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Math 123: Algebra II
Course Taught by Sebastien Vasey
Notes Taken by Forrest Flesher
Spring 2020
Welcome to Math 123: Algebra II. Here’s some important information:
- The course webpage is: http://people.math.harvard.edu/~sebv/123-spring-2020/
- The syllabus is located at: http://people.math.harvard.edu/~sebv/123-spring-2020/syllabus.pdf Office hours are at the following times over Zoom: - Tuesday, Thursday, 4-5pm ET.
- Your CAs will also hold office hours at the following times over Zoom: - Garrett: Sunday 10-11am ET, Monday 8-10pm ET, - Forrest: Tuesday 9:30-11:59pm, Friday 3-4pm.
- The text for the course is Dummit and Foote “Abstract Algebra, 3rd edition.” If you don’t have a copy, don’t worry: all the problem sets will include copies of the problems, and the course notes will contain all the course material.
- Relevant emails are [email protected] , [email protected], [email protected], Email with any questions, comments, or concerns, especially if you find a mistake in these notes.
- We will use the Canvas site for submitting/grading problem sets.
- It is okay if your psets are legibly non-latexed, but latex is much preferred.
- The prerequisites are math 122 or math 55a. If you haven’t taken either of these but are comfortable enough with algebra, that is also okay.
Contents
§1 January 29, 2020
§1.1 Review
Over the next few classes, we’ll be doing a quick review of the prerequisites for the course. A binary operation on a set X is a function ⋅ ∶ X × X → X, typically written a ⋅ b instead of ⋅(a, b). The familiar operations of addition, multiplication, subtraction, etc. on Z, R, etc. are examples of binary operations. A binary operation ⋅ is associative if for all a, b, c then (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c). An identity for a binary operation on a set X is an element e such that for all a ∈ X, then a⋅e = e⋅a = a. The identity is always unique, since if e and e′^ are identities, then e = e ⋅ e′^ = e′^ ⋅ e = e′. An inverse for an element a is an element b such that a ⋅ b = e, the identity. A group is a set together with an operation that is associative, has an identity, and has inverses. Some examples are (Z, +), (R − { 0 }, ⋅), and Sn, the set of functions from { 1 ,... , n} to itself, with the operation of addition. These first two examples are abelian (or commutative ) groups, meaning a ⋅ b = b ⋅ a for all a, b, and the symmetric group Sn is not. We now define the primary objects of study for this course.
Definition 1.1 — A ring is a set R with two operations, +, ⋅, such that (R, +) is an abelian group with identity 0 , and (R, ⋅) is associative with identity 1. The two operations must also satisfy distributivity , meaning (a + b) ⋅ c = a ⋅ c + b ⋅ c and a ⋅ (b + c) = a ⋅ b + a ⋅ c for all a, b, c in the ring. If the operation ⋅ is commutative, then the ring is called a commutative ring. NB: we will always assume rings have a multiplicative identity.
Example 1.
- The zero ring ({ 0 }, +, ⋅), with 1 = 0.
- The familiar R, C, Q, Z with addition and multiplication.
- The set of n × n matrices with entries in R, with matrix addition and multipli- cation (note this ring is not commutative unless n = 1 ).
- The quaternions a + ib + jc + kd, with a, b, c, d ∈ R. Addition is component by component, and multiplication is defined by the relations i^2 = j^2 = k^2 = − 1 , ij = k, ji = −k, jk = i, kj = −i, ki = j, ik = −j. You should check that these actually form a ring. Notice that this ring is not commutative, but it is an example of a division ring (or skew field), where 1 ≠ 0 and every nonzero element has a multiplicative inverse:
(a + ib + jc + kd)−^1 = a − ib − jc − kd a^2 + b^2 + c^2 + d^2
- The ring Z~nZ, integers modulo n, defined as the set { 0 ,... , n − 1 } together with the operations of addition and multiplication modulo n. You should check that this is indeed a ring.
Definition 1.3 — A field is a commutative division ring ( 1 ≠ 0 ). The sets Q, R, C are fields, but Z is not.
If we are working in a general ring R, a unit is an element a ∈ R such that a has a multiplicative inverse. We write R×^ for the set of unites. As an exercise, you should prove that R×^ is always a group with the operation of multiplication. Note that 0 is never a unit.
Example 1.
- Z×^ = { 1 , − 1 }, since the only integers with integer inverses are 1 and − 1.
- Q×^ = Z − { 0 }, since the only non-unit is 0.
A zero divisor a in a ring R is a nonzero element such that there exists some b ≠ 0 with ab = 0 or ba = 0. Note, zero divisors are never units, since zero is not a unit. However, units aren’t always nonzerodivisors, for example there are no zerodivisors in Z, but many non-units.
Definition 1.5 — A integral domain is a commutative ring with 0 ≠ 1 and with no zero divisors. Note that any field is an integral domain.
The ring Z~nZ is an integral domain if and only if n is prime. More generally, if a ≠ 0 is an element of Z~nZ, then a is a zero divisor if and only if gcd(a, n) ≠ 1. Note that in this case, Z~nZ is also a field if n is prime. In fact, we have the following theorem.
Theorem 1. Any finite integral domain R is a field.
Proof. Let a ∈ R nonzero. Consider the function f ∶ R → R defined as x ↦ ax. The function f is injective, since in an integral domain ab = ac implies b = c, for a ≠ 0 (check this). Since f is an injective map from a finite set to itself, it is also surjective. Then there exists some r ∈ R such that f (r) = 1 , so ar = 1 and thus a is invertible. Since the arbitrary nonzero element a is invertible, then R is a field.
§1.2 Course Overview
There are many applications of rings and fields in math. For example it is very useful for algebraic geometry (Math 137) and number theory (Math 129). There are also many applications outside of pure math in physics, computer science, and other fields. One famous problem we will cover in this class is the problem of a formula for roots of 5th degree polynomials. There is the familiar quadratic formula for finding all the roots of 2nd degree polynomials, and also a less-familiar formula for 3rd and 4th degree polynomials. However, there is no such formula for 5th degree polynomials. We will prove that such a formula cannot exist using Galois theory in class. Although we don’t have a formula to find roots of such polynomials, we can prove that roots exist, which is the content of the Fundamental Theorem of Algebra. Another application of abstract algebra is to ruler and compass constructions (you might have done these in high school geometry). The Greeks were very into ruler and
first two properties, but ϕ( 1 ) = 0 ≠ 1. However, if we instead consider the zero map into the zero ring, ϕ ∶ Z → { 0 } defined by ϕ ∶ x ↦ 0 , this is a ring homomorphism. In fact, for any ring R, there is a unique homomorphism R → { 0 }.
Example 2.
- The map ϕ ∶ Z → Z~ 2 Z defined by
n ↦
0 , n even 1 , n odd
is a ring homomorphism.
- For any ring S, there is a unique map ϕ ∶ Z → S given by ϕ( 1 ) = (^1) S , and ϕ(n) = (^1) S + ⋯ + (^1) S ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ n-times
- The evaluation at zero map, Z[x] → Z, p ↦ p( 0 ), is a ring homomorphism.
We say that a ring homomorphism is an isomorphism if it is bijective. If ϕ ∶ R → S is a ring homomorphism, then the kernel of ϕ is
ker(ϕ) ∶= {r ∈ R ∶ ϕ(r) = 0 }.
The image of ϕ is im(ϕ) ∶= {s ∈ S ∶ ϕ(r) = s, for some S}. Observe that the image im(ϕ) ⊆ S of any ring homomorphism is a subring of S. Note however that the kernel is not always a subring, since it does not always contain 1. However, the kernel does have many interesting properties. It is closed under addition: if a, b ∈ ker(ϕ), then ϕ(a + b) = ϕ(a) + ϕ(b) = 0 + 0 = 0. It is also closed under multiplication by any element of the ring: if r ∈ R, and a ∈ ker(ϕ), then ϕ(ar) = ϕ(a)ϕ(r) = 0 ⋅ ϕ(r) = 0. These properties make the kernel into an ideal, which we now defined.
Definition 2.2 — An ideal in a ring R is a subset I ⊆ R such that
- 0 ∈ I,
- if a, b ∈ I then a + b ∈ I,
- and if a ∈ I, and r ∈ R (any element of the ring), then ra ∈ I.
Example 2.
- The kernel ker(ϕ) is always an ideal, as discussed above.
- If R is a ring, then { 0 } is an ideal.
- What are the ideals of Q? Certainly { 0 } and Q are ideals inside Q. In fact, these are the only ideals. Since Q, and any ideal is closed under multiplication by arbitrary elements of the ring, then the multiplicative identity 1 must be in the ideal. This implies that every element of Q is in the ideal. So in Q, the only ideals are { 0 } and Q. In fact, if R is any division ring, then the only
ideals are { 0 } and R.
- What are the ideals of Z? The ideals must be additive subgroups, so we should first find these. The additive subgroups of Z are those of the form nZ = (n), the set of multiples of n ∈ Z. Any additive subgroup is of this form, and additionally all these subgroups are ideals: if we multiply any r ∈ Z by x ∈ (n), then rx ∈ (n), since it is still a multiple of n.
- In the polynomial ring, the set of polynomials with zero constant and linear coefficient I = {anxn^ + ⋯ + a 0 ∶ a 0 = a 1 = 0 } is an ideal of Z[x].
§2.2 Quotients, Isomorphism Theorems, and Ideals
Now, if R is a ring and I is an ideal, then I is a normal subgroup of the group (R, +), so it makes sense to define a quotient. The quotient group R~I with elements being equivalence classes of the form r +I, with multiplication defined by (a+I)⋅(b+I) ∶= a⋅b+I. This makes R~I into a ring, which is called the quotient ring. In your homework, you will show that in order for this to be a ring, we do need that I is an ideal and not just some arbitrary additive subgroup. The definition of the quotient ring R~I leads us to the definition of an important ring homomorphism. For any ring R and any ideal I, there is a homomorphism
ϕ ∶ R → R~I ϕ ∶ r ↦ r + I.
This map is surjective, with kernel I. It is often called the quotient map. Now that we have defined quotients, there are several theorems, similar to those for groups, about isomorphisms between quotient rings, called the isomorphism theorems.
Theorem 2.4 (First Isomorphism Theorem) If ϕ ∶ R → S is a ring homomorphism, then R~ ker(ϕ) ≃ im(ϕ).
Proof. By the first isomorphism theorem for groups, it is enough to check that the map
R~ ker(ϕ) → im(ϕ) a + ker(ϕ) ↦ ϕ(a)
preserves multiplication. You should verify this as an exercise.
There are more ‘isomorphism theorems’ for rings, but the first is the most important.
Example 2. Let R = Z[x], and I = {anxn^ + ⋯ + a 1 x + a 0 ∶ a 0 = a 1 = 0 }. In R~I, the [x^3 + 5 x + 2 ] = [x^3 + x^2 + 5 x + 2 ] = [ 5 x + 2 ] ≠ [ 5 x + 3 ], where the [] denotes the equivalence class in R~I. That is, in R~I, we can think of x^2 , x^3 ,... as being
NB: the assumption that R is commutative is essential, as you can see in the following exercise.
Exercise 3.2. The ring Mn(R) of n × n real-valued matrices has only { 0 } and Mn(R) as its ideals.
Corollary 3. If ϕ ∶ F → S is a homomorphism from a field F to any set S is either constantly zero or injective.
Proof. The kernel ker(ϕ) is an ideal of F , so it is either { 0 } or F. If it is F , then the map ϕ is the zero map, and if it is { 0 }, then ϕ is injective.
Now we’ll cover some important types of ideals. An ideal M is a maximal ideal if M ≠ R, and M ⊆ I ⊆ R implies I = M or I = R for any ideal I. That is, an ideal is maximal if it is not properly contained in any other ideal aside from the ideal R. NB: The ring R is not a maximal ideal.
Example 3.
- In Z, the ideal (a) is maximal if and only if a is prime
- In a field, { 0 } is a maximal ideal.
- The zero ring has no maximal ideal.
The following theorem is very useful for proving that rings are fields, or for proving that ideals are maximal.
Theorem 3. In a commutative ring R, an ideal M is maximal if and only if R~M is a field.
Proof. First we note the following fact: the ideals containing an ideal I in a ring R are in bijection with the ideals of R~I. This follows from one of the isomorphism theorems, and a proof can be found here. Using this, suppose that M is maximal. Then there are only two ideals containing M , namely M itself and R. Since the ideals of R containing M are in bijection with the ideals of R~M , this tells us that there are only two ideals in R~M , which must be { 0 } and R~M. By proposition 3.1, this implies R~M is a field. The other direction of the proof uses the same fact and is similar.
Remember that this only works for commutative rings. For noncommutative rings, all we can say is that if M is maximal, then R~M has no proper nontrivial two sided ideals (R~M is a simple ring).
Example 3. Consider the ring Z[x], and the ideal ( 2 , x). Then Z[x]~( 2 , x) ≃ Z~ 2 Z is a field, so ( 2 , x) is maximal.
Another important type of ideal is the prime ideal. An ideal P ⊂ R is a prime ideal if a ⋅ b ∈ P implies a ∈ P or b ∈ P. In the integers Z, the prime ideals are (p) for p prime or zero. For example, ( 4 ) is not prime, since 2 ⋅ 2 ∈ I, but 2 ~∈ I.
Proposition 3. In a commutative ring R, an ideal P is prime if and only if R~P is an integral domain.
The proof is left as an exercise to the reader. As an example application, note that (x) in Z[x] is prime, since Z[x]~(x) ≃ Z, and Z is an integral domain. However, Z is not a field, so (x) is not maximal. Since fields are integral domains, we have the following corollary.
Corollary 3. Any maximal ideal is prime.
Proof. If M is maximal, then R~M is a field. Since fields are integral domains, then R~M is an integral domain, so M is prime.
§3.2 Fields of Fractions and Operations on Ideals
We now define another object which is useful in the study of rings: the field of fractions of a ring. This construction turn a ring into a field by adding inverses, as equivalence classes of elements of the ring.
Definition 3.9 — Given any integral domain R, the field of fractions of R is defined as follows. Let
F = {(a, b)Sa ∈ R, b ∈ R, b ≠ 0 }.
Define an equivalence relation (a, b) ∼ (c, d) if ad = bc (intuitively a~b = c~d). Then the field of fractions is Q = F ~ ∼. Usually, for the equivalence class [(a, b)]∼ of the element (a, b), we just write ab. We define addition and multiplication as
a b
c d
ad + bc bd
a b
c d
ac bd
You can check that this is well defined, and that Q is a field. In addition, R embeds into Q via the map r ↦ 1 r.
The field of fractions of a ring R is the smallest field containing R. A slightly more general construction than the field of fractions is localization. We will not define localizations in this class, but you can read about them here.
Example 3.
- The field of fractions of Z is Q.
As a consequence, let R = Z, and let n = pα 1 1 ⋯pα k k. By the above, we have
Z~nZ = Z~pα 1 1 Z × ⋯ × Z~pα k kZ.
Then (Z~nZ)×^ = (Z~pα 1 1 )×Z × ⋯ × (Z~pα k kZ)×,
where the superscript × denotes the group of units. We also have ϕ(n) = ϕ(pα 1 1 )⋯ϕ(pα k k), where ϕ is the Euler Totient Function.
Proof. We proceed by induction on k, where A 1 ,... , Ak are comaximal. Suppose k = 2. Since A 1 and A 2 are comaximal, there exist x ∈ A 1 and y ∈ A 2 with x + y = 1 , so ϕ(x) = ( 0 , 1 ), ϕ(y) = ( 1 , 0 ). So given (r, s) ∈ R~A 1 × R~A 2 , then
ϕ(sx + ry) = (r, s),
so ϕ is surjective, as desired. Also, if c ∈ A 1 ∩ A 2 , then we have c = c ⋅ 1 = c(x + y) = xc + cy ∈ A 2 A 1 + A 1 A 2 , so c ∈ A 1 ⋅ A 2 (the ring is commutative), thus A 1 ∩ A 2 ⊆ A 1 ⋅ A 2. Since A 1 ⋅ A 2 ⊆ A 1 ∩ A 2 for any ideals, then we have A 1 ⋅ A 2 = A 1 ∩ A 2 , as desired. The induction step is straightforward.
February 12, 2020
§3.4 Integral Domains and PIDs
Last time we discussed different types of ideals, quotients, and integral domains. Recall that a maximal ideal is a proper ideal not contained in any other proper ideal. We now show that these actually exists
Proposition 3. For any ideal I ⊊ R, there is a maximal ideal M such that I ⊆ M ⊊ R.
Proof. If I is maximal, we are done. If I is not maximal, then there exists some I 2 ⊋ I, with I 2 ≠ R. If I 2 is maximal, we are done. Else, we can take I 3 ⊋ I 2 with I 3 ≠ R, and continue. Let IN = ∪In, which is an ideal. We can continue this process, and use Zorn’s lemma to prove the proposition.
Recall that in Z, all ideals are of the form (a), for some a ∈ Z. That is, all the ideals are principal. The integers are an example of a special type of ring which has this property.
Definition 3.14 — An integral domain in which all ideals are principal is called a principal ideal domain (PID).
Definition 3.15 — A Euclidean domain is an integral domain R, together with a norm N ∶ R → Z≥ 0 , where
- N ( 0 ) = 0 , and
- q = qb + r for some q, r with r = 0 or N (r) < N (b). Note that the codomain of N is Z≥ 0 , so norm are nonnegative.
Example 3.
- The ring of integers Z is a Euclidean domain, with N (a) = SaS, the ordinary absolute value.
- Let F be a field. There are multiple ways to choose a norm to make F into a Euclidean domain. In fact, any norm N with N ( 0 ) = 0 works, since every element a is invertible and thus can be written a = qb + r, with r = 0. If F is a field, then F [x], the polynomial ring with coefficients in F , is also a field. Take
N (p) =
deg(p), p ≠ 0 0 , p = 0
You can use polynomial division to verify that any element a can be written in the desired a = qb + r form. Note, we need F to be a field in order that F [x] be a Euclidean domain.
- The Gaussian integers Z[i] = {a + bi ∶ a, b, ∈ Z} are a Euclidean domain. Take N (a + bi) = a^2 + b^2.
Theorem 3. Any Euclidean domain is a PID.
Proof. Let I be an ideal, and 0 ≠ b ∈ I have least norm. If a ∈ I, then a = qb + r. By minimality of the norm of b, then N (a) = 0 , and thus a = qb. That is, any element of the ideal is a multiple of b, so I = (b). Thus, every ideal is principal.
Proposition 3. In a PID, any nonzero prime ideal is maximal.
Proof. Let (p) be a prime ideal, with p ≠ 0. Let (p) ⊆ I ⊆ R, some ideal I. Then I = (m) for some m ∈ R. Thus, p = rm for some r ∈ R. Since (p) is prime, one of r or m must be in (p). If m ∈ (p), then (p) = (m), since (p) ⊆ (m) and (m) ⊆ (p), and thus (p) is maximal. If r ∈ (p), then r = sp for some s. Then p = rm = spm. Since R is an integral domain, then sm = 1 , so m is a unit (has an inverse). Thus I = (m) = ( 1 ) is the entire ring, so (p) is maximal.
Corollary 3. For R a commutative ring, then R[x] is a PID if and only if R is a field.
Proof. We’ve already shown that if R is a field, then R[x] is an integral domain. For the other direction, suppose R[x] is a PID. Since R[x] is an integral domain and R ⊂ R[x], then R is an integral domain. Since R[x]~(x) ≃ R and R is an integral domain, then (x) is prime. Since every prime ideal in a PID is maximal, then (x) is maximal, and R[x]~(x) ≃ R is a field, as desired.
- this product is unique up to associates. That is, if r = p 1 ⋯pn = q 1 ⋯qm, then n = m and after reordering, then pi and qi are associates.
Exercise 3.24. In a UFD, an element is prime if and only if it is irreducible.
Theorem 3. Any principal ideal domain is a unique factorization domain.
Proof. We use that in a PID, and element is prime if and only if it is irreducible. We first prove uniqueness of factorization. Suppose p 1 ⋯pn = q 1 ⋯qm, for irreducibles pi, qi. Then the pi, qi are prime, so p 1 must divides some qj , and without loss, say it divides q 1. Since q 1 is irreducible, then q 1 = up 1 for some unit u. Continue this reasoning for p 2 ⋯pn = q 2 ⋯qm. To prove existence of factorization, suppose that I 1 ⊆ I 2 ⊆ I 3 ⊆ ⋯ is a chain of ideals in a PID. Then for some n, In = Im for all m ≥ n. To see why this is true, consider I = (^) ⋃n In. Since we are in a PID, then I = (a) for some a. But a ∈ In for some n ∈ N, by definition of I. Then for each m ≥ n, Im ⊇ In, so a ∈ Im, and thus Im = In. Now, let r be an element of our PID. We show that r = r 1 b, for some irreducible r 1. If not, then r = s 1 t 1 , and s 1 = s 2 t 2 , and s 2 = s 3 t 3 , etc. Consider the chain of primes (r) ⊊ (s 1 ) ⊊ (s 2 ) ⊊ ⋯. This contradicts the fact about chains of prime ideals discussed above. Continue this reasoning for b, and we obtain that r is a product of irreducibles.
§4 February 14, 2020 ♡
§4.1 Factorization in Polynomial Rings
Throughout the class today, all rings are commutative, and 0 ≠ 1. We begin with some definitions. The degree of a polynomial p(x) = anxn^ + ⋯ + a 0 ∈ R[x] is the largest m such that am ≠ 0. If p = 0 , then the degree is not defined. A polynomial p is monic if adeg(p) = 1. You have probably used the following proposition before, and indeed it is very important.
Proposition 4.
- If p(x), q(x) ∈ R[x] are nonzero, then
deg(p(x)q(x)) = deg(p(x)) + deg(q(x)).
- The units of R[x] are the units of R.
- The ring R[x] is an integral domain
- If I is an ideal in R, then (I) ⊆ R[x] is an ideal of R[x], and is the set of polynomials with coefficients from I. Also, R[x]~I ≃ (R~I)[x]. In particular, if I is prime, then (I) is prime.
Exercise 4.2. Prove proposition 4.1.
We will prove later that R is a unique factorization domain if and only if R[x] is. One direction is easy, and you should do it as an exercise.
Example 4. The polynomial ring Z[x] is a UFD. For example, x^2 − 1 = (x − 1 )(x + 1 ). But Q[x] is not a UFD, since for example x^2 − 1 = (x − 1 )(x + 1 ) = ( 12 x − 12 )( 2 x + 2 ).
In the above example, we were able to factor a polynomial in Q[x] and in the ring Z[x]. We can ask the question in general: if R is a UFD, F its field of fractions, and p(x) is a polynomial which is reducible in F [x], is it also reducible in R[x]. The answer to this is the content of Gauss’ Lemma.
Lemma 4.4 (Gauss’ Lemma) Let R be a UFD, and F the field of fractions of R. Let p(x) be a polynomial which is reducible in F [x]. Then p(x) is also reducible in R[x]. Moreover, if p(x) = A(x)B(x), where A, B ∈ F [x], then p(x) = a(x)b(x), where a, b ∈ R[x] and A(x) = ra(x), B(x) = sb(x), for some constants r, s. That is, the factorization in R[x] is the same as the factorization in F [x], except for possibly multiplication by some constant factors.
It might be helpful in the proof below (and in general when dealing with polynomial rings) to keep in mind the example of R = Z and F = Q.
Proof. Say p(x) = A(x)B(x), where A, B ∈ F (x). Let d ∈ R be a common denominator of all the coefficients of A and B. Then dp(x) = a 1 (x)b 1 (x), where now a 1 , b 1 ∈ R[x]. We are not quite done yet, since the left hand side, dp(x), contains a factor of d, and we want just p(x). To accomplish this, we use the fact that we are working in a unique factorization domain to write d = p 1 ⋯pn (assuming d is not a unit). Let’s divide both sides of dp(x) = a 1 (x)b 1 (x) by p 1. That is, we reduce this expression modulo the ideal (p 1 ), to get 0 = a¯ 1 (x) b¯ 1 (x).
Since (p 1 ) is prime, then R[x]~(p 1 ) is an integral domain, which means either a 1 = 0 or b 1 = 0. Thus, p 1 divides a 1 or b 1. Thus, after dividing both sides of dp(x) = a 1 (x)b 1 (x) by p 1 , the right hand side will still have coefficients in R. Continuing similarly for p 2 ,... , pn, we end up with p(x) = an(x)bn(x), where an and bn have coefficients in R, as desired.
For the converse, we can ask: if p is reducible in R[x], is it also reducible in F [x]? This might seem to be obviously true because R ⊆ F , but it’s actually not true, due to a subtlety in the definitions. As an example, 7 x ∈ Z[x] is reducible, since 7 x = 7 ⋅ x. However, in Q[x], it is irreducible, since 7 is a unit in Q. So the converse of the above is not true. However, if we impose some additional hypotheses, we can get a useful converses.
Corollary 4.11 (Small degree reducibility check) If p(x) ∈ F [x] has degree 2 or 3 , then p(x) is reducible if and only if p has a root in F.
Proof. We know p(x) is reducible if and only if p has a linear factor. Then apply corollary 4.10.
Example 4. In Q[x], then (x^2 + 1 )(x^2 + 1 ) = x^4 + 2 x^2 + 1. Since x^2 + 1 is irreducible, this tells us that x^4 + 2 x^2 + 1 has no roots in Q.
Proposition 4.13 (Rational Roots Theorem) If p(x) = anxn^ + ⋯ + a 0 ∈ Z[x] is a polynomial with a root of the form rs ∈ Q, and gcd(r, s) = 1 , then s divides an and r divides a 0.
Proof. Since p( rs ) = 0 , then an( rs )n^ + ⋯ + a 0 = 0 , which implies
anrn^ + an− 1 srn−^1 + ⋯ + sna 0 = 0 ,
by multiplying both sides by sn. So
−anrn^ = s(an− 1 rn−^1 + ⋯ + sn−^1 a 0 ),
and similarly, −a 0 sn^ = r(anrn−^1 + ⋯ + a 1 ).
Because s and r have no common factors, then s divides an, and r divides a 0.
Example 4. Let p(x) = x^2 − 2 , and suppose that rs is a root. Then we must have s = ± 1 and r = ± 1 , ± 2. Checking each of these, we see that none of these are a root, and therefore x^2 − 2 has no roots in the rationals. Since
2 is a root of x^2 − 2 , we have another proof that
2 is irrational.
More generally, x^2 − p, x^3 − p are irreducible in Q[x] for any prime p. Now, for higher degree polynomials, Eisenstein’s criterion will be very useful.
Proposition 4.15 (Eisenstein’s Criterion) let R be a UFD, and let P be a prime ideal in R. Let f (x) = anxn^ + an− 1 xn−^1 + ⋯ + a 0 , n > 0 be a polynomial such that an− 1 , an− 2 ,... , a 0 ∈ P , but a 0 ~∈ P 2 and an ~∈ P. Then f is irreducible in R[x].
Proof. Suppose that f is reducible. By Gauss lemma, there exists a factorization f (x) = A(x)B(x) in R[x]. Since the leading coefficient of f is not in P , then reducing modulo P , we have anxn^ = A¯(x) B¯(x).
Since P is prime, then R~P is an integral domain. The only way that two polynomials over an integral domain can multiply to anxn^ is if each of the polynomials A¯(x) and B^ ¯(x) is in fact a monomial. However, the degrees A¯ and B¯ must be positive, which means the constant terms of A¯ and B¯ are zero (in (R~P )[x]). The only way for this to be true is if the constant coefficients of A and B are both in P to begin with. But this implies that the constant term of A(x)B(x) = f (x) is in P 2 , a contradiction. Thus, f is irreducible.
Exercise 4.16. In the above proof of Eisenstein’s criterion, where did we use the fact that R is a UFD?
Eisenstein’s criterion is very useful for showing that polynomials are irreducible, as the following example shows.
Example 4. The polynomial x^4 + 1 is irreducible in Q[x]. To show this, first note that by making the change of variable x → x + 1 , it is equivalent to show that (x + 1 )^4 + 1 is irreducible. But (x + 1 )^4 + 1 = x^4 + 4 x^3 + 6 x^2 + 4 x + 2 , which satisfies Eisenstein’s criterion, with P = ( 2 ), and is therefore irreducible. So x^4 + 1 is irreducible. This same trick can be used to show that cyclotomic polynomials of prime degree are irreducible. Note: P = ( 2 ) is in Z, not in Q, and then the fact that the polynomial is irreducible in Z[x] implies it is irreducible in Q[x], by the contrapositive to Gauss’ Lemma.
§5 February 19, 2020
§5.1 Modules
You are likely familiar with vector spaces over a field. The structure of a module over a ring is very similar, except that the field is replaced with a ring, and there are some slight changes in the definitions.
Definition 5.1 — Let R be a ring. A (left) R -module is an Abelian group (M, +), together with a map ψ ∶ R × M → M , which satisfies the following properties (where we write rm of r ⋅ m for ψ(r, m)):
- (r 1 + r 2 )m = (r 1 m) + (r 2 m), where r 1 , r 2 ∈ R and m ∈ M ,
- r(m + n) = rm + rn, where r ∈ R and m, n ∈ M ,
- r(sm) = (rs)m, for r, s ∈ R and m ∈ M , and
- 1 m = m, for any m ∈ M.
Example 5. If R = F is a field, then an F -module is the same as a vector space over F.
