Dr. Marques Sophie
Office 519
Number theory
Fall Semester 2013
marques@cims.nyu.edu
Practice for the MIDTERM
ADVISE
• Please make sure that your understanding of the question has a mathematical
sense.
• Ask yourself if you have a complete answer to a question before moving to another
one.
• Think about all the ways at your disposal to answer the question, try to
find out, which one has most chances to succeed.
• Make sure all the assumption of the theorem you want to use are in the exercise
before using it.
• Do not forget that there can be connection between the questions. this can make
you win a lot of time.
• Start with a problem that you know how to solve to build confidence.
• Remember that partial answers , as long as they make sense, can earn you come
points.
Exercise 1: (?) √
√
Show that R := Z[ 3] = {a+b 3|a, b√∈ Z} is an integral domain (you may use without
proof that C is a field) and that 2 + 3 is a unit in R.
Solution: R ⊂ C which is a field. You can prove easily that R is a subring of C (easy).
Now, C is a field, in particular
an integral domain thus also R is an integral domain.
√
We want to find a + b 3 ∈ R such that
√
√
(a + b 3)(2 + 3) = 1
but this is equivalent to
√
a+b 3=
√
1
√ =2− 3∈R
2+ 3
√
√
√
Thus, 2 + 3 admit an inverse 2 − 3 in R and 2 + 3 is a unit.
Exercise 2: (??) Construct a field with 16 elements and show that it is a field. (Hint:
Consider (Z/2Z[x])/(x4 + x + 1).)
Solution: The polynomial p(x) = x4 + x + 1 in Z/2Z[x] since you can check that it
has no roots in Z/2Z[x] since 0̄ and 1̄ are no root. To see that it does not factor as
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quadratics use the fact that since the coefficient must be or 0̄ or 1̄, it is easy to list all
these polynomials and check that they are not factor of x4 + x + 1 using the method of
indeterminate coefficients, since the only possibilities for the coefficient are 0̄ or 1̄. Thus
p(x) is irreducible. Put, F = Z/2Z[x]/ < p(x) > is a field since p(x) is irreducible. All
the elements of F can be represented by polynomials of degree strictly less than 4, there
is exactly 24 = 16 such a polynomials, so there is 16 elements in F .
Exercise 3: (??) Let F be a field and A ∈ Mn (F ) an n × n matrix. Show that
the map θ : F [x] → Mn (F ), sending p(x) to p(A) ( the polynomial evaluated at the
matrix) is a ring homomorphism. Conclude that there is a polynomial mA (x) ∈ F [x],
such that every polynomial p(x) ∈ F [x], such that every polynomial p(x) ∈ F [x] for
which p(A) = 0n×n is a multiple of mA . (mA is called the minimal polynomial of A.)
Solution: The fact that θ is a ring homomorphism is easy to check. The kernel I of
this morphism is an ideal of F [x], defined I = {p(x) ∈ F [x]|p(A) = 0}. Since F [x]
is a principal ideal domain then I is principal. So, there is a polynomial such that
I =< mA (x) > and then we get the property we want.
Exercise 4: (?)
Let R be a commutative ring with 1. Show that R is a field if and only if {0} and R
are the only ideals of R.
Solution: Done in class
Exercise 5: (??)
1. Let R be a ring and I, J be ideal of R. Show that the map
θ : R → R/I × R/J
r 7→ (r + I, r + J)
is a ring homomorphism with kernel I ∩ J. Show that if I + J = R, then θ is
surjective.
2. Show that for gcd(m, n) = 1 we have an isomorphism
Z/mnZ ' Z/mZ × Z/nZ
(Hint: Let I = mZ and J = nZ, use 1.)
3. Let m, n as in 2) Let m̃ be the multiplicative inverse of m modulo n and ñ the
multiplicative inverse of n modulo m. Show that the map
φ : Z/mZ × Z/nZ
→ Z/mnZ
(a + mZ, b + nZ) 7→ (a.ñ.n + b.m̃.m)
is well define and it is an inverse to
θ̄ : Z/mnZ → Z/mZ × Z/nZ
r + nmZ 7→ (r + mZ, r + nZ)
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4. Determine all the integers x such x ≡ 3 mod 1000 and x = 5 mod 1001.
Solution:
1. It is easy to show that it is a homomorphism (left to you!). Let’s compute the
kernel
ker(θ) = {r ∈ R|θ(r) = 0}
= {r ∈ R|r + I = 0 and r + J = 0}
= {r ∈ R|r ∈ I and r ∈ J}
= I ∩J
Clearly, I +J ⊆ R. If θ is surjective, then in particular for any r ∈ R, (r+I, 0+J)
have an antecedent by θ which means that r + I = 0 + J this means that r ∈ I + J.
Thus I + J = R.
2. By Bezout’s theorem, we have that gcd(n, m) = 1 is equivalent to the existence
of k and l integers such that nk + ml = 1, that is equivalent to Z = nZ + mZ.
Moreover, mZ ∩ nZ = nmZ since m and n coprime a multiple of both m and n
is a multiple of mn. Then by the previous question
Z/nmZ ' Z/nZ × Z/mZ
3. To prove that φ is an inverse of θ we have to show that φ ◦ θ = IdZ/mnZ and
θ ◦ φ = IdZ/mZ×Z/nZ . We notice that ñn ≡ 1 mod m then there is an integer k
such that ñn = 1 + mk we have then that −mk ≡ 1 mod n then −k ≡ m̃ modn
then ñn + m̃m ≡ 1 mod nm. For any r ∈ Z,
φ(θ(r + nmZ)) = φ(r + mZ, r + nZ) = rñn + rm̃m + nmZ = r + nmZ
For any r ∈ Z and s ∈ Z,
θ(φ(r + mZ, s + nZ)) = θ(rñn + sm̃mZ)
= (rñn + sm̃m + mZ, rñn + sm̃m + nZ)
= (r + mZ, s + nZ)
4. We take n = 1000 and m = 1001, they are coprime (1001−1000 = 1, Bezout’s theorem), 1000 ≡ −1 mod 1001, then we can choose ñ = −1 and 1001 ≡ 1 mod 1000,
then we can choose m̃ = 1. Then we know that we have a unique congruence class
mod 1001000 by the isomorphism, which is the one of 3 − 5 = 2.
Exercise 6: (?) Consider the polynomial f (x) = x2 + 2x + 3 in Z/5Z[x]
1. Is f (x) irreducible in Z/5Z[x]? If yes, explain why it is so, if not determine a
proper factorization of f (x) in Z/5Z[x].
2. Consider the ring F = Z/5Z[x]/ < f (x) >. Prove that x̄+ < f (x) > is invertible
in F and determine the order of x̄ in the multiplicative group F× of units of F .
3. Find, if exists, an element of order 3 in F × .
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Solution:
1. Easy to check to show that f (x) has no roots in Z/5Z, so since it has degree 2, it
is irreducible in Z/5Z[x].
2. Since f (x) is irreducible in Z/5Z, the ideal < f (x) > is prime and maximal, thus
F is a field and using the Euclidean algorithm one sees that F = {a + bx̄|a, b ∈
Z/5Z}, hence |F | = 25. Thus F × = F \{0} is a cyclic group, with 24 elements.
One checks the powers of x̄ running through all possible divisors of 24: x̄ 6= 1,
x̄2 = 3x̄ + 2 6= 1. Similarly one tests that x̄3 , x̄4 , x̄6 , x̄8 , x̄12 6= 1. So the order of
x̄ is 24.
3. It follows from 2. that x̄8 = 4x̄ + 1 has order 3.
Exercise 7: (??) Let F be a field and let F [x] be the ring of polynomials of one
variable x with coefficients in F . Prove that there are infinitely many irreducible monic
polynomials in F [x].
Solution: Assume there is only a finite number of irreducible monic polynomials, say
p1 (x),..., pn (x). Now, irreducible
= prime, so each pi (x) is prime.
Qn Now consider the
Qn
monic polynomial q(x) = i pi (x) + 1. By construction, q(x) = i pi (x) + 1. By construction, q(x) is not divisible by any of the pi (x) hence it is either prime= irreducible
itself, or divisible by another prime polynomial greater then pn (x), contradicting the
assumption of a finite number of prime=irreducible monic polynomials.
Exercise 8: (?) Let p ∈ Z be a prime number.
1. Does f (x) = xp − px − 1 have a proper factorization in Z[x]? If yes, show an
explicit factorization of f (x), if not justify your answer.
2. Show that the polynomial g(x) = x4 + 5x2 + 3x + 2 is irreducible in Q[x].
Solution:
1. Consider f (x) ∈ Q[x]. Since Eisenstein’s criterion does not apply directly to f (x),
one looks at the image of f (x) via the ring automorphism x 7→ x + 1. Then f (x)
transforms into
f (x + 1) = xp + pxp−1 +
p(p − 1) p−2
p(p − 1)(p − 2) 3 p(p − 1) 2
x
+ ... +
x +
x −p
2!
3!
2!
which is now irreducible by Eisentein’s criterion using the prime p, thus f (x) is
also irreducible in Q[x] and by applying Gauss Lemma f (x) ∈ Z[x] has no proper
factorization.
2. The rational roots test gives no roots, no g(x) does not split into linear factors in
Q[x]. Then by comparing coefficientts of the product of two degree 2 polynomials,
one sees that g(x) cannot factor into any such product, thus g(x) is irreducible in
Q[x].
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Exercise 9: (?) Let A = C([0, 1], R) be the ring of continuous functions (for the
Euclidean topology) f : [0, 1] → R and let I ⊂ A be the subset of functions which
vanish at 1/2/
1. Show that I an ideal of A.
2. An ideal I is said prime if for any x, y ∈ I xy ∈ I is equivalent to x ∈ I or y ∈ J.
Is I a prime ideal? Prove it or disprove it.
3. Is I a maximal ideal? Prove it or disprove it.
Solution:
1. Let f (x) and g(x) be two elements of A which vanish at 1/2; it follows that
f (x) ± g(x) also vanish at 1/2 and that the zero function vanishes at 1/2. Thus I
is a subgroup of the additive group (A, +). Let h(x) ∈ A; then h(1/2)f (1/2) = 0
thus h(x)f (x) ∈ I and I is an ideal of the commutative ring A (Equivalently one
can observe that I is the kernel of the ring homomorphism f (x) 7→ f (1/2).
2. I is prime, since if f (x)g(x) vanishes at 1/2, then necessarily either f or g (or
both) vanish at 1/2.
3. The ring homomorphism φ : A → R, f 7→ f (1/2) is surjective (evident), thus
A/ker(φ) ' R, with ker(φ) = I. Thus, since R is a field, I is maximal.
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(?) = easy , (??)= medium, (???)= challenge
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