Given a field $F\subset R$ contained in a ring which is not an integral domain. How can R fail to be a vector space over F?

Question

My algebra notes mention being an integral domain as a requirement, but I can't figure out which vector space axioms are dependent on it.

Possible the result is the zero vector. Why is this a problem? — Othman El Hammouchi, Mar 17 '21 at 00:36
I guess not, but then I find it strange that it's mentioned if it's not required. — Othman El Hammouchi, Mar 17 '21 at 00:41
Are you sure you're understanding what your professor wrote? Can you post the part in your notes that say this? — Saegusa, Mar 17 '21 at 00:43
@Saegusa it's in Dutch, but the translation is: 'if a field is contained in an integral domain, then we can consider this domain as a vector space over the field'. The implication I derived was that this would not be possible for rings that are not integral domains. — Othman El Hammouchi, Mar 17 '21 at 00:48
Well take $F^2$ with pointwise multiplication. We have $(1,0)(0,1)=(0,0)$ so it's not an integral domain but it's very clearly a vector space over $F$ right. — Saegusa, Mar 17 '21 at 00:54

score 4 · Accepted Answer · answered Mar 17 '21 at 00:36

4

It isn't necessary. $R$ is indeed a vector space over $F$ where the scalar multiplication is obtained by restricting the multiplication of $R$ to $F\times R$.

answered Mar 17 '21 at 00:36

Mark

39,605

Any idea why the requirement of R being integral domain is mentioned everywhere, e.g.: https://math.stackexchange.com/questions/146857/if-a-an-integral-domain-contains-a-field-k-and-a-over-k-is-a-finite-dime – Othman El Hammouchi Mar 17 '21 at 00:39
Why should that scalar multiplication give a vector space if $R$ annihilates $F$? See my answer. – Rob Arthan Mar 17 '21 at 00:42
@othi In your linked post they are not trying to prove that $R$ is a vector space over $F$, they are trying to prove that $R$ itself is a field. The latter proof does use finite dimensionality and the domain assumption, but it is completely different from what you are asking. – paul blart math cop Mar 17 '21 at 00:42
Not only did you answer 4 minutes faster than I did, you had the gall to also be named Mark. But you are absolutely correct. +1. – Mark Saving Mar 17 '21 at 00:42
@Rob Arthan A trivial vector space is also a vector space. – Mark Mar 17 '21 at 00:44
@paulblartmathcop I see. Well, in my notes it's just mentioned as a general requirement, which is why I found it strange. – Othman El Hammouchi Mar 17 '21 at 00:45
The ring $\Bbb{R}\times \Bbb{Z}_2$ is not in any useful way a vector space over $\Bbb{R}$. – Rob Arthan Mar 17 '21 at 00:46

score 1 · Answer 2 · answered Mar 17 '21 at 00:41

There is no reason $R$ has to be an integral domain at all. I'm assuming you're saying that $F$ is a subring of $R$ when you write $F \subset R$; it's irrelevant whether $F$ is a subset of $R$ when determining whether $R$ has an $F$-vector space structure.

The most general result is that whenever we have a ring homomorphism $i : F \to R$, $R$ can be given an $F$-vector space structure, where the additive structure is $R$'s additive structure and addition is given by

$f \cdot r = i(f) r$

We must check the axioms:

$f \cdot (r + s) = i(f) (r + s) = i(f) r + i(f) s = f \cdot r + f \cdot s$
$(f + q) \cdot r = i(f + q) r = (i(f) + i(q)) r = i(f) r + i(q) r = f \cdot r + q \cdot r$
$1 \cdot r = i(1) r = 1r = r$
$(fq) \cdot r = i(fq) r = i(f) i(q) r = i(f) (q \cdot r) = f \cdot (q \cdot r)$

So $R$ is an $F$-vector space.

In the special case that $F$ is a subring of $R$, we take $i : F \to R$ to be the inclusion ring homomorphism. We see that there is no reason at all to demand $R$ be an integral domain.

Given a field $F\subset R$ contained in a ring which is not an integral domain. How can R fail to be a vector space over F?

2 Answers2