Math 547
Review Exam #2
Be able to define these terms:
Evaluation homomorphism. Ideal
Kernel of a homomorphism
Maximal Ideal
Prime Ideal
Polynomial over a ring R
Principal ideal
Irreducible polynomial over
a field
Irreducible element of an
integral domain
Prime element of an integral
domain
Principal Ideal Domain
Ring homomorphism
Be able to state and use:
1. If f (x) !Z[x] , then if f(x) factors as the product of two smaller degree
polynomials in Z[x] , then it factors as the product of two polynomials of the
same degrees in ![x] .
2. The division Algorithm for F[x].
3. For a field F, a non-zero polynomial p(x) !F[x] of degree in has at most n zeros.
R
4. If R1 ,!R2 are rings and ! : R1 " R2 is a homomorphism, then 1 ker ! " Image(! ) .
5. The only irreducible polynomials over the reals are the irreducible quadratics.
6. If R1 ,!R2 are rings and ! : R1 " R2 is a homomorphism, and A is a subring of R1 ,
B is an ideal of R1 , then ! (A) is a subring of R2 and if ! is onto, then ! (B) is an
ideal in R2 .
7. If D is an integral domain, then so is D[x] .
8. For a field F, a non-zero polynomial p(x) !F[x] of degree in has at most n zeros.
9. If f (x), g(x) !F[x] , then deg ( f (x)g(x)) = deg ( f (x)) + deg ( g(x)) .
Be able to state, use and prove:
1. If A is a maximal ideal then A is a prime ideal.
2. Every ideal of a ring R is the kernel of some ring homomorphism.
3. The Factor Theorem.
4. A polynomial of degree at most 3 in F[x] is irreducible over F if and only if it has
no zeros in F.
5. The Rational Root Theorem.
6. Eisenstein’s Irreducibility Criterion.
7. Z is a principal ideal domain.
![x]
! ".
< x2 + 1 >
8. Every ideal in a ring R is the kernel of some homomorphism.
Be able to do problems assigned in class, on previous problem sets, or those below.
Extra Problems
1. In an arbitrary integral domain D, a non-zero element a is said to be reducible in
D if a can be written as the product of two non-units of D. Otherwise a is
irreducible in D.
Check that this definition is consistent with the definition of an irreducible
polynomial. Let F be a field. Explain why, using this definition, we may define a
non-constant polynomial in F[x] to be irreducible over F if it is not the product of
two smaller degree polynomials in F[x]. Show that this is not the case for nonfields by exhibiting an example of a polynomial that is reducible over Z , but is
not the product of smaller degree polynomials in Z[x] .
2. For a and b elements of a ring R, we say that a divides b if b = ac for some c in R.
If a divides b we write a b .
An element p of an integral domain D is prime if p ab ! p a or p b .
(a). Show that if p is an prime element of the integral domain D, then p
is irreducible..
Solution: Suppose that p is a prime element of D and p = ab for some a and b.
We must show that one of a or b is a unit. Then p ab and so WLOG p a . Hence,
a = pd for some element d. Thus p = pdb and so p(1 ! db) = 0 " db = 1 and thus
b is a unit.
(b). Show that if p is a prime in an integral domain D, then <p> is a prime ideal.
3. Show that x 4 + 8x 3 + 17x 2 + 8x + 1 factors in Z[x] .
Solution: x 4 + 8x 3 + 17x 2 + 8x + 1 = (x 2 + 3x + 1)(x 2 + 5x + 1) .
4. Prove or disprove: For some ring R with 1, R[x] has a non-constant unit.
Solution: In Z 9 [x] , (3x + 1)(6x + 1) = 1.
5. (a). Show that in any field F there are exactly two elements a, such that a 2 = 1 .
The polynomial f (x) = x 2 ! 1 can not have more than two zeros, but it does have
x = 1 and x = !1 as zeros.
(b). Use (a) to prove Wilson’s Theorem below:
If p is any prime integer, then ( p ! 1)! " !1mod p .
Solution: Consider the product ( p ! 1)! = 1 " 2 " 3 " !( p ! 2)( p ! 1) .
Then except for 1 and p–1 (which is –1 mod p), the elements of this product can
be paired with their multiplicative inverses in the field Z p . Thus the product
reduces to –1 mod p.
6. (a). Is A =!< x 2 + x + 1 > a maximal ideal in ![x] ?
Solution: Yes, because x 2 + x + 1 is irreducible in ![x] .
(b). Is B =!< x 4 + x + 1 > a maximal ideal in ![x] ?
Solution: No, every polynomial of degree larger than 2 is reducible in ![x] .
7. Show that p(x) = x 3 + 17x + 36 is irreducible in ![x] .
Hint: Reduce coefficients modulo 5.
8. (a). Show that x 4 + 1 has x 2 + 2 as a factor in ! 5 [x] . What’s the other one?
(
)(
)
Solution: x 4 + 1 = x 2 + 2 x 2 + 3 .
(b). Show that x 4 + 1 is reducible in !17 [x] . What are its factors?
Solution: This polynomial has x = 2 as a zero and so it has the linear factor x – 2.
So, x 4 + 1 = ( x ! 2 ) x 3 + 2x 2 + 4x + 8 . Can you do better?
(
(
)
)
[Yes, x + 1 = ( x ! 2 ) ( x + 2 ) x + 4 ]
4
2
9. What condition on a, b and c guarantees that ax 2 + bx + c is irreducible over the
real numbers?
Solution: b 2 ! 4ac < 0 .
10. (a). Suppose that f (x) !![x] has degree 4 and f (1) = f (2) = f (3) = 0, and f (0) = 48 .
What is f (x) ?
(b). Suppose that f (x) !![x] and f (1) = f (2) = f (4) = f (5) = 7 . What is f (x) ?
Problems from Your text:
Page 207 – 209 #1, 3, 5, 7, 9, 15, 18, 22, 23, 24, 27.
Page 218 – 220 #3, 10, 14, 15, 17, 18, 19, 21, 25, 26, 27, 28, 34, 35, 36.
Page 243 – 245 #10, 22, 29, 30.
Page 252 – 254 #5, 7, 18, 19.
Solutions:
Page 207:
#18. All but a finite number of the coefficients must be 0.
#22. 2x + 1
Page 218: #26. p = 3 and p = 5.