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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. 1 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 . 2 (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. Page 2 (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 . Page 3