Download Midterm 2 solutions

Math 258 Midterm #2
Name:
Question Points Score
1
15
2
25
3
30
4
20
Total:
90
Answer the questions in the spaces provided on the question sheets. If you
run out of room for an answer, continue on the back of the page.
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1. (a) (10 points) Give an example of a commutative ring R and two prime ideals in R,
one of which is maximal and one of which isn’t. Prove that each of your examples
satisfies the desired condition.
Solution: Let R = Z[x], and consider the ideals (x) and (2, x). Z[x]/(x) ∼
= Z,
∼
∼
which is an integral ideal, so (x) is prime. Z[x]/(2, x) = Z/(2) = Z/2, which is
a field; thus (2, x) is maximal.
(b) (5 points) Give an example of a commutative ring R with no nonzero prime ideals.
Solution: Let R = R. Since R is a field the only proper ideal is the zero ideal,
so it has no nonzero prime ideals.
2. Let F = R and V be the vector space of polynomials in x over R of degree at most n.
(a) (5 points) What is dim V ⊗ V ? dim V ∧ V ?
Solution: V has a basis 1, x, . . . , xn ; thus V ⊗V has basis xi ⊗xj for 0 ≤ i, j ≤ n.
Thus V ⊗ V has dimension (n + 1)2 . On the other hand, V ∧ V has basis xi ∧ xj
where i < j, so dim V ∧ V = n+1
.
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(b) (10 points) Write down the definition of the determinant of a linear transformation
V →V.
Solution: For any basis v1 , . . . , vn of V , the determinant of T is the constant
which defines the linear transformation
dim
^V
T :
dim
^V
V →
dim
^V
V.
(c) (10 points) Let T be the linear transformation that takes p(x) to p0 (x). What is
det T ?
Solution: Since T sends all constants to 0, it has a nontrivial kernel. T is not
invertible if and only if det T = 0. Since T is not invertible, det T = 0.
3. All vector spaces in this problem are Q-vector spaces.
(a) (10 points) State the universal property for tensor products.
Solution: The tensor product V ⊗ W is the vector space such that bilinear
maps V × W → Z are in bijection with linear maps V ⊗ W → Z for all vector
spaces Z.
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(b) (10 points) Consider R as a Q vector space, and give an example of a nonzero
bilinear map R × R → R.
Solution: Take the map (a, b) 7→ ab. This is bilinear because multiplication is
commutative and distributes over addition. This map is nonzero as, for example,
(1, 1) 7→ 1.
(c) (10 points) Use the universal property to prove that R ⊗ R ∼
6= 0.
Solution: There are at least two bilinear maps R × R → R: the one given in
part (b) and the zero map. Thus the set of linear maps R ⊗ R → R has size at
least 2. However, the set of linear maps 0 → R has exactly one map. If R ⊗ R
were zero then the set of linear maps R ⊗ R → R would have size 1; however,
as we just established this set has at size at least 2. Thus R ⊗ R ∼
6= 0.
4. Let V be a finite dimensional vector space over F .
(a) (10 points) Write down an isomorphism V → V ∗ . Prove that it is, indeed, an
isomorphism.
Solution: Let {v1 , . . . , vn } be a basis for V . Write vi∗ for the linear transformation defined by vi∗ (vj ) = δij . Then we have an isomorphism V → V ∗ given
by vi 7→P
vi∗ . This map is clearly injective, since the vi∗ are linearly independent:
n
∗
∗
if f =
Pn i=1 vi is∗ zero then f (vi ) = ai = 0. Now let f ∈ V , and consider
g =
i=1 f (vi )vi . Then g(vi ) = f (vi ), so g = f , and we see that this is a
bijective linear transformation, and thus an isomorphism.
(b) (10 points) Let T : V → W be a linear transformation. If T has matrix (aij ) with
respect to bases {v1 , . . . , vn } of V and {w1 , . . . , wm } of W , what is the matrix of
the linear transformation T ∗ which takes a linear transformation S : W → F to the
linear transformation S ◦ T : V → F ?
Solution: It suffices to compute T ∗ on the basis wj∗ of W . Note that
(T
∗
(wj∗ ))(vi )
=
wj∗ (T (vi ))
=
wj∗
m
X
aki wk = aji .
k=1
Thus
T ∗ wj∗ =
n
X
aji vi∗ ,
i=1
∗
and we see that the matrix for T is (aji )—the transpose of the matrix for T .
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