Download Ma 5b Midterm Review Notes

Ma 5b Midterm Review Notes
February 4, 2017
1
Rings.
Throughout this section, let R denote a commutative ring.
Recall. An ideal P ⊂ R is prime if it is proper and its complement is closed under multiplication, i.e. for any ab ∈ P, either a ∈ P or b ∈ P. An ideal is maximal if it is proper and
is not properly contained in any other proper ideal. Let I ⊂ R be an ideal. I is prime ⇔
R/I is an integral domain; I is maximal ⇔ R/I is a field. Hence a maximal ideal is prime.
The converse is not true in general. The zero ideal is prime if and only if R is an integral
domain.
Proposition. The nilradical of R, denoted N (R), is equal to the intersection of all prime
ideals of R.
Proof. Let A denote the intersection of prime ideals of R. By a homework problem (7.4.26.
in Dummit and Foote), N (R) ⊂ A. Suppose there exists some element a ∈ A which is not
in N (R). We use Zorn’s lemma to show that there exists some prime ideal that does not
contain a. Let S be the set of ideals of R that do not contain any element of the form an .
Since (0) ∈ S, S 6= ∅. We order S by containment, then by Zorn’s Lemma, S has a maximal
element I. If I is not a prime ideal, then there exist elements x, y ∈
/ I for which xy ∈ I.
Then the set of elements z for which xz ∈ I is an ideal that properly contains I; it therefore
contains an for some integer n. Now the set of elements z for which an z ∈ I is an ideal
that properly contains I, so this set contains am for some integer m, which implies am+n ∈ I
but this is impossible. Thus I is a prime ideal that does not contain a, contradicting the
definition of a.
Proposition. The following are equivalent:
• R is a field;
1
• the only ideals in R are (0) and (1);
• every nonzero ring homomorphism from R into a nonzero ring S is injective.
Definition. A ring R with exactly one maximal ideal m is called a local ring. The field
k = R/m is called the residue field of R.
Proposition. (1) Suppose m is a proper ideal of R such that every x ∈ R − m is a unit in
R, then R is a local ring and m is its maximal ideal; (2) Suppose m is a maximal ideal of R
such that every element of 1 + m is a unit in R, then R Is a local ring.
Proof. (a) Every proper ideal consists of non-units and is therefore contained in m. Thus m
is the only maximal ideal of R.
(b) Let x ∈ R − m. Since m is maximal, the ideal generated by x and m is (1), so there exists
y ∈ and z ∈ m such that xy + z = 1. Thus xy = 1 − z ∈ 1 + m and therefore is a unit. By
part (1) we conclude R is a local ring.
2
ED, PID, UFD and Polynomial Rings.
One has the following class inclusions: commutative rings ⊃ integral domains ⊃ unique
factorization domains ⊃ principal ideal domains ⊃ Euclidean domains ⊃ fields ⊃ finite
fields.
Recall the following definitions:
(1) A Euclidean domain R is an integral domain which can be endowed with at least on e
Eculidean function. A Euclidean function (also called “degree function”, “norm function”
etc.) on R is a function f : R − {0} → Z≥0 such that for any a, b ∈ R with b 6= 0, there exist
q, r ∈ R such that a = bq + r where either r = 0 or f (r) < f (b).
(2)A principal ideal domain is an integral domain where every ideal is principal.
(3)A unique factorization domain R is an integral domain in which every non-zero element
a ∈ R can be written as a product of irreducible elements pi of R and a unit u : x = up1 · · · pn ,
n ≥ 0. (If x is a unit, x is an empty product.)
Examples. (1) Euclidean domains: Z; Z[i]; K[x] where K is a field; K[[x]] where K is a
field.
(2) The ring Q[x, y]/(x2 + 2y 2 + 1) is a UFD, but Q(i)[x, y]/(x2 + 2y 2 + 1) is not.
Examples. (1) Let K be a field and let R = K[x1 , · · · , xn ]. Let f ∈ A be an irreducible
polynomial. By unique factorization, the ideal (f ) is prime. The ideal m of all polynomials
in R with zero constant term is maximal; it is the kernel of the homomorphism R → k which
maps f ∈ R to f (0). If n > 1, m is not a principal ideal and it requires at least n generators.
2
(2) In a PID, a non-zero element is prime if and only if it is irreducible; correspondingly, a
nonzero ideal is prime if and only if it is maximal. (3) Any two elements of a UFD have
a greatest common divisor and a least common multiple. The same holds for finitely many
elements of a UFD. (4) Any UFD is integrally closed, i.e. if R is a UFD and K = F rac(R),
and if k ∈ K is a root of a monic polynomial f [x] ∈ R[x], then k ∈ R. (5) Let S be
a multiplicative closed subset of a UFD R. Then the localization S −1 R is a UFD. (The
idea of localization is to “add multiplicative inverses” to a ring. When S = R − {0},
S −1 R = F rac(R). If S = R − P where P ⊂ R is a prime ideal, then the localization is
denoted RP . Note that RP has a unique maximal ideal.)
For a polynomial ring generated by finitely many elements over a UFD, we have the following
variation of the Eisenstein’s Criterion.
Proposition. Let R be a UFD and let P ⊂ R be a prime ideal. Let f (x) = an xn + · · · +
a1 x + a0 ∈ R[x] for some integer n ≥ 1. Suppose an ∈
/ P, an−1 , · · · , a0 ∈ P and a0 ∈
/ P 2.
Then f (x) is irreducible in F [x] where F is the fraction field of R.
Lemma. Let D be an integral domain. If q, s, t ∈ D[x] are polynomials such that q = st
and q = cxk has only one nonzero term, then both s and t also only have one nonzero term.
In particular, s = axn and t = bxm where k = n + m and c = ab.
Proof of Lemma. Let a1 xn1 (resp. b1 xm1 ) and a2 xn2 (resp. b2 xm2 ) be the nonzero terms of
highest and lowest degree in s (resp. t). Then in q we have cxk = a1 b1 xn1 +m1 = a2 b2 xn2 +m2 ,
so (n1 − n2 ) + (m1 − m2 ) = 0. But n1 − n2 and m1 − m2 are both nonnegative, so both are
zero and we obtain the desired result.
Proof of Proposition. Suppose p(x) is reducible in R[x], then we can write f (x) = s(x)t(x)
for some nonunits s, t ∈ R[x]. Consider the reduction mod P, then f (x) ≡ cn xn ≡ s(x)t(x)
in (R/P )[x]. Since P is prime, R/P is an integral domain, so by the lemma s(x) and t(x)
both have one nonzero term. Suppose both s and t have positive degree. Then in particular
s0 , t0 ∈ P, so a0 = s0 t0 ∈ P 2 , which is a contradiction. Thus one of s and t has degree 0. We
assume s has degree 0, which then implies s(x) also has degree 0. Thus f (x) = s0 t(x) with
x0 ∈ R, and t(x) is irreducible over R.
Now suppose f (x) is reducible in F [x], say f (x) = a(x)b(x). Clearing denominators, we have
df (x) = a0 (x)b0 (x) where a0 , b0 ∈ R[x]. Then ds0 t(x) = a0 (x)b0 (x). Considering degrees, one
of a0 (x), b0 (x) must be constant, say WLOG a0 (x). Then a(x) is constant and thus a unit in
F [x], a contradiction. Therefore f (x) is irreducible in F [x].
3