Math 150C ­ Algebra, Spring 2015

27 Pages·2015·1.02 MB·English

by  Brian Osserman

Math 150C ­ Algebra
Spring 2015
Instructor: Brian Osserman
Lectures: MWF 10:00­10:50, Olson 223.
CRN: 39518
Office: MSB 3218, e­mail:
Office Hours: T 2:10­3:00, W 11:00­11:50
Prerequisites: Math 150AB
Textbook: Michael Artin, Algebra (2nd edition)
Syllabus: The main topics for the quarter will be ring theory and field theory. Note
that, in coordination with this year's 150B, this deviates from the usual department
syllabus. The ring theory is roughly weeks 6­10 of the 150B department syllabus,
while the field theory is roughly weeks 5­10 of the 150C department syllabus.
TA: Nathaniel Gallup
Discussion: R 7:10­8:00, Cruess 107
TA Office Hours: M 3:10­4:00, R 4:10­5:00, MSB 3217
Grading: 30% homework, 25% midterm, 45% final exam
Homework: Assigned weekly, due each Friday in class
Welcome to Math 150C: Algebra
Algebra is a core branch of mathematics. It is important in and of itself, and also
to a wide range of other fields, including number theory, algebraic geometry, and
algebraic topology. In addition, it has applications ranging from cryptography to
crystallography. We will spend the first half of the quarter on the theory of rings,
including factorization and modules. In the second half of the quarter, we will
discuss fields and field extensions, with applications to topics such as ruler and
compass constructions.
Lecture notes
I will post notes here to supplement Artin as seems appropriate.
Summary of rings material from 150B: this is a summary of the material on
rings which I expect you to already be familiar with from the end of 150B.
Unique factorization domains: this is a presentation of the material from
Artin's Section 12.3, rephrased for polynomials over general unique
factorization domains (rather than just over the integers).
Review of definitions for midterm: A summary of all the definitions you will
be expected to know for the midterm, as well as some of the basic results
relating the definitions to one another.
Splitting fields: this is the presentation of splitting fields as given in lecture.
Finite fields: this is the presentation of finite fields as given in lecture.
Galois theory: this is the presentation of Galois theory as given in lecture. It
will be updated further as more is covered.
Problem sets
Problem sets will be posted here each Friday, due the following Friday in class.
You are encouraged to collaborate with other students, as long as you make sure
you understand your answers and they are in your own words. You are not, under
any circumstances, allowed to get answers to problems from any outside sources.
A selection of problems will be graded from each problem set, and some points
will be assigned based on the number of problems completed. To minimize
resulting randomness of scores, your lowest problem set score will be dropped
when calculating your grade.
Problem set #0, "due" 4/3: do Exercises 1.6 (a), 2.1, 3.2, and 4.1 of Chapter
11. Also do Exercises 3.3 and 3.4 if you didn't do them last quarter. 
Note: these are suggested review problems for last quarter's material, and
are not actually to be turned in.
Problem set #1, due 4/10: do Exercises 11.1.3, 11.3.5(b) (take as the
definition of α being a multiple root that f is a multiple of (x­α)2), 11.3.9,
11.3.11 (prove or disprove each direction separately), 11.4.2, 11.5.3 and
11.5.5. 
Grading: 10 points for 11.3.5(b), 10 points for 11.5.3, 10 points for
completeness of remaining problems.
Problem set #2, due 4/17: do Exercises 11.5.6, 11.5.7, 11.7.1, 11.7.2, 11.7.3,
11.8.1 and 11.8.3. 
Grading: 10 points for 11.5.6, 6 points for 11.7.1, 6 points for 11.7.2, and 8
points for completeness of remaining problems.
Problem set #3, due 4/24: do Exercises 12.2.1, 12.2.3, 12.2.5, 12.2.6,
12.3.1, 12.3.2 and 12.3.4. 
Grading: 10 points for 12.2.1, 10 points for 12.3.1, 10 points for
completeness of remaining problems.
Problem set #4, due 5/1: do Exercises 12.3.6, 12.5.1, 12.5.5, 13.1.3, 13.2.1,
13.4.3, and 13.1.2. 
Grading: 10 points for 12.3.6, 10 points for 13.2.1, 10 points for
completeness of remaining problems.
Problem set #5, due 5/8: do Exercises 13.1.4, 13.6.6, 15.1.1, 15.2.2, 15.3.1,
15.3.7, and 15.3.10. 
Grading: 10 points for 15.2.2, 10 points for 15.3.7, 10 points for
completeness of remaining problems.
Problem set #6, due 5/15: do Exercises 15.1.2, 15.2.1, 15.2.3, 15.3.2,
15.3.6, 15.3.9, and 15.6.1. 
Grading: 10 points for 15.2.3, 10 points for 15.3.2, 10 points for
completeness of remaining problems.
Problem set #7, due 5/22: do Exercises 16.3.1, 16.3.2, 16.3.3, 15.M.4,
15.7.1, 15.7.10 (hint: show that if F has pr elements, then the Frobenius
map sending each element to its pth power is surjective). 
Grading: 10 points for 16.3.2, 10 points for 15.7.10, 10 points for
completeness of remaining problems.
Problem set #8, due 5/29: do Exercises 15.7.5, 15.7.7, 15.7.9, 15.M.1,
16.4.1 (the first part of (a) was already done in class), 16.6.2, and 16.7.1. 
Grading: 6 points for 15.7.7, 8 points for 16.4.1, 8 points for 16.7.1, and 8
points for completeness of remaining problems.
Problem set #9, due 6/5 (you may turn in the assignment in the reader box
marked for the class on the first floor of MSB): do Exercises 16.5.1 (you
should assume these automorphisms keep C fixed, otherwise they are not
uniquely determined by Artin's description), 16.5.3, 16.6.3, 16.7.2, 16.7.6
and 16.7.8. 
Grading: 10 points for 16.5.1, 10 points for 16.7.6, and 10 points for
completeness of remaining problems.
Exams
There will be one in­class midterm exam, on Wednesday, May 6. It will cover all
material up through and including the lecture on May 1, corresponding to the first
five problem sets.
The final exam is scheduled for Monday, June 8 6:00­8:00 PM.
Anonymous Feedback
If you have any feedback on the course, regarding lecture, discussion section,
homework, or any other topic, you can provide it anonymously with the below
form.
RINGS: SUMMARY OF MATERIAL FROM LAST QUARTER
BRIAN OSSERMAN
ThisisasummaryofthematerialonringsIexpectyoutoknowfromtheendof150B,consisting
of most of 11.1-11.4 of Artin. This does not include the new material (also from these sections)
which we have covered in the first week of 150C, so make sure to study both in preparation for
homework and exams.
In order to highlight some basic and/or important facts, I have included them as propositions
below even when Artin has buried them in running text.
1. Definitions
From §11.1.
Definition. A ring R is a set with two binary operations +,× called addition and multiplication,
that satisfy:
(1) R is an abelian group under the operation +.
(2) Multiplication is commutative and associative, and has an identity denoted 1.
(3) For all a,b,c ∈, we have (a+b)c = ac+bc.
Definition. IfRisaring,asubringofRisasubsetwhichisclosedunderaddition,multiplication,
and subtraction, and which contains 1.
Definition. An element a ∈ R is a unit if it has a multiplicative inverse. We denote the set of
units of R by R×.
From §11.2.
Definition. Let R be a ring. Given a formal symbol x, a polynomial in x with coefficients in R
is a finite formal sum
f(x) = a xn+a xn−1+···+a x+a ,
n n−1 1 0
where each a ∈ R. The a are called the coefficients of f(x). Polynomials of the form xi are
i i
called monomials.
The polynomial ring R[x] is the set of all polynomials in x with coefficients in R, with the
usual rules for polynomial addition and multiplication: if f(x) is as above, and
g(x) = b xm+b xm−1+···+b x+b ,
m m−1 1 0
then
max{m,n}
(cid:88)
f(x)+g(x) = (a +b )xk,
k k
k=0
where we set a = 0 if k > n, and b = 0 if k > m, and
k k
 
m+n
(cid:88) (cid:88)
f(x)·g(x) =  aibjxk,
k=0 i+j=k
where we restrict i,j to be nonnegative and at most n,m respectively.
1
Definition. If f(x) = a xn +···+a ∈ R[x] is not equal to 0, let i be maximal so that a (cid:54)= 0.
n 0 i
Then the degree of f(x) is equal to i, and the leading coefficient of f is equal to a .
i
We say f(x) is constant if it is equal to 0, or has degree 0.
Alsofamiliarizeyourselfwiththemultivariablecounterpartsoftheabovedefinitions,asdeveloped
at the end of §11.2 of Artin.
Definition. A polynomial f(x) is monic if (it is not 0 and) its leading term is 1.
From §11.3.
Definition. A map ϕ : R → R(cid:48) is a homomorphism if ϕ(1) = 1, and for all a,b ∈ R, we have
ϕ(a+b) = ϕ(a)+ϕ(b) and ϕ(ab) = ϕ(a)ϕ(b).
Definition. A homomorphism is an isomorphism if it is bijective.
Definition. Given ϕ : R → R(cid:48) a homomorphism, and α ∈ R(cid:48), the homomorphism R[x] → R(cid:48)
mapping x to some α and mapping r ∈ R to ϕ(r) is called the evaluation homomorphism
determined by α. If f(x) is a polynomial, we write f(α) for the image of f(x) under the homomor-
phism corresponding to α.
Definition. Given a homomorphism ϕ : R → R(cid:48), the kernel is {α ∈ R : ϕ(α) = 0}.
Definition. An ideal of a ring R is a nonempty subset I such that:
(1) For all a,b ∈ I, we have a+b ∈ I.
(2) For all a ∈ R and b ∈ I, we have ab ∈ I.
Definition. If a ∈ R, we have the ideal consisting of the multiples of a. This is denoted (a), or
aR. Ideals of this form are called principal ideals.
The ideal which is all of R is called the unit ideal.
The principal ideal (0) contains only 0, and is called the zero ideal.
An ideal is called proper if it is not the unit ideal or the zero ideal.
Definition. Given a ,...,a ∈ R, the ideal generated by a ,...,a is the smallest ideal of R
1 m 1 m
containing all the a .
i
The above definition makes sense also for infinite sets of generators.
Definition. The characteristic of a ring R is the smallest positive integer n such that
1+···+1n times = 0
(cid:124) (cid:123)(cid:122) (cid:125)
in R, or if no such n exists, is equal to 0.
From §11.4.
Definition. Given a ring R and an ideal I, the quotient ring R/I is the ring structure on the
set of cosets of I in R induced by addition and multiplication in R.
Results.
From §11.1.
Proposition. If R is a ring, and a ∈ R, then a·0 = 0.
0 = 1 in R if and only if R = {0}.
Proposition. If R is a ring, and a ∈ R, if there exists b ∈ R such that ab = 1, then b is the unique
element of R with this property.
2
From §11.2.
Proposition. IfRisaring,thenR[x]isaring,andRisimbeddedasthesubringofR[x]consisting
of constant polynomials.
From §11.3.
Proposition. Let R,R(cid:48) be rings, and ϕ : R → R(cid:48) a homomorphism.
(1) For any α ∈ R(cid:48), there is a unique homomorphism Φ : R[x] → R(cid:48) that is equal to ϕ on the
constant polynomials, and sends x to α. It is given by the formula
a xn+···+a x+a (cid:55)→ a αn+···+a α+a .
n 1 0 n 1 0
(2) Moregenerally,foranyα ,...,α ∈ R(cid:48),thereisauniquehomomorphismΦ : R[x ,...,x ] →
1 n 1 n
R(cid:48) that is equal to ϕ on the constant polynomials, and sends x to α for each i = 1,...,n.
i i
It is given by the formula
(cid:88) (cid:88)
a xi1···xin (cid:55)→ a(i ,...,i )αi1···αin.
(i1,...,in) 1 n 1 n 1 n
(i1,...,in) (i1,...,in)
Proposition. Givenvariablesx ,...,x andy ,...,y andaringR,thereisauniqueisomorphism
1 m 1 n
(R[x ,...,x ])[y ,...,y ] → R[x ,...,x ,y ,...,y ]
1 m 1 n 1 m 1 n
which is the identity on R and sends the variables to themselves.
Proposition. The kernel of a ring homomorphism ϕ : R → R(cid:48) is an ideal of R.
Proposition. A ring R is a field if and only if it has no proper ideals.
Proposition. TheidealsofZarepreciselynZforn (cid:62) 0. Inparticular, everyidealofZisprincipal.
From §11.4.
Theorem. Given an ideal I of a ring R, the quotient group R/I has a ring structure induced by
multiplication in R. The canonical map π : R → R/I is a ring homomorphism, and its kernel is I.
Theorem. Let ϕ : R → R(cid:48) be a ring homomorphism with kernel K, and let I be another ideal of
R. Let π : R → R/I be the canonical map.
(1) If I ⊆ K, there is a unique homomorphism ϕ¯: R/I → R(cid:48) such that ϕ¯◦π = ϕ.
(2) (First Isomorphism Theorem) If ϕ is surjective and I = K, then ϕ¯ is an isomorphism.
Theorem (Correspondence theorem). Let ϕ : R → R(cid:48) be a surjective ring homomorphism with
kernel K. Then there is a bijective correspondence between ideals of R(cid:48), and ideals of R containing
K. This bijection is induced simply by taking images and preimages of ideals under ϕ.
In addition, if I ⊆ R is an ideal containing K, and I(cid:48) its image in R(cid:48), then ϕ induces an
isomorphism R/I →∼ R(cid:48)/I(cid:48).
3
UNIQUE FACTORIZATION DOMAINS
BRIAN OSSERMAN
In this note, we prove the following theorem:
Theorem. If R is a UFD, then R[x] is also a UFD.
We follow the definitions and arguments in §12.3 (Gauss’s Lemma) of Artin Algebra, generalizing
from the case R = Z which is given in the book.
Definition. Let R be a UFD, and f(x) ∈ R[x]. We say f(x) is primitive if there is no irreducible
element p ∈ R which divides f(x).
Since R has factorization into irreducibles, the definition is equivalent to requiring that no non-
unit x ∈ R divides f(x). Also, recall that since R is a UFD, irreducible is the same as prime for
elements of R.
The basic tools we will use are reduction modulo p, and inclusion into the field of fractions.
The first statement we want is the following:
Lemma (Gauss). Let p ∈ R be irreducible, and suppose for some f(x),g(x) ∈ R[x], the product
f(x)g(x) is a multiple of p. Then either f(x) or g(x) is a multiple of p.
In particular, the product of primitive polynomials is primitive.
Note that the first statement of the lemma is just saying that p, which we already know is prime
in R, is also prime in R[x].
Proof. The second statement follows immediately from the first, if f(x) and g(x) are the primitive
polynomials in question.
The proof of the first statement uses reduction modulo p. Given f(x),g(x) ∈ R[x], suppose that
p divides f(x)g(x). Now, since p is prime, R/(p) is an integral domain, and them so is (R/(p))[x].
We have a natural homomorphism
R[x] → (R/(p))[x]
obtained by reducing each coefficient modulo p, and it is clear that a polynomial in R[x] is in the
kernel of this homomorphism if and only if it is a multiple of p. But now we are done: if f¯(x) and
g¯(x) are the images of f(x) and g(x), then f¯(x)g¯(x) is the image of f(x)g(x), so by hypothesis,
f¯(x)g¯(x) = 0. Since (R/(p))[x] is an integral domain, we must have either f¯(x) = 0 or g¯(x) = 0,
which means that one of f(x) nor g(x) must be a multiple of p, as desired. (cid:3)
Next, let K be the fraction field of R. Our next result is the following:
Lemma. If f(x) ∈ K[x] has positive degree, it can be written as cf (x), where c ∈ K, and
0
f (x) ∈ R[x] is primitive. Moreover, this is unique up to replacing c by cu and f (x) by u−1f (x),
0 0 0
for u ∈ R×.
Proof. To check existence, first we can multiply f(x) by the product of the denominators of the
coefficients to get a polynomial with coefficients in R. We can then factor the coefficients into
irreducibles, and remove any common factors to obtain f (x). The first step gives the denominator
0
of c, while the second step gives the numerator. In this way, we write f(x) = cf (x), as desired.
0
To see uniqueness, suppose also f(x) = dg (x), where d ∈ K and g (x) ∈ R[x] is primitive. Then
0 0
we have cf (x) = g (x), so we want to show that for any two primitive polynomials f (x),g (x) in
d 0 0 0 0
1
R[x], if αf (x) = g (x) for some α ∈ K, then we must have α ∈ R×. Write α = a for a,b ∈ R, and
0 0 b
assume that a,b have no common factors. We will prove that for any irreducible p ∈ R, neither a
nor b can be a multiple of p. Multiplying our equation through by b, we have af (x) = bg (x). If
0 0
p divides a, then p doesn’t divide b, since we assume a,b have no common factors. But p divides
bg (x), so p divides every coefficient of bg (x). Since R is a UFD, we conclude that p must divide
0 0
every coefficient of g (x), contradicting that g (x) is primitive. Thus, p cannot divide a. Similarly
0 0
p cannot divide b, so we conclude that a and b, and hence also α, are units, as desired. (cid:3)
Putting the lemmas together, with some additional work we obtain the desired result.
Proof of the Theorem. First, we verify that factorizations into irreducibles exist in R[x]. This is
more or less clear: certainly, each time we factor out a positive-degree polynomial, the degree goes
down, so this can only happen finitely many times. On the other hand, since factorizations into
irreducibles exist in R, if we first factor out the common irreducible factors of the coefficients of a
polynomial, then there are no further constant factors to remove, so the only further factorizations
have to involve polynomials of positive degree.
Knowing that factorizations into irreducibles exist, in order to prove that R[x] is a UFD, by the
lemma we proved earlier (in the process of showing that every PID is a UFD) we just need to prove
that irreducible elements are prime. Thus, let f(x) be an irreducible element of R[x]. If f(x) is
constant, then it is prime by the first statement of Gauss’ Lemma. If f(x) is nonconstant, then
it clearly must be primitive, and we claim that in fact, it must be irreducible in K[x]. Suppose
we have written f(x) = g(x)h(x) for g(x),h(x) ∈ K[x] nonconstant. By the previous lemma, we
can write g(x) = cg (x) and h(x) = dh (x) for c,d ∈ K and g (x),h (x) ∈ R[x] primitive. Then
0 0 0 0
f(x) = cd(g (x)h (x)). By Gauss’ Lemma, g (x)h (x) is primitive, so by the uniqueness in the
0 0 0 0
previous lemma, we conclude that f(x) is a unit times g (x)h (x), so we get a factorization of f(x)
0 0
in R[x], contradicting irreducibility.
Now, since K[x] is a PID, we know that f(x) is prime in K[x]. It then suffices to prove that if a
primitive polynomial f(x) ∈ R[x] is prime in K[x], then it is prime in R[x]. Suppose f(x) divides
g(x)h(x) for some g(x),h(x) ∈ R[x]. Then since f(x) is prime in K[x], it must divide either g(x)
or h(x) in K[x]. Without loss of generality, suppose that it divides g(x), so that g(x) = f(x)q(x)
for some q(x) ∈ K[x]. We will prove that q(x) ∈ R[x], so that f(x) divides g(x) in R[x], and f(x)
is prime in R[x], which will complete the proof of the theorem. By the previous lemma, we can
write g(x) = cg (x), and q(x) = dq (x), where c,d ∈ K and g (x),q (x) ∈ R[x] are primitive. Note
0 0 0 0
that since g(x) ∈ R[x], we must have c ∈ R, as otherwise the primitivity of g (x) would create
0
denominators in g(x). Then cg (x) = df(x)q (x), and f(x)q (x) is primitive in R[x] by Gauss’
0 0 0
Lemma, so by the uniqueness in the previous lemma, for some unit u ∈ R we have d = cu. But
then d ∈ R, so q(x) = dq (x) ∈ R[x], as desired. (cid:3)
0
Note that the proof of the theorem also gives a description of the irreducible elements of R[x]:
they consist of constants which are irreducible in R, and nonconstant primitive polynomials which
are irreducible in K[x].
Applying the theorem inductively, we conclude that
Corollary. Z[x ,...,x ] is a UFD, and F[x ,...,x ] is a UFD for any field F.
1 n 1 n
Example. Z[x] is not a PID: for instance, the ideal (2,x) (which is the ideal of polynomials with
evenconstantterm)isnotthesetofmultiplesofanyoneelement. Similarly, ifF isafield, F[x ,x ]
1 2
is not a PID: the ideal (x ,x ) is not principal. Thus, we now have many examples of UFDs which
1 2
are not PIDs.
2
SUMMARY OF DEFINITIONS FOR MIDTERM
BRIAN OSSERMAN
This includes all the definitions you will be expected to know, as well as the fundamental results
relating them. However, it does not include a comprehensive summary of the results we have
covered – you should be sure to look through your notes and/or Artin for these as well as for
examples.
Rather than follow the order covered in class, I have grouped the definitions by subject, while
still keeping them in a logically consistent order.
Basic definitions relating to rings
Definition. A ring R is a set with operations addition and multiplication, which are commutative
and associative, and which together are distributive. Both operations must have identity elements,
and every element must have an additive inverse.
Definition. The characteristic of a ring R is the smallest positive integer n such that
1+···+1 = 0
(cid:124) (cid:123)(cid:122) (cid:125)
ntimes
in R, or if no such n exists, the characteristic of R is defined to be 0.
Definition. A nonzero ring R is an integral domain if for all nonzero b,c ∈ R, the product bc is
also nonzero.
Definition. IfRisaring,asubringofRisasubsetwhichisclosedunderaddition,multiplication,
and subtraction, and which contains 1.
Definition. If R is a ring, and R(cid:48) ⊆ R a subring, and g ,...,g ∈ R, the ring generated by the
1 n
g over R(cid:48), denoted R(cid:48)[g ,...,g ] is the smallest subring of R which contains R(cid:48) together with the
i 1 n
g .
i
Definition. An element a ∈ R is a unit if it has a multiplicative inverse.
Definition. We will say a divides b in R, and write a|b, if b is a multiple of a in R.
Definition. An integral domain R is a Euclidean domain if there exists a ‘size’ function
σ : R(cid:114){0} → Z(cid:62)0
such that we can carry out division with remainder with respect to σ. That is, for any a,b ∈ R
with b (cid:54)= 0, there exist q,r ∈ R such that
a = bq+r
and either r = 0 or σ(r) < σ(b).
Definition. Let z be an element of an integral domain R. We say z is irreducible if it is not a
unit, and for any factorization z = xy in R, either x or y must be a unit. We say z is prime if z
is not a unit, and if for all x,y ∈ R such that z divides xy, then z divides x or y.
Definition. We say elements x,y ∈ R are associates if x = yu for some unit u ∈ R×.
1
Definition. We say that (an integral domain) R is a unique factorization domain (or UFD) if
every nonzero, non-unit z ∈ R can be written as a product p ···p of irreducible elements, and if
1 n
further this factorization is unique, in the sense that if p(cid:48) ···p(cid:48) is another factorization of z, then
1 m
m = n, and the p(cid:48) can be reordered so that each p(cid:48) is an associate of p .
i i i
Definitions relating to polynomials
Definition. Let R be a ring. Given a formal symbol x, a polynomial in x with coefficients in R
is a finite formal sum
f(x) = a xn+a xn−1+···+a x+a ,
n n−1 1 0
where each a ∈ R. The a are called the coefficients of f(x). Polynomials of the form xi are
i i
called monomials.
The polynomial ring R[x] is the set of all polynomials in x with coefficients in R, with the
usual rules for polynomial addition and multiplication: if f(x) is as above, and
g(x) = b xm+b xm−1+···+b x+b ,
m m−1 1 0
then
max{m,n}
(cid:88)
f(x)+g(x) = (a +b )xk,
k k
k=0
where we set a = 0 if k > n, and b = 0 if k > m, and
k k
 
m+n
(cid:88) (cid:88)
f(x)·g(x) =  aibjxk,
k=0 i+j=k
where we restrict i,j to be nonnegative and at most n,m respectively.
Definition. If f(x) = a xn +···+a ∈ R[x] is not equal to 0, let i be maximal so that a (cid:54)= 0.
n 0 i
Then the degree of f(x) is equal to i, and the leading coefficient of f is equal to a . We say
i
that f(x) is monic if the leading coefficient is equal to 1.
We say f(x) is constant if it is equal to 0, or has degree 0.
Definition. Let R be a UFD, and f(x) ∈ R[x]. We say f(x) is primitive if there is no irreducible
element p ∈ R which divides f(x).
Definitions relating to homomorphisms
Definition. A map ϕ : R → R(cid:48) is a homomorphism if ϕ(1) = 1, and for all a,b ∈ R, we have
ϕ(a+b) = ϕ(a)+ϕ(b) and ϕ(ab) = ϕ(a)ϕ(b).
Definition. A homomorphism is an isomorphism if it is bijective.
Definition. Given a homomorphism ϕ : R → R(cid:48), the kernel is {α ∈ R : ϕ(α) = 0}.
Definition. Given ϕ : R → R(cid:48) a homomorphism, and α ∈ R(cid:48), the homomorphism R[x] → R(cid:48)
mapping x to some α and mapping r ∈ R to ϕ(r) is called the evaluation homomorphism
determined by α. If f(x) is a polynomial, we write f(α) for the image of f(x) under the homomor-
phism corresponding to α.
We say that α ∈ R(cid:48) is a root of f(x) if f(α) = 0.
2