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Given a vector space $(K,V)$ where $K$ is the scalar field and $V$ is the abelian group, then another vector space $(K,W)$ is a subspace of $(K,V)$ if $W\subseteq V$. My question is, is there a special name for a vector space $(F,V)$ where $F$ a sub-field of $K$?

Ralph B.
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  • Every $K$-vector space is naturally also a vector space over every subfield of $K$. There's no special name for this situation, as far as I know. – egreg Sep 26 '16 at 21:46
  • There is no decided name for it, i'd coin "underspace" for it just becausr i am alazy bastard – Zelos Malum Sep 27 '16 at 03:37

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It is common to prefix the name of the vector space with the field it's over, instead of using pairs. That could give you want you want here. So for example:

  • Let $V$ be a $K$-vector space. Consider $V$ as an $F$-vector space, for any $F \subset K$. Then...

  • Every $K$-vector space is an $F$-vector space, therefore...

This seems like it could be sufficient for your purposes. Anyway, I don't know of any special term other than this and I'm not sure one is needed.

Caleb Stanford
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  • 6005, is Canis Lupus' answer correct please? (Please see in particular the '$(K, V)=K \otimes_F (F,V)$') – BCLC Feb 03 '20 at 03:58
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    @JohnSmithKyon That's a very interesting question but I don't immediately know the answer, and don't have the time to look into it right now. It certainly seems true to me for when $F = \mathbb{R}$ and $K = \mathbb{C}$. – Caleb Stanford Feb 03 '20 at 04:19
  • 6005, thanks, but anyway it's wrong if Canis Lupus means $(\mathbb C_{\mathbb R})^{\mathbb C} = \mathbb C$, but right if Canis Lupus means $\mathbb C = (\mathbb R^2$, with almost complex structure $J(v,w):=(-w,v)$) and then the time consuming part is what is meant? (I seem to notice in studying maths that the time consuming parts are sometimes understanding what the author means rather than proving already understood claims of authors. Sad.) – BCLC Feb 03 '20 at 04:38
  • @JohnSmithKyon Sorry, I didn't get that, I am not familiar with your notation $(F_K)^L$. – Caleb Stanford Feb 03 '20 at 13:11
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Operations $(F, V)\mapsto (K, V)$ and $(K,V)\mapsto (F,V)$ where $F$ is a subfield of $K$ are called respectively the extension and restriction of scalars. In case of $\Bbb R$ and $\Bbb C$ they are called respectively complexification and realification. It's OK to say "$(\Bbb R,V)$ is the realification of $(\Bbb C,V)$".

Restriction and extension of scalars can be naturally extended to homomorphisms and actually are functors $\mathsf{Vec}_K\to\mathsf {Vec}_F$ and $\mathsf{Vec}_F\to\mathsf {Vec}_K$. It also can be defined for more general modules. Restriction is just a "forgetting" the multiplication by some elements of the base ring. Extension is "forcing" the multiplication by elements of the extended ring, i. e. $(K, V)=K \otimes_F (F,V)$.

Note also, that notation $(K, V)$ is very rarely used, usually people just say "$K$-vector space", or "over $K$".

See more details in Kostrikin, Manin "Linear Algebra and Geometry".

Canis Lupus
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    Nitpick: Extension $K \otimes_F (F,V)$ generally won't give you the same $(K,V)$. – arkeet Sep 27 '16 at 03:10
  • @arkeet Are you referring the concepts of external and internal complexifications (and generalisations from $\mathbb R, \mathbb C$ to $F,K$)? It seems the tensor one is external while $(K,V)$ is internal. Either this or Canis Lupus has made a mistake in that Canis Lupus implies $(\mathbb C_{\mathbb R})^{\mathbb C} = \mathbb C$ when actually $(\mathbb C_{\mathbb R})^{\mathbb C} \cong \mathbb C \bigoplus \overline{\mathbb C}$...? – BCLC Feb 03 '20 at 03:54
  • @JohnSmithKyon I think the latter is what I meant (that generally $K \otimes_F V_F \not\cong V$ unless $K=F$ - restriction followed by extension raises the dimension by a factor of $[K:F]$). – arkeet Feb 06 '20 at 18:19
  • @arkeet So not really a nitpick but an actual mistake? – BCLC Feb 07 '20 at 10:49