Showing that the quotient space is isomorphic to the field it's over

[jdinatale]

I wasn't

 

[jbunniii]

What if you start by defining $\phi : V/Ker(f) \rightarrow F$ by $\phi(v + Ker(f)) = f(v)$. Can you show that (1) $\phi$ is well defined, (2) $\phi$ is a linear map; (3) $\phi$ is an injection; (4) $\phi$ is a surjection.

 

[Dick]

If V is really a one dimensional space then there is a basis consisting of a single vector {v}. If f is a linear function then either ker(f)={0} (if f(v) is nonzero) or ker(f)=V (if f(v)=0). Since f is specified to be nonzero ker(f)=V and the quotient space is {0}. I think they likely mean two dimensional.

 

[jdinatale]

What if you start by defining $\phi : V/Ker(f) \rightarrow F$ by $\phi(v + Ker(f)) = f(v)$. Can you show that (1) $\phi$ is well defined, (2) $\phi$ is a linear map; (3) $\phi$ is an injection; (4) $\phi$ is a surjection.

I'm not sure that it's true that it's well defined because it's not necessarily true that f(u) = f(v):

 

[Dick]

I'm not sure that it's true that it's well defined because it's not necessarily true that f(u) = f(v):

If u+ker(f)=v+ker(f) then u-v is in ker(f).

 

[jdinatale]

If V is really a one dimensional space then there is a basis consisting of a single vector {v}. If f is a linear function then either ker(f)={0} (if f(v) is nonzero) or ker(f)=V (if f(v)=0). Since f is specified to be nonzero ker(f)=V and the quotient space is {0}. I think they likely mean two dimensional.

I'm pretty sure the f \not= \mathbf{0}_{V \rightarrow f} means that the function isn't f(v) = 0 for all v, but I'm pretty sure that some vector v could have f(v) = 0, right? My book defines f \not= \mathbf{0}_{V \rightarrow f} as the zero map. Couldn't the single basis vector v map to zero?

If u+ker(f)=v+ker(f) then u-v is in ker(f).

I'm really not sure why this is true.

 

[jbunniii]

I'm really not sure why this is true.

What does it mean if $u + ker(f) = v + ker(f)$? It means that $u$ and $v$ are in the same coset of $ker(f)$. Therefore, $u \in v + ker(f)$, so there exists some $k \in ker(f)$ such that $u = v + k$. Thus $u - v = k \in ker(f)$.

 

[Dick]

I'm pretty sure the f \not= \mathbf{0}_{V \rightarrow f} means that the function isn't f(v) = 0 for all v, but I'm pretty sure that some vector v could have f(v) = 0, right? My book defines f \not= \mathbf{0}_{V \rightarrow f} as the zero map. Couldn't the single basis vector v map to zero?

If the single basis vector maps to zero then all vectors map to zero. In a one dimensional space every vector is a multiple of the basis vector.

 

[jdinatale]

I think I proved it using Dick's ideas!

 

[jbunniii]

Since $\mathbb{F}$ has dimension one, let $\{v\}$ be a basis for $V$.

Nowhere does it say that $V$ has dimension one, so "let $\{v\}$ be a basis for $V$" is an invalid step. All it says is that you can consider $\mathbb{F}$ as a one-dimensional vector space over itself, but this is always true for any field. They're just saying that so it's clear that they are looking for an isomorphism between vector spaces.

 

[jbunniii]

Did you try the approach I listed earlier? It gives you an explicit construction for an isomorphism between $V/ker(f)$ and $F$:

What if you start by defining $\phi : V/Ker(f) \rightarrow F$ by $\phi(v + Ker(f)) = f(v)$. Can you show that (1) $\phi$ is well defined, (2) $\phi$ is a linear map; (3) $\phi$ is an injection; (4) $\phi$ is a surjection.

 

[Dick]

Nowhere does it say that $V$ has dimension one, so "let $\{v\}$ be a basis for $V$" is an invalid step. All it says is that you can consider $\mathbb{F}$ as a one-dimensional vector space over itself, but this is always true for any field. They're just saying that so it's clear that they are looking for an isomorphism between vector spaces.

Ohhhh. That's the part I was reading wrong. Now the question makes more sense.

 

[jdinatale]

Did you try the approach I listed earlier? It gives you an explicit construction for an isomorphism between $V/ker(f)$ and $F$:

Yes, and thank you for the advice. I'm now at the part where I show phi is one-to-one.

I'm not quite sure that it is. We can show one-to-one in three ways.

Suppose \phi(u + ker(f)) = \phi(v + ker(f)). Then f(u) = f(v). We have to show that u = v, which is only true if f is one-to-one, which we don't necessarily know.

Or we can suppose that u + ker(f) != v + ker(f). Then we have to show that f(u) != f(v)

Or we can show that Ker(phi) = {0}. But I'm not even sure that phi(0 + Ker(f)) = 0, because f(0) might not be 0.

 

[jbunniii]

Yes, and thank you for the advice. I'm now at the part where I show phi is one-to-one.

I'm not quite sure that it is. We can show one-to-one in three ways.

Suppose \phi(u + ker(f)) = \phi(v + ker(f)). Then f(u) = f(v). We have to show that u = v, which is only true if f is one-to-one, which we don't necessarily know.

No, we don't need $u = v$. If $f(u) = f(v)$, then $f(u) - f(v) = 0$. Now $f$ is linear, so this means that $f(u - v) = 0$. Therefore what can you say about $u - v$?

 

[jdinatale]

No, we don't need $u = v$. If $f(u) = f(v)$, then $f(u) - f(v) = 0$. Now $f$ is linear, so this means that $f(u - v) = 0$. Therefore what can you say about $u - v$?

yes, I understand now, thanks. But the last part is tricky - showing that phi is onto. That requires that f(v) be onto, which we don't know. I let w be an arbitrary element of the field F. We have to find a coset (v + Ker(f)) such that phi[(v + Ker(f)] = w.

 

[Dick]

yes, I understand now, thanks. But the last part is tricky - showing that phi is onto. That requires that f(v) be onto, which we don't know. I let w be an arbitrary element of the field F. We have to find a coset (v + Ker(f)) such that phi[(v + Ker(f)] = w.

You know there is some vector such that f(v) is nonzero. What kinds of values can f(cv) take as c ranges over the elements of your field?

 

[jbunniii]

You know there is some vector such that f(v) is nonzero. What kinds of values can f(cv) take as c ranges over the elements of your field?

Or, what amounts to the same thing, what can you say about the dimension of the image of $f$?

 

[jdinatale]

Thanks for the hints jbunniii and Dick, I've now completed the problem