Math 3510 Spring 2013 -- Homework #1

Suggested Due Date: Thursday, January 31st

Homework #1

  1. Explain why \( \mathbb{Z}_{10} \) and \( \mathbb{Z} \) are not fields.

  2. Show \( \mathbb{Q}[\sqrt{7}] \) is a field (just check closure under +, -, x, and /. The rest of the axioms are taken care of since \(\mathbb{Q}[\sqrt{7}] \subseteq \mathbb{R}\)).

  3. Let \( \omega = \sqrt[4]{5} \) so that \( \omega^2 = \sqrt{5} \) and \( \omega^4 = 5 \). Next, let \( \mathbb{F}=\mathbb{Q}[\sqrt{5}] \) and \( \mathbb{K}=\mathbb{Q}[\sqrt[4]{5}] \) (obviously \( \mathbb{Q} \subseteq \mathbb{F} \subseteq \mathbb{K} \)). Find a basis for \( \mathbb{F} \) viewed as a vector space over \( \mathbb{Q} \) and prove your basis is actually a basis. Then do the same for \( \mathbb{K} \) over \( \mathbb{Q} \) and also for \( \mathbb{K} \) over \( \mathbb{F} \). You may use the fact that \( f(x)=x^4-5 \) has \( \omega \) as a root and there is no polynomial (over \( \mathbb{Q} \)) of lower degree with this property and \( \omega^2 \) is a root of \( g(x) = x^2-5 \) and there is no lower degree polynomial with this property. Finally state the dimensions of each field over its subfield(s): \( \dim_\mathbb{Q}(\mathbb{F}), \dim_\mathbb{Q}(\mathbb{K}),\) and \( \dim_\mathbb{F}(\mathbb{K}) \).

  4. Let \( \mathbb{F} \subseteq \mathbb{K} \subseteq \mathbb{L} \) be fields. Suppose that \( \dim_\mathbb{F}(\mathbb{K})=n < \infty \) and \( \dim_\mathbb{K}(\mathbb{L})=m < \infty \). Show that \( \dim_\mathbb{F}(\mathbb{L})=mn < \infty \). Hint: Form a basis for \( \mathbb{L} \) over \( \mathbb{F} \) by forming all products \( a_ib_j \) where \( a_i \) and \( b_j \) are elements of bases for \( \mathbb{K} \) over \( \mathbb{F} \) and \( \mathbb{L} \) over \( \mathbb{K} \).

  5. Spence Insel Friedberg (2nd and 3rd editions) Section 1.6 Problem #8.

  6. Spence Insel Friedberg (2nd and 3rd editions) Section 1.6 Problem #13.

  7. Let \( W_1 \) and \( W_2 \) be subspaces of some finite dimensional vector space \( V \) (over some field \(\mathbb{F}\)).
    1. Prove that \( \dim(W_1 \cap W_2) \leq \mathrm{min} \{ \dim(W_1), \dim(W_2) \} \)
    2. Prove that \( \dim(W_1 + W_2) \geq \mathrm{max} \{ \dim(W_1), \dim(W_2) \} \)
    Note: \( W_1 + W_2 = \{ u+v \,|\, u \in W_1 \mathrm{\ and\ } v \in W_2 \} \)
    Suggestion: For problems like this assume (without loss of generality: WLOG) that \( \dim(W_1)=m \leq \dim(W_2)=n \) so \( \mathrm{min} \{ \dim(W_1), \dim(W_2) \} = m \) etc.