# When is a vector space (over field $K$) also a ring (with subring $K$)?

(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.)

• @ChristophPegel: I don't even know what an algebra is, so if that's what I'm asking, it's by sheer accident.
– kjo
Jan 11, 2014 at 13:14
• 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 $K \hookrightarrow V$. Jan 11, 2014 at 13:28
• @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?
– kjo
Jan 11, 2014 at 14:30
• @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". Jan 11, 2014 at 14:34
• 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 ($cd = 0$). Jan 11, 2014 at 21:29

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.