Math 361, Spring 2014, Assignment 5

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Carefully define the following terms, then give one example and one non-example of each:[edit]

  1. Vector space (over a field $F$).
  2. Spanning set.
  3. Linearly independent set.
  4. Basis.
  5. Dimension.
  6. Algebraic extension.
  7. Algebraically closed field.
  8. Algebraic closure.

Carefully state the following theorems:[edit]

  1. Theorem concerning the existence of bases.
  2. Dimension formula.
  3. Theorem relating finite-dimensional extensions to algebraic extensions.
  4. Theorem concerning the existence of algebraic closures.

Solve the following problems:[edit]

  1. Section 30, problems 1, 4, 5, 7, and 9.
  2. Section 31, problems 1, 3, 5, 23, and 24.
--------------------End of assignment--------------------

Questions:[edit]

Solutions:[edit]

Definitions:[edit]

  1. Vector space (over a field $F$).

    Let $F$ be a field. A vector space over F is a triple $(V,+,\mu)$ where $(V,+)$ is an abelian group (with identity $\vec{0}$) and $\mu:F\times V\rightarrow V$ satisfies:

    $$(c_1+c_2)\vec{v} = c_1\vec{v} +c_2\vec{v}$$ $$c(\vec{v_1}+\vec{v_2}) = c\vec{v_1} + c\vec{v_2}$$ $$ c_1(c_2\vec{v}) = (c_1c_2)\vec{v}$$ $$1\cdot\vec{v} = \vec{v}$$

    Note: we abbreviate $\mu (c,\vec{v})$ by $c\vec{v}$

  2. Spanning set.

    Let $S$ be any subset of the vector space $V$. $S$ spans $V$ if every element of $V$ is a linear combination.

  3. Linearly independent set.

    $S$ is a linearly independent if no element of $S$ is a linear combination of other elements.

  4. Basis.

    $S$ is a basis for $V$ is if it spans and is linearly independent.

  5. Dimension.

    The dimension of $V$ over $F$ is the cardinality of any basis written $dim_F V$ or $\left[V:F\right]$

  6. Algebraic extension.

    The extension $F\rightarrow E$ is algebraic is every element in $E$ is algebraic over $F$.

    Example:

    $\mathbb{C}$ is algebraic $/\mathbb{R}$

    Non-example

    $\mathbb{R}$ is not algebraic $/\mathbb{Q}$

  7. Algebraically closed field.

    A field $F$ is algebraically closed if it satisfies any of the following:

    1. Every non-constant polynomial in $F\left[ x\right]$ has a root in $F$.
    2. Every non-constant polynomial in $F\left[ x\right]$ splits over $F$.
    3. Every irreducible polynomial in $F\left[ x\right]$ has degree 1.
    4. Every algebraic extension of $F$ is trivial.
    Note: These are equivalent statements, so if $F$ satisfies one of them, it satisfies all of them,

    Example:

    $\mathbb{C}$ <\p>

    Non-example:

    $\mathbb{R}$ is not algebraically closed since $x^2+1$ is not constant but has no roots in $\mathbb{R}$.

  8. Algebraic closure.

    An extension $F\rightarrow\Omega$ is said to be an algebraic closure of $F$ if:

    1. it is algebraic
    2. $\Omega$ is algebraically close

    Example:

    $\mathbb{R}\rightarrow\mathbb{C}$ is an algebraic closure of $\mathbb{R}$.

    Non-example

    $\mathbb{Q}\rightarrow\mathbb{C}$ is not an algebraic closure of $\mathbb{Q}$, since it is not an algebraic extension.

Theorems:[edit]

  1. Theorem concerning the existence of bases.

    Every vector space has a basis

    or

    Any linearly independent set can be grown to a basis and any spanning set can be shrunk to a basis.

  2. Dimension formula.

    Suppose we have a chain of extensions

    $$F\rightarrow K\rightarrow E$$

    Then $\left[E:F\right] = \left[E:K\right]\left[K:F\right]$.

  3. Theorem relating finite-dimensional extensions to algebraic extensions.

    The extension $F\rightarrow E$ is a finite extension $\Leftrightarrow$ it is algebraic and finitely generated.

  4. Theorem concerning the existence of algebraic closures.
    1. Every field has an algebraic extension
    2. Universal mapping property of an algebraic closure: If $F\rightarrow \bar{F}$ is an algebraic closure, and $F\rightarrow E$ is any algebraic extension, then there exists a monomorphism $E\rightarrow\bar{F}$.
    3. If $\Omega$ is any algebraically closed extension of $F$, then there exists a unique $\phi:\bar{F}\rightarrow\Omega$