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Problem Set 3. Solutions.
Math 323, Winter 2017
(1) A + B is a subring of R containing B. So we take ϕ : A → R/B the restriction of
the quotient morphism R → R/B to A + B. Since quotient morphisms are ring
morphisms, so is ϕ.
Note that if a ∈ A, then ϕ(a) = a + B. So ϕ(a) = 0 + B if and only if a ∈ B,
which is to say that Ker(ϕ) = A ∩ B. In particular, this proves that A ∩ B is an
ideal of A.
Im(ϕ) = A + B/B : For the inclusion ⊆, if a ∈ A, then ϕ(a) = a + B =
a + 0 + B ∈ A + B/B. Conversely, if a + b + B = a + B ∈ A + B/B with a ∈ A
and b ∈ B, we get a + b + B = ϕ(a) ∈ Im(ϕ).
By the first isomoprhism theorem, we get A/Ker(ϕ) ∼
= Im(ϕ) so A/(A ∩ B) ∼
=
A + B/B.
(2) Define the moprhism ϕ : R/I → R/J the morphism defined by ϕ(r + I) = r + J.
• ϕ is well defined : If r + I = r0 + I, then r − r0 ∈ I ⊂ J hence ϕ(r + I) =
r + J = r0 + J = ϕ(r0 + I).
• ϕ is a ring homomorphism : This is straightforward since
ϕ((r + I) + (r0 + I)) = ϕ(r + r0 + I) = r + r0 + J = (r + J) + (r0 + J) = ϕ(r + I) + ϕ(r0 + I),
and
ϕ((r + I)(r0 + I)) = ϕ(rr0 + I) = rr0 + J = (r + J)(r0 + J) = ϕ(r + I)ϕ(r0 + I).
• ϕ is surjective : If r + J ∈ R/J then r + J = ϕ(r + I).
• Ker(ϕ) = J/I : Let r + I ∈ R/I. Then ϕ(r + I) = 0 + J if and only if
r + J = 0 + J, in other words, if and only if r ∈ J. Hence Ker(ϕ) = J/I.
P
(3) (a) Let I, J be two ideals in R. Let x = ni=1 αi βi with αi ∈ I, βi ∈ J for all
i ∈ {1, . . . , n}. Then for each i, we have that αi βi ∈ I since αi ∈ I and
αi βi ∈ J since βi ∈ J therefore αi βi ∈ I ∩ J, since both ideals are stable
under +, we get that x ∈ I ∩ J as desired.
(b) Take R = Z, I = J = 2Z. We get IJ = 4Z ⊂ 2Z = I ∩ J but it is not an
equality.
• We claim that I is an ideal. When a = 0, b = 0 we get 0 ∈ I, √so I is
nonempty.√ Now, assume x, y ∈ I. Then, by definition, x = a + b 5 and
y = c + d 5 for some a, b, c, d ∈ Z. Now, we have 4 | (a − b) and 4 | (c − d),
so 4 | (a − b) + (c − d) = (a + c) − (b + d). Thus,
√
√
√
x + y = a + b 5 + c + d 5 = (a + c) + (b + d) 5 ∈ I
√
Also, given r ∈ R where r = m + n 5 we get that
√
√
√
rx = (m + n 5)(a + b 5) = (am + 5bn) + (na + mb) 5
2
But now, we observe that
(am+5bn)−(na+mb) = 4bn+(am+bn)−(na+mb) = 4bn+a(m−n)−b(m−n) = 4bn+(a−b)(m−n)
Since 4 | 4bn and 4 | (a − b)(m − n) as 4 | a − b, we get that 4 | 4bn + (a −
b)(m − n), or equivalently, 4 | (am + 5bn) − (na + mb). Therefore, rx ∈ I and
we conclude that I is an ideal.
To find the√quotient R/I, we first prove that the map φ : R → Z/4Z defined
by φ(a + b 5) = a − b + (4) (where a − b + (4) is the congruence class of the
number a − b in Z/4Z) is a homomorphism.
We need to check that φ((a+I)+(b+I)) = φ(a+b+I), and
√ φ((a+I)(b+I))
√=
0
0
φ(ab + I), where a, b are elements of R, so a = x + y 5, b = x + y 5,
x, x0 , y, y 0 ∈ Z. √
√
√
Checking: (x + y 5) + (x0 + y 0 5) = (x + x0 ) + (y + y 0 ) 5 by definition of
addition in R. So, φ(a+b+I) = (x+x0 )−(y+y 0 )+(4) = (x−x0 )+(y−y 0 )+(4) =
φ((a + I) + (b + I)). Note that we used the definition of addition in Z/4Z
and the fact that it is well-defined. The fact that φ preserves multiplication
is proved similarly.
Now it is easy to prove that R/I is isomorphic to Z/4Z. All one needs
to do here is use the First Isomorphism Theorem: We proved above that
φ : R → Z/4Z is a homomorphism. Note that
√ it is surjective: for every
element in c ∈ Z/4Z there exist elements a + b 5 in R with a − b ≡ c mod 4.
By definition of I and φ, the ideal I is the kernel of φ: it consists of the
elements r such that φ(r) = 0. Then by the Isomorphism Theorem, R/I is
isomorphic to Z/4Z.
(4) Let x2 + x + 1 be an element of the polynomial ring E = F2 [x] and use the bar
notation to denote passage to the quotient ring F2 [x]/(x2 + x + 1).
(a) Prove that E has 4 elements: 0, 1, x, x + 1.
Solution: The elements of F2 [x] are of the form an xn + · · · + a1 x + a0 where
ai ∈ F2 , so in particular ai = 0 or ai = 1 for each 1 ≤ 1 ≤ n. Now, when
we take the quotient by x2 + x + 1, we are essentially setting x2 + x + 1 = 0, or
equivalently x2 = −(x + 1), but −(x+1) = x+1 in F2 [x], so x2 = x + 1. Imposing
this condition, we see that every polynomial an xn + · · · + a1 x + a0 can be reduced
to a linear polynomial after repeated applications of x2 = x + 1. More precisely,
since F2 is a field, we can use long division for polynomials, and the remainder
of any polynomial, when dividing by x2 + x + 1, has to be of degree less than 2,
that is, linear. Therefore, the elements of E are precisely of the form a + bx where
a, b ∈ F2 .
3
Since a, b ∈ {0, 1}, we see that there are exactly 4 choices.
a = 0, b = 0 =⇒ 0 + 0x = 0
a = 1, b = 0 =⇒ 1 + 0x = 1
a = 0, b = 1 =⇒ 0 + 1x = x
a = 1, b = 1 =⇒ 1 + 1x = 1 + x
It is also clear that above 4 elements are distinct (indeed, to have two elements
equal in the quotient we need their difference to be a multiple of x2 + x + 1, but
here all the difference have degree at most 1 and are not 0). We conclude that
E = {0, 1, x, 1 + x}
(b) To calculate the entries in the addition table, first observe that 0 acts as the
additive identity. Since addition operation in E is commutative, we only need to
compute the following expressions: 1+1 = 0, 1+x = 1 + x, 1+1 + x = 2 + x = x,
x + x = 2x = 0, x + x + 1 = 2x + 1 = 1, 1 + x + 1 + x = 2 + 2x = 0. For the
simplicity of display, we will omit the bar notation. So we will write 1 + x in the
table to mean the element 1 + x in the quotient ring F2 [x]/(x2 + x + 1). The 4 × 4
addition table for E is as follows: (Klein 4-group also given on the right)
+
0
1
x
1+x
0
0
1
x
1+x
1
1
0
1+x
x
x
x
1+x
0
1
1+x 1+x
x
1
0
∼
=
·
1
a
b
c
1
1
a
b
c
a
a
1
c
b
b
b
c
1
a
c
c
b
a
1
Note that Klein 4-group is written multiplicatively (page 68 in Dummit & Foote).
The isomorphism between additive group E and Klein 4-group is clear if we identify
0 with 1, 1 with a, x with b, and 1 + x with c.
Remark: A shorter, but more abstract, solution is to say that Klein group
(written additively) is Z/2Z × Z/2Z, and since the addition of polynomials is
coefficient-wise mod 2, we immediately obtain the isomorphism ax + b 7→ (a, b).
(c)
Solution: To calculate the entries in the multiplication table, observe that
0 times anything is zero. This fills out the first column and first row right
away. Similarly, the element 1 acts as the multiplicative identity. The only
non-trivial computations we need to do are the following: (x)(x) = x2 = x + 1;
(1 + x)(1 + x) = 1 + 2x + x2 = 1 + x2 = 1 + 1 + x = 2 + x = x, and (1 + x)(x) =
(x)(x + 1) = x2 + x = x + 1 + x = 2x + 1 = 1. For the simplicity of display, we
will omit the bar notation. So we will write 1 + x in the table to mean the element
1 + x in the quotient ring F2 [x]/(x2 + x + 1). The 4 × 4 multiplication table for E
4
is as follows:
·
0
1
x
1+x
0
0
0
0
0
1
0
1
x
1+x
x
0
x
1+x
1
1+x 0 1+x
1
x
Looking at nonzero terms, namely 1, x and 1 + x, we see that they form cyclic
group of order 3:
·
1
x
1+x
1
1
x
1+x
x
1+x
1
x
1+x 1+x
1
x
∼
=
· 1 a a2
1 1 a a2
a a a2 1
a2 a2 1 a
We can identify generator a of the cyclic group with the element x. We conclude
×
×
that E is isomorphic to the cyclic group of order 3. Since E \ {0} = E , we
deduce that E is a field.