\mathbb{F}_2\) and is thus irreducible in \(\mathbb{F}_2\). The quotient field \(\mathbb{F}_2[T]/(p(t))\) has \(4=2^2\) elements, and thus is a \(2\)-dimensional vector space over \(\mathbb{F}_2\).

• Let \(\alpha:=T+(p(T))\), which is in \(K\).
• We have \(\alpha\not\in \mathbb{F}_2\), so the set \(\set{1, \alpha}\) is linearly independent in \(K\), and is thus a basis of \(K\)
• Thus, \(K\) consists of \(\set{0, 1, \alpha, \alpha+1}\).

• Proof (existence): We consider polynomial \(f(T)=T^{p^{n}}-T\in \mathbb{F}_p[T]\). We note that \(f'(T)\ne 0\), so \(f(T)\) is separable and thus has different roots. Let \(K\) be a splitting field for \(f(T)\) with roots \(0, a_1, \dots, a_{p^{n}-1}\). Clearly, these roots are closed under addition and multiplication, so \(\set{0, a_1, \dots, a_{p^{n}-1}}\subseteq K\) is a subfield containing all the roots of \(f(T)\). But, since \(K\) is a splitting field of \(f(T)\) by construction, i.e. is generated by the roots of \(f(T)\), we have \(K=\set{0, a_1, \dots, a_{p^{n}-1}}\). So, \(K\) contains \(p^{n}\) elements.

By corollary, for any \(n\in \mathbb{N}^+\), there exists an irreducible polynomial \(h(T)\in \mathbb{Z}_p[T]\) of degree \(n\); any such polynomial divides \(f(T)=T^{p^n}-T\).

• Proof