Download Math 4320 - Homework #6 - Solutions

Math 4320 - Homework #6 - Solutions
Due: Friday, May 2nd
1. Let F be a field and K be an extension. If α, β ∈ K are algebraic over F , i.e. they are roots of
polynomials in F [x], then show α ± β, αβ and α/β (when β 6= 0) are algebraic.
If α and β are algebraic then [F (α) : F ] = a and [F (β) : F ] = b are finite. Since the irreducible
polynomial with root β over F (α) divides the irreducible polynomial for β over F , it follows that
[F (α, β) : F (α)] ≤ b and so [F (α, β) : F ] is finite. Since F (αβ), F (α/β) and F (α ± β) are all subfields
of F (α, β) it follows that they all have finite degree over F (since their degree over F must divide
[F (α, β) : F ]). But all elements of a finite degree extension over F are algebraic over F , so the given
elements are algebraic.
2. An element of a ring x ∈ R is nilpotent if there exists n ∈ N with xn = 0. Show that N(R), the set of
nilpotent elements of R is an ideal. More over, show that R/N(R) has only one nilpotent element 0.
Let r ∈ R and x ∈ N(R), there exists n with xn = 0, so (rx)n = rn xn = rn 0 =P
0 and rx
∈ N(R).
n+m i n+m−i
If x, y ∈ N(R), then there exists n, m with xn = y m = 0, but then (x + y)n+m =
xy
.
i
If i < n then n + m − i > m and if n + m − i < m then i > n, so it follows that xi y n+m−i = 0.
Finally 0 is clearly nilpotent. Thus N(R) is an ideal. Let x + N(R) ∈ R/N(R), and suppose that
(x + N(R))n = 0 + N(R), then xn ∈ N(R). But this means that there exists some m such that
(xn )m = 0, so xnm = 0 and x ∈ N(R). Thus any nilpotent element is 0 + N(R).
3. Consider the polynomial x2 + x + 1 ∈ F2 [x]. Let E = F2 [x]/(x2 + x + 1). First show that E has 4
elements, represented by 0, 1, x, x + 1. Next, compute the multiplication table with respect to these
representatives and conclude that E is a field. This shows that x2 + x + 1 is irreducible over F2 .
Consider an element f + (x2 + x + 1) in the ring E. Consider dividing f by x2 + x + 1. This gives
f = g(x2 + x + 1) + r where r = 0 or deg r < 2. Now f + (x2 + x + 1) = g(x2 + x + 1) + r + (x2 + x + 1) =
r + (x2 + x + 1), so any class is represented by an polynomial over F2 of degree less than 2. There are
4 such polynomials 0, 1, x and x + 1. It is clear that each of these gives a unique class in E, since none
of their differences are in (x2 + x + 1). Finally the multiplication table:
0
1
x
x+1
0
0
0
0
0
1
0
1
x
x+1
x
0
x
x+1
1
x+1 0 x+1
1
x
Notice that the rows corresponding to non zero entries each contain a 1 so each element is invertible.
Thus E is a field.
4. Show that x3 − 2x − 2 is irreducible over Q and let θ be a root. Compute (3 + θ)(5θ2 − 2θ + 10) and
1
5θ 2 −2θ+10 in Q(θ).
Since 2 divides all the coefficients but the leading coefficient and 22 = 4 does not divide the constant
term, by Eisenstein’s criterion, this polynomial is irreducible. Now expanding the first expression gives
us 15θ2 − 6θ + 30 + 5θ3 − 2θ2 + 10θ, but θ satisfies θ3 = 2θ + 2, so we have 13θ2 + 14θ + 40. Let
(aθ2 + bθ + c) be the inverse of (5θ2 − 2θ + 10), then (aθ2 + bθ + c)(5θ2 − 2θ + 10) = (20a − 2b + 5c)θ2 +
(6a + 20b − 2c)θ + (−4a + 10b + 10c) = 1. Thus we must solve the system of equations 20a − 2b + 5c = 0,
35
103
−24 2
θ + 2602
θ + 1301
.
6a + 20b − 2c = 0 and −4a + 10b + 10c = 1. Solving this linear system we get 1301
5. Let F be a field of characteristic 6= 2, and D1 , D2 ∈ F elements
√
√that are not squares in F (i.e. there
does not exists an f ∈ F with f 2 = D1 ). Prove that F ( D1 , D2 ) is a degree 4 extension of F if
1
D1 D2 is not a square in F or is degree 2 over F otherwise.
Since there does not exists an f ∈ F with f 2 = D1 , it follows that
x2 −D1 is irreducible.
√the polynomial
√
2
As is x − D2 . Thus, as in question number 2√we see that [F ( D1 , D2 ) : D] ≤ 2 ∗ 2√= 4. Suppose
D1 x −√
f is a linear √
polynomial
over F ( D1 ) with root
that
with f 2 = D1√
D2 , then
√ there exists
√ f ∈ F√
√
√
D2 . Thus D2 ∈ F ( D1 ) and F ( D1 , D2 ) √
= F ( D1 ) so [F ( D1 , D2 ) : F ] = 2. On the other
quadratic, there must be an
hand suppose x2 − D2 is not irreducible over F ( D1 ), then, since it is a √
element √
of the field which is a root,
i.e.
there
exists
a,
b
∈
F
with
(a
+
b
D1 )2 = D2 , but this gives
√
2
2
2
2
a + 2ab D1 + b D1 = D2 , or 2ab D1 = D√
2 − a − b D1 . Notice that the right side is composed of
elements that are in F , thus we see that √
2ab D1 ∈ F . Since the characteristic of F is not 2, we have
2 6= 0, we can cancel the 2 and we get ab√ D1 ∈ F , which is impossible unless a = 0 or b = 0. If b = 0,
then a2 = D2 which is impossible since D2 6∈ F . Thus we must have a = 0 and we have b2 D1 = D2 ,
or (bD1 )2 = D1 D2 , i.e. D1 D2 is a square!
√
√
√ √
√
√
6. Prove that Q( 2 + 3) = Q( 2, 3) and then find an irreducible polynomial satisfied by 2 + 3.
√ √
√
√
Here
use the minimality of extensions.
Since
Q( 2, 3) contains Q and 2 + 3 it√must√contain
√
√
√ we √
minimality
(since
2 + 3)
Q( 2 + 3) by √
√
√ Q( √
√ is the
√ smallest
√ field√containing
√ Q and
√ 2 +√ 3). √On
the other √
hand ( 2 + 3)3 = 11 2 + √
9 3√so ( 2 +√ 3)3 √
− 9( 2 + 3) = 4 2 thus 2 ∈ Q( 2 + 3)
Q( 2, 3) ⊂ Q( 2 + 3).
and so is 3. Thus by minimality
√ √
By
the
previous
problem,
[Q(
2,
3) : Q] = 4 since 2 ∗ 3 = 6 is not a square in Q. It follows that
√
√
2 + 3 will
be
the
root
of
an
irreducible
polynomial of degree 4. On the other hand, working over
√ √ √
√ √
the basis 1, 2, 3, 6 of Q( 2, 3) is most convenient.
√
√ In order to find the polynomial we have to
find
a
linear
dependence
amongst
the
powers
of
2
+
3. We know that the dependence must contain
√
√
(√2 + √3)4 with coefficient
one,
thus
the
polynomial
must
contain a √
term of degree x4 . Now we have
√
√
√ 2
√
4
( 2 + 3) = 49 + 20√ 6 and
5 + 2 6. To cancel the 6 coming from the 4-th power,
√ ( 2 + 3) = √
we must subtract √
10( 2√+ 3)2 = 10(5 + 2 6). Thus we have x4 − 10x2 in our polynomial. But
evaluating this at 2 + 3 gives −1, so the irreducible polynomial is x4 − 10x2 + 1.
7. Prove that it is not possible to construct a regular 7-gon. Hint: 2 cos(2π/7) satisfies x3 +x2 −2x−1 = 0.
We from lecture that an n-gon is constructable if and only if cos(2π/n) is constructable. We are given
that 2 cos(2π/7) is a root of x3 + x2 − 2x − 1. This polynomial is irreducible because its residue modulo
2 is x3 + x2 + 1 which is an irreducible polynomial over F2 (just need to check that 0 and 1 aren’t roots
over F2 ). But this means that [Q(cos(2π/n)) : Q] = [Q(2 cos(2π/n)) : Q] = 3 which is not a power of
2. Thus cos(2π/n) is not constructable.
8. Given a circle along with its center and a point on the circle itself, is it possible to inscribe a regular
hexagon in the circle using straight edge and compass? If so, describe such a construction otherwise,
show it is not possible.
Let p1 and p2 be the two starting points. Draw a circle centered at p1 through p2 . Then draw a circle
centered at p2 through p1 . The two intersection points p3 and p4 form equilateral triangles with p1
and p2 . Thus the angle at p1 are 2π/6. Now draw a circles centered at p3 and p4 through p1 , each
adds one new intersection point with the circle centered at p1 . Call these p5 and p6 respectively. The
triangles (p1 , p3 , p5 ) and (p1 , p3 , p6 ) are also equilateral and have angles at p1 of 2π/6 radians. Finally,
draw a line through p1 and p2 , it intersects the circle centered at p1 in the point p7 . The two new
triangle (p1 , p5 , p7 ) and (p1 , p6 , p7 ) are also equilateral. Thus p2 , p3 , p5 , p7 , p6 , p4 form a regular hexagon.
9. Let p be a prime and φp = xp−1 + xp−2 + · · · + 1 the p-th cyclotomic polynomial. We showed in class
that this is irreducible over Q. Show that φp splits over Q(ζ) where ζ 6= 1 is a p-th root of unity.
2
Since ζ is a p-th root of unity, that is not 1, it follows that each of the p − 1 non trivial p-th roots of
unity are of the form ζ k for some k. Since ζ k ∈ Q(ζ) it follows that all p distinct roots of xp − 1 are
in Q(ζ) and so xp − 1 splits. But (x − 1)φp = xp − 1, so φp splits as well.
10. What is the splitting field for x4 + x2 + x + 1 ∈ F2 [x].
First note that 1 is a root of this polynomial over F2 . Thus x4 + x2 + x + 1 = (x + 1)(x3 + x2 + 1).
Now since x3 + x2 + 1 is cubic, if it has no roots in F2 , then it is irreducible. If one evaluates at both
0 and 1, we get 1, so x3 + x2 + 1 is irreducible over F2 . Thus the splitting field for x4 + x2 + x + 1
must at least contain θ, a root of x3 + x2 + 1. Factoring x3 + x2 + 1 over F2 (θ), we must at least have
the linear factor (x + θ). Factoring gives (x + θ)(x2 + (θ + 1)x + (θ2 + θ)) = x3 + x2 + 1 (this is true
over the field F2 (θ)). At this point we must check if x2 + (θ + 1)x + (θ2 + θ) is irreducible over F2 (θ).
It suffices to check if any of the 8 elements of F2 (θ) are roots. Evaluating the quadratic at θ2 gives
θ4 + (θ + 1)θ2 + θ2 + θ = θ4 + θ3 + θ = θ(θ3 + θ2 + 1) = 0. Thus the quadratic factors into linear terms
over F2 (θ) and this is the splitting field.
3