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(Apologies in advance for the very naive question. I'm just learning about all this. Also, for the sake of expedience, below I use the word "ring" when it would more correct for me to use "commutative ring".)

If a (commutative) ring $R$ has a subring $K$ that happens to be a field, then $R$ can be "reinterpreted" as a vector space over $K$. I want to read more on the converse "reinterpretation", as it were, namely: Given a vector space $V$ over a field $K$, can $V$ be "reinterpreted" as a ring?

OK, I realize that the situation is not symmetric: viewing a ring $R$ as a vector space over subring/field $K$ entails "forgetting" the multiplication between elements of $R\backslash K$, whereas viewing a vector space $V$ over $K$ as a ring would require "conjuring up" an embedding of the field $K$ in $V$, together with a multiplication between vectors in $V$ consistent with the scalar multiplication in $V$, and such that the ("newly-embedded") $K$ becomes a subfield of $R$.

I can imagine a couple of possible resolutions to this question. The first one is that, for any arbitrary vector space $V$ over a field $K$ there always exists a way to make $V$ into a ring having $K$ as a subring. (The proof of this may even give a canonical construction that yields the embedding of $K$ and product of vectors alluded to above.) The second (and IMO more likely) resolution is a counterexample (or at least a proof of the existence) of a vector space $V$ over field $K$ that admits no such ring structure, along with a characterization of those vector spaces that admit such ring structure.

My question here boils down to pointers to the relevant mathematics: what terms, theorems, authors, etc. should I search for to learn more about the issues sketched above?

(By way of coda, I imagine that removing commutativity from the picture would lead to interesting counterparts to the questions above; I hope that the pointers to the commutative case will be enough for me to find about the non-commutative case as well.)

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  • $\begingroup$ @ChristophPegel: I don't even know what an algebra is, so if that's what I'm asking, it's by sheer accident. $\endgroup$
    – kjo
    Jan 11, 2014 at 13:14
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    $\begingroup$ Choose a non-zero vector $e \in V$ and choose a vector space complement $C$ for the subspace $Ke$ of $V$ spanned by $e$. Define a multiplication on $V$ as follows: $(\lambda e + c) (\mu e + d) := \lambda \mu e$. Here $\lambda,\mu$ are scalars in $K$ and $c,d$ are arbitrary elements of $C$. You can check that this multiplication turns $V$ into a ring. Furthermore, $\lambda \mapsto \lambda e$ is an injective ring homomorphism $K \hookrightarrow V$. $\endgroup$ Jan 11, 2014 at 13:28
  • $\begingroup$ @KonstantinArdakov: Thanks! I can "fill the gaps" (I think) in your description for the case where $V$ is finite dimensional, but not in general. How does one construct the complement of $Ke$ when one cannot assume that $V$ is finite-dimensional? $\endgroup$
    – kjo
    Jan 11, 2014 at 14:30
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    $\begingroup$ @kjo You're welcome. One can construct a complement of $Ke$ in the infinite-dimensional case by "waving ones hands". Precisely, you can use some form of the Axiom of Choice, such as Zorn's Lemma. The relevant keywords here are "Hamel basis". $\endgroup$ Jan 11, 2014 at 14:34
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    $\begingroup$ Incidentally, if one wants a unital ring structure, one simply needs to modify the multiplication to yield $$ (\lambda e + c)(\mu e + d) := \lambda\mu e + \lambda d + \mu c, $$ in which case, $e$ becomes the multiplicative identity. This construction realises $V$ as the unitalisation of $C$ endowed with the trivial product ($cd = 0$). $\endgroup$ Jan 11, 2014 at 21:29

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For any infinite cardinality $\kappa$, there is a field extension of $K$ of degree $\kappa$. So any infinite-dimensional vector space over $K$ can be made into a field, not just a ring.

For finite cardinalities, the situation is slightly more complex, since there may not exist field extensions of some degrees (for example, there are no finite extensions of $\mathbb{C}$). If we insist that $R$ is an integral domain, then it is a field, so there exists a ring structure on $V$ exactly when there exists a degree $\dim_K V$ extension of $K$.

On the other hand, if we ignore the requirement of being an integral domain, the problem is trivial, for we can set $R = K^{\dim_K V}$.

Note that there is no canonical choice for $R$. Even if there is only one field extension up to isomorphism (for example, a finite extension of a finite field), we still have automorphisms to worry about, with no structure to help us decide between them.

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