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If the triple is defined over the integers, why would you allow 1/3 as a scalar in the first place. By this logic, should R not be a vector space, as it is not

If the triple is defined over the integers, why would you allow 1/3 as a scalar in the first place. By this logic, should R not be a vector space, as it is not closed under scalar multiplication by i, or Q not be a vector space, as it is not closed under scalar multiplication by sqrt(2)?

skh 8y ago
Something being sidestepped in the post you responded to is that what is being talked about is a valid algebraic object with lots of structure to it. It’s called a module which you can think of as a sort of vector space. It’s just that the scalars may not have the property that they have multiplicative inverses. (I’m deliberately focusing on rings that are integral domains for the nitpickers.). When talking about these objects you have to include the underlying scalar set. For instance the real numbers are a vector space over the rationals. They are a different vector space over the reals. They are not a vector space over the complex numbers and are not a vector space over the integers. But they are a module over the integers. But not a module over the complex numbers.
- throwawaymath 8y ago
  Everything you've said is true, but circling back to the example given, we still can't choose a scalar 1/n for integral n. Yes the integers are a ring, and yes you can define a module over a ring which generalizes a vector space. But the point being spoken to here is that the explanation is backwards: you can't choose 1/n from Z. Therefore you can't use it as a scalar, so you'd never even break closure in the vector space. The hypothesis doesn't work before you can engage that contradiction.
  - skh 8y ago
    You are correct and I wasn't trying to criticize what you wrote. gizmo686's post (the one I responded to) indicated a sense of insight into these issues. I wanted gizmo686 to feel justified in his/her thoughts. Namely, that what we call modules are natural objects and they look at feel like vectors spaces on the surface.
ajross 8y ago
> why would you allow 1/3 as a scalar in the first place. Because it's a definitional thing. A "scalar" is routinely defined as a real number, not an integer. And you're absolutely right that it makes no sense, which is the whole point of the multiple-choice question. Four of those answers are plausible, the other requires you to make assumptions (like a redefinition of scalar) not in the question as posed.
- gizmo686 8y ago
  Except that we have vector spaces with scalars that are not the reals all the time. For instance, consider this excerpt from the article: "Or perhaps they wouldn’t like A because the scalar field [the complex numbers] is the same as the set of vectors (unless, that is, they thought that the obvious scalars were the real numbers)." In this case, while there is an an acknowledgement that you could take the reals as your scalars, it is regarded as the secondary of the "natural" choices. Or, in my example, example, there is no way to view Q as a vector space over R, but it is clearly a vector space. There is an entire field of algebra (field theory), that relies on the fact that, for example, Q(sqrt(2)) is a 2 dimensional vectorspace over Q.
- skh 8y ago
  Consider the possibility that it does make sense but that you aren’t aware of why it makes sense. A vector space has much more structure than a module and the distinction is not unimportant. Also, scalars are not defined as a real number. Scalars are elements of the base field. When talking about a vector space one must always specify the base field. This is important and is the point of the problem in question. For instance the real numbers are a vector space over the real numbers and that vector space structure is different than the vector space structure of the real numbers as a vector space over the rational numbers.
- throwawaymath 8y ago
  Huh? A scalar is very specifically a field element by definition. This is why it's important to specify the field you're working with when you talk about a vector space - a scalar is not going to be a real or a complex if your field isn't R or C. If you've seen someone define a scalar as a real number, that's really only because they're informally stating their underlying field is R.

skh 8y ago

Something being sidestepped in the post you responded to is that what is being talked about is a valid algebraic object with lots of structure to it. It’s called a module which you can think of as a sort of vector space. It’s just that the scalars may not have the property that they have multiplicative inverses. (I’m deliberately focusing on rings that are integral domains for the nitpickers.). When talking about these objects you have to include the underlying scalar set. For instance the real numbers are a vector space over the rationals. They are a different vector space over the reals. They are not a vector space over the complex numbers and are not a vector space over the integers. But they are a module over the integers. But not a module over the complex numbers.

throwawaymath 8y ago
Everything you've said is true, but circling back to the example given, we still can't choose a scalar 1/n for integral n. Yes the integers are a ring, and yes you can define a module over a ring which generalizes a vector space. But the point being spoken to here is that the explanation is backwards: you can't choose 1/n from Z. Therefore you can't use it as a scalar, so you'd never even break closure in the vector space. The hypothesis doesn't work before you can engage that contradiction.
- skh 8y ago
  You are correct and I wasn't trying to criticize what you wrote. gizmo686's post (the one I responded to) indicated a sense of insight into these issues. I wanted gizmo686 to feel justified in his/her thoughts. Namely, that what we call modules are natural objects and they look at feel like vectors spaces on the surface.

ajross 8y ago

> why would you allow 1/3 as a scalar in the first place. Because it's a definitional thing. A "scalar" is routinely defined as a real number, not an integer. And you're absolutely right that it makes no sense, which is the whole point of the multiple-choice question. Four of those answers are plausible, the other requires you to make assumptions (like a redefinition of scalar) not in the question as posed.

gizmo686 8y ago
Except that we have vector spaces with scalars that are not the reals all the time. For instance, consider this excerpt from the article: "Or perhaps they wouldn’t like A because the scalar field [the complex numbers] is the same as the set of vectors (unless, that is, they thought that the obvious scalars were the real numbers)." In this case, while there is an an acknowledgement that you could take the reals as your scalars, it is regarded as the secondary of the "natural" choices. Or, in my example, example, there is no way to view Q as a vector space over R, but it is clearly a vector space. There is an entire field of algebra (field theory), that relies on the fact that, for example, Q(sqrt(2)) is a 2 dimensional vectorspace over Q.
skh 8y ago
Consider the possibility that it does make sense but that you aren’t aware of why it makes sense. A vector space has much more structure than a module and the distinction is not unimportant. Also, scalars are not defined as a real number. Scalars are elements of the base field. When talking about a vector space one must always specify the base field. This is important and is the point of the problem in question. For instance the real numbers are a vector space over the real numbers and that vector space structure is different than the vector space structure of the real numbers as a vector space over the rational numbers.
throwawaymath 8y ago
Huh? A scalar is very specifically a field element by definition. This is why it's important to specify the field you're working with when you talk about a vector space - a scalar is not going to be a real or a complex if your field isn't R or C. If you've seen someone define a scalar as a real number, that's really only because they're informally stating their underlying field is R.