Math 574 — Practice Exam 2
1. (a). What is meant by an evaluation homomorphism?
Solution:
!a : F[x] " E , defined by !a ( f (x)) = f (a) where E is an extension field of F.
(b). A Principal Ideal Domain is _____________________________________.
Solution: An integral domain in which every ideal is a principal ideal.
(c). Show that the polynomial x 6 + 21x 3 + 105x 2 + 42x + 28 is irreducible
over the rationals.
Solution: Use Eisenstein’s Criterion with p = 7.
(d). Every ideal in F[x], where F is a field, has the form _______________.
Solution: < p(x) > where p(x) is irreducible.
(e). Give an example of a cubic polynomial in ![x] that is irreducible over the
rationals but can be written as the product of two non-units in ![x] .
Solution: p(x) = 2x 3 + x + 2 = 2 x 3 + 2x + 1
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2. Prove that if A is a maximal ideal of the ring R, then A is a prime ideal of R.
Solution: If A is a maximal ideal then R A is a field. Thus R A is an integral
domain. But R A is an integral domain only when A is a prime ideald.
3. (a). Is x 4 + 3x 3 + 3x 2 ! x + 1 irreducible over ! 7 ?
If not, then factor it as completely as possible.
Solution: first note that 1 and –2 are zeros. So we have that
(x ! 1)(x + 2) = x 2 + x ! 2 is a factor. Now divide x 4 + 3x 3 + 3x 2 ! x + 1 by the
later polynomial to get the remaining factor (which is irreducible).
So we get x 4 + 3x 3 + 3x 2 ! x + 1 = (x ! 1)(x + 2)(x 2 + 2x + 3) .
(b). According to the rational root theorem, what are the possible zeros of
2x 5 + 8x 4 + 5x 3 + 2x 2 + 9 ?
9
1
3
Solution: ± ,!± ,!± ,!±1,!±3 .
2
2
2
(c). Show that x 3 + 9x 2 + 5x + 7 is irreducible over the rationals.
Hint: Consider the reduced version of this polynomial in some ! p [x] .
Solution: In ! 3[x] , f (x) = x 3 + 2x + 1 is irreducible because this is a cubic
polynomial with no zeros.
4. Complete the details of the following proof of Eisenstein’s Criterion.
Theorem. Let p(x) = a0 + a1 x + a2 x 2 +! + an x n be a polynomial in ![x] . Suppose
that p is a prime and p ai !0 ! i ! n " 1,!! p /| an , and p 2 /| a0 .
Then p(x) is irreducible over ![x] .
Solution: [See your notes]
Proof. Suppose that p(x) = a0 + a1 x + a2 x 2 +! + an x n = f (x)g(x) ,
r
s
where f (x) = ! bi x ,!!g(x) = ! ci x i .
i
i=0
i=0
Now argue why it must be that p divides exactly one of b0 and c0 .
So, without loss of generality, p divides b0 but!p does not divide c0 .
Thus there exists a largest integer t such that p divides bi !i ! t < r ! n .
but … [now argue that p does not divide at and get a contradiction.]
5. Let f (x) !F[x] for some field F. Show that if f (x) is irreducible then < f (x) > is
a maximal ideal.
Solution: Suppose that < f (x) > is not a maximal ideal. Then < f (x) > mustbe
properly contained in some other ideal B. But since F[x] is a Principal Ideal
Domain, B =< g(x) > for some g(x) .
But < f (x) >!!!< g(x) > implies that f (x) !!< g(x) > and so f (x) = g(x)h(x) for
some h(x). But since f (x) is irreducible and F is a field, it follows that g and h
cannont have smaller degree than f. But then one of g and h must be a constant
polynomial. If g is a constant, then B = F[x] , and if h is a constant, then
< f (x) >!=!< g(x) > . Both of these assertions contradict the assumption that
< f (x) > is a maximal ideal.
6. (a). Show that the polynomial x 4 + x 2 + 1 has no rational root.
Solution: Apply the Rational Root Theorem.
(b). Is x 4 + x 2 + 1 irreducible over the rationals? Justify your answer.
Solution: Yes. If x 4 + x 2 + 1 factors over the rationals, then it also factors over
the integers. But since it has no rational root, then it must factor as the product of
two quadratics. Thus it must be that either x 4 + x 2 + 1 = (x 2 + ax + 1)(x 2 + bx + 1)
or x 4 + x 2 + 1 = (x 2 + ax ! 1)(x 2 + bx ! 1) for some integers a and b. Neither case is
possible. For if x 4 + x 2 + 1 = (x 2 + ax + 1)(x 2 + bx + 1) , then comparing coefficients
we see that b = –a and !a 2 + 2 = 0 which is impossible since
number. The other case fails even more easily.
2 is not a rational
(c). Is x 4 + x 2 + 1 irreducible over the reals? Justify your answer.
Solution: No. No polynomial of degree larger than 2 is irreducible over the reals.
(d). Is x 4 + x 2 + 1 irreducible over ! 2 ?
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)
2
Solution: No. In ! 2 , x 4 + x 2 + 1 = x 2 + x + 1 x 2 + x + 1 = x 2 + x + 1 .
7. Show that < x 2 + 1 > is not a prime ideal in ! 5 [x] and in fact find two
polynomials f (x) and g(x) in ! 5 [x] such that f (x)g(x) !< x 2 + 1 >
but f (x) !< x 2 + 1 > and g(x) !< x 2 + 1 > .
Solution: (x + 2)(x + 3) !!< x 2 + 1 > .
8. Let A =!< x 2 + x + 1 > be an ideal in ! 5 [x] .
! [x]
(a). Justify the claim that F = 5 A is a field.
Solution: A is a maximal ideal since x 2 + x + 1 is irreducible.
(b). How many elements are there in F ?
Solution: 25.
__________
(c). What is the inverse of ( 2x + 1) = ( 2x + 1) + A in F ?
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)
Solution: Solve (2x + 1)(ax + b) = 1 + k x 2 + x + 1 in ! 5 for a and b.
We get k = 2a, 2b + a = 2a, and 2a = b – 1 in ! 5 . We get a = 1 and b = 3.
__________
_________
So, the inverse of ( 2x + 1) = ( 2x + 1) + A is (x + 3) !=!(x + 3) + A .