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

1. Feb 10, 2013

### jdinatale

I wasn't

Last edited: Feb 10, 2013
2. Feb 10, 2013

### jbunniii

Re: Showing that the quotient space is isomorphic to the field it's ov

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.

3. Feb 10, 2013

### Dick

Re: Showing that the quotient space is isomorphic to the field it's ov

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.

Last edited: Feb 10, 2013
4. Feb 10, 2013

### jdinatale

Re: Showing that the quotient space is isomorphic to the field it's ov

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

5. Feb 10, 2013

### Dick

Re: Showing that the quotient space is isomorphic to the field it's ov

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

6. Feb 10, 2013

### jdinatale

Re: Showing that the quotient space is isomorphic to the field it's ov

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?

I'm really not sure why this is true.

7. Feb 10, 2013

### jbunniii

Re: Showing that the quotient space is isomorphic to the field it's ov

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 $u - v = k \in ker(f)$.

Last edited: Feb 10, 2013
8. Feb 10, 2013

### Dick

Re: Showing that the quotient space is isomorphic to the field it's ov

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.

9. Feb 10, 2013

### jdinatale

Re: Showing that the quotient space is isomorphic to the field it's ov

I think I proved it using Dick's ideas!

10. Feb 10, 2013

### jbunniii

Re: Showing that the quotient space is isomorphic to the field it's ov

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.

Last edited: Feb 10, 2013
11. Feb 10, 2013

### jbunniii

Re: Showing that the quotient space is isomorphic to the field it's ov

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

12. Feb 10, 2013

### Dick

Re: Showing that the quotient space is isomorphic to the field it's ov

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

13. Feb 10, 2013

### jdinatale

Re: Showing that the quotient space is isomorphic to the field it's ov

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.

14. Feb 10, 2013

### jbunniii

Re: Showing that the quotient space is isomorphic to the field it's ov

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 $u - v$?

15. Feb 10, 2013

### jdinatale

Re: Showing that the quotient space is isomorphic to the field it's ov

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.

16. Feb 10, 2013

### Dick

Re: Showing that the quotient space is isomorphic to the field it's ov

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?

17. Feb 10, 2013

### jbunniii

Re: Showing that the quotient space is isomorphic to the field it's ov

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

18. Feb 10, 2013

### jdinatale

Re: Showing that the quotient space is isomorphic to the field it's ov

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