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In my notes, I have the following definition:

"A field $(F,*)$ consists of set $F$ combined with a binary operation $+$ or $\cdot$, where

  • The operation $+$ turns $F$ into an additive group with identity element $\underline{0}$.
  • For all $a,b \in F$ we also have $ab \in F$ and $a \cdot b$ is commutative. It turns $F^{*}=\{x \in F | x \neq 0 \}$ into a multiplicative (abelian) group.
    • $\cdot$ is distributive"

This is very similar to the definition of a group $(G,*)$. The only difference I can see in these definitions is that the operator $*$ used by a group is not restricted to just $+$ or $\cdot$.

I'm currently trying to understand fields in the context of vector spaces. Could someone also explain what is meant by "a vector space over $F$" (where $F$ is a field)? Does it mean the collection of vectors who's elements are all contained in $F$?

EDIT: I've just asked a professor a similar question and have been told that there are in fact 2 definitions of a field. The first is the definition being discussed in this thread, which is the "group theory definition". Similarly, there is the "vector calculus definition" which simply states that a field is a "function of position". Unfortunately, this has just confused me further, but at least I understand the reason behind my confusion.

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    $\begingroup$ That's not a complete definition... $\endgroup$ – Zhen Lin Nov 26 '15 at 11:35
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Travis Willse Nov 26 '15 at 11:41
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    $\begingroup$ For one, the definition of a group only entails one operation, whereas that of a field entails two. $\endgroup$ – Travis Willse Nov 26 '15 at 11:42
  • $\begingroup$ This link provides a concise summary of how certain algebraic objects are related. $\endgroup$ – MPW Nov 26 '15 at 11:43
  • $\begingroup$ A field has two binary operations. A group has only one. $\endgroup$ – Lee Mosher Nov 26 '15 at 13:15
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The first thing you should bear in mind is that fields have two operations, whereas groups only have one. (In your notes, the word "or" is misleading, as it should really be "and"). These two operations are more often than not called "addition" and "multiplication", so they are customarily denoted by $"+"$ and "$\cdot$", because the most natural examples of fields -- as the rationals and the reals, have precisely these two operations.

A vector space over a field is yet another algebraic structure. It is similar to a group, in the sense that you can add together its elements, but it differs from a group inasmuch as you can multiply its elements by scalars. These scalars come from the underlying field $F$. So the vectors are not contained in the underlying field $F$. Vectors are not scalars.

A good way to internalize these concepts is to have a precise definition of a field and a vector space, and then check -- in full detail -- how these definitions are implemented in particular examples.

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The binary operations for a field are not restricted to just $+$ or $\cdot$. Like $*$, the symbols $+$ and $\cdot $ are just generic symbols for arbitrary binary operations. However, a field needs two operations (i.e., the definition should read "$+$ and $\cdot$" instead of "$+$ or $\cdot$", and accordingly the tuple should be $(F,+,\cdot)$ instead of $(F,*)$). But to repeat: You might as well speak of a field $(F,*,\#)$ with two operations $*$ and $\#$, it need not be $+$ and $\cdot$. The use of the common symbols $+$ and $\cdot$ is motivated by the improved mnemonic. Or would you recall after a while which of $(a*b)\#c=(a\#c)*(b\#c)$ or $(a\#b)*c=(a*c)\#(b*c)$ holds in $(F,*,\#)$?

Of course there are relations to group theory:

  • $(F,+)$ is a group (even an abelian group)
  • $(F\setminus\{0\},\cdot)$, where $0$ is the neutral element of $+$, is a group (again, abelian)

And of course these are intertwined by the distributive property. Note however that it is not possible to turn an arbitrary abelian group $(F,+)$ into a field $(F,+,\cdot)$ be finding a suitable operation $\cdot$.

A vector space over a field $(F,+,\cdot)$ is an abelian group $(V,\color{red}+)$ together with an action of $F$ on $V$, i.e., a ring homomorphism $\color{red}\cdot\colon F\to\operatorname{End}(V)$ (or spelled out a map $F\times V\to V$ such that $(a+b)\color{red}\cdot v= a\color{red}\cdot v\color{red}+b\color{red}\cdot v$, $(a\cdot b)\color{red}\cdot v=a\color{red}\cdot(b\color{red}\cdot v)$, $a\color{red}\cdot(v\color{red}+ w)=a\color{red}\cdot v \color{red}+a \color{red}\cdot w$ and $1\color{red}\cdot v=v$) - colours only used to distinguish different operations using the same symbol for mnemonic reasons. So once again we see how much groups play a role, but also that it is necessary to specify the field when talking about vector spaces. For example, $\Bbb C$ is a two-dimensional vector space over $\Bbb R$, or a one-dimensional vector space over $\Bbb C$, or an infinite-dimensional vector space over $\Bbb Q$.

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  • $\begingroup$ (+1) Just a friendly note that for accessibility reasons, you might consider using blue rather than red (or perhaps even better, \oplus or \boxplus (etc.) instead of colours, though I do understand your pedagogical reasons for using $+$ and $\cdot$). :) I'm not colourblind, but the red is a tad hard to read. $\endgroup$ – Andrew D. Hwang Nov 26 '15 at 12:15
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Uniquesolution gives a good answer. The answer to your second question "is a vector space over $F$ just a collection of vectors with coordinates in $F$?" is yes.

A particular vector space has dimension, which is just the number of different entries in a (column) vector drawn from the space. Two columns of different length have to live in different vector spaces. But the columns are allowed to be infinite.

A pure vector space doesn't know whether the coordinates of its vectors are columns, rows, matrices or whatever: a vector space of $n$-by-$m$ matrices is exactly the same as (isomorphic to) a vector space of column vectors of length $mn$. The difference between row and column is only visible when you want to define an operation between two vector spaces (like matrix multiplication, say) or some other structure.

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  • $\begingroup$ Not downvoting, but in a fundamental sense, a vector space is not a collection of vectors with coordinates in $F$. To express a vector space in coordinates amounts to picking a basis; the impossibility of doing this naturally is connected to concrete, substantial phenomena in geometry and topology (e.g., the non-existence of a continuous, non-vanishing vector field -- much less an orthonormal frame -- on an even-dimensional sphere). $\endgroup$ – Andrew D. Hwang Nov 26 '15 at 12:23
  • $\begingroup$ It depends on your attitude to the Axiom of Choice, obviously, but an inability to pick a smoothly varying collection of bases for the tangent bundle is quite a long way from the concerns of the question, which evidently starts, more or less, from the school-maths concept of a vector as a column of numbers with brackets each side. (At school a vector is usually described as an object with "magnitude and direction", which takes even more for granted.) My aim was to write the least wrong answer that could help the questioner. $\endgroup$ – HTFB Nov 27 '15 at 14:53
  • $\begingroup$ The pedagogical issue I had in mind was lower-tech than my comment suggested: A tangent plane to a surface in $3$-space is a two-dimensional vector space consisting of ordered triples (subject to a constraint). (Moreover, the zero vector isn't generally the triple $(0,0,0)$.) I think this seeming dichotomy (are tangent vectors pairs because they're elements of a $2$-dimensional space, or triples?) can be confusing to students at the OP's level. In other words, I find that treating vectors as tuples from the outset creates an incorrect fundamental presumption that must be broken later. $\endgroup$ – Andrew D. Hwang Nov 27 '15 at 15:29
  • $\begingroup$ I hope the OP will meet the notion of a vector subspace next week. I suspect their first class in differential geometry is a way off. (I note you're allowing $3$-space to come with a canonical basis, and even assigning a canonical origin, which is at least as contentious as for the vector space where $0$ is genuinely uniquely defined...) $\endgroup$ – HTFB Nov 27 '15 at 16:30
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Your definition should read "combindend with a binary operation $+$ and $\cdot$", also it should better say that $+$ turns $F$ into an abelian group. Thereby the difference become more apparent:

  • The set with the operator $+$ is not only a group, it's abelian (or commutative)
  • The set (excluding the zero) with the operator $\cdot$ is also not only a group, it's abelian too.
  • The distributive law holds.
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