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The set of smooth real functions $C^{\infty}=$ {$f: R\to R| \text{f is smooth}$} constitutes a vector space $C:$ {$C^{\infty}, +^C, •^S$} over the field {$R, +, •$}, where $$(v+^C w)(x):= v(x)+^R w(x)$$, and $$(a \in R •^S v)(x) := a•^R v(x)$$

But it seems like $C^{\infty}$ is also by itself a ring: Addition $+^C$ inherits the group properties from $R$. And defining an additional multiplication $C^{\infty} \times C^{\infty} \to C^{\infty}$ as follows $$(v •^C w)(x) := v(x) •^R w(x)$$ This multiplication constitutes a monoid on the set of smooth real functions. I am not sure if it is also a group on that set.

In any case, this set therefore constitutes a Ring (and possibly also a Field, I am not sure). Is this correct?

Are there generalizations of this? Are there many vector spaces that are also separately Fields or Rings (i.e. apart from the trivial way in which all fields are also vector spaces)?

Thanks for the answers so far. A lot of different concepts are falling into place.

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    It is indeed a ring, but not a field (think about $x \mapsto x$ or any function that is zero somewhere). – 2017-02-05
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    For a function $f$ with $0 \in Rng(f)$ there won't exist a multiplicative inverse, so it won't be a field.. – 2017-02-05
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    You get many different rings and fields which are also (non-trivial) vector spaces. Note that even $\mathbb{R}$ is a non-trivial vector space over $\mathbb{Q}$. Another standard example would be polynomial rings. – 2017-02-05
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    @DRF, Ah the $R-Q$ example is interesting, I didnt think of it. I can't think of how polynomials are a ring (except as a sub-ring of the ring I defined above). – 2017-02-05
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    You can also get non commutative examples if you look say at $m\times m$ matrices over a field $K$. With the usual addition and multiplication of matrices. – 2017-02-05
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    You can take any field $K$ and define $K[X]$ to be the ring of polynomials in one variable over $K$. It is easy to see it is a ring with the usual multiplication and addition. (Note that's not exactly what you define above in the sense that multiplication and addition in $K[X]$ is not defined via evaluation but abstractly). I can't really think of a situation this specifically matters currently though. You can also define a ring $K[X,Y]$ of polynomials in two variables etc. – 2017-02-05
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    Of course. I checked it and indeed you mean that the set of $n \times m$ matrices constitute both a ring by itself, and a vector field over $R$, correct? – 2017-02-05

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A vector space is actually a module structure .In your case , yes it also has a ring structure. We called such structures as algebras. It can't be a field as $ x^3$ is a smooth function but its inverse isn't.

Yes in the case of field extensions $L /K$ ,L is also a field and has a basis over K. For example $ \mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$

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    So it is a commutative ring, but one Axiom short of being a Field (the inverse axiom)? – 2017-02-05
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    yes absolutely . – 2017-02-05
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    Could you clarify the part about $L/K$? – 2017-02-05
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    We define extension of fields, suppose we have $ \mathbb{Q}(\sqrt{2})$ over $ \mathbb{Q}$. we see that both are fields and the former one is a vector space over the latter one . Also you can see that $ 1 ,\sqrt{2}$ is a basis of it. – 2017-02-05