2
$\begingroup$

Let $L/K$ be a field extension.

I want to show that $$[L:K]=1 \Leftrightarrow L=K$$

$$$$

I have done the following:

For the direction $\Rightarrow \ : $

Since $[L:K]=1=\text{dim}_KL$ we have that there exist $a\in L$ with $\langle a\rangle$ a $K$-basis of $L$.

So, let $\ell\in L$, then $$\ell=ak, k\in K$$

To get the desired result, can we just take $a=1$ ?

$$$$

Could you ive me a hint for the other direction?

$\endgroup$
1
$\begingroup$

To get the desired result, can we just take $a = 1$?

Yes, you are using that in a one-dimensional vector space, any non-zero vector gives a basis.

Can you give me a hint for the other direction?

You have to show that $K$ is one-dimensional as a vector space over itself.

$\endgroup$
  • $\begingroup$ Why does it hold that in a one-dimensional vector space any non-zero vector gives a basis? $\endgroup$ – Mary Star Jul 8 '16 at 10:11
  • 1
    $\begingroup$ The definition of "dimension" is the number of vectors in any basis. These are non-zero, because any set that contains 0 is linearly dependent, and a basis is a linearly independent set. $\endgroup$ – David Wheeler Jul 8 '16 at 11:13
  • $\begingroup$ So, in this case the basis consists of any one element of $L$ ? @DavidWheeler $\endgroup$ – Mary Star Jul 8 '16 at 11:28
  • $\begingroup$ Any non-zero element. $\endgroup$ – David Wheeler Jul 8 '16 at 22:27
0
$\begingroup$

Show first arrow:
$\Rightarrow)$ If $L/K $ has dimension $1$ we know that $K/(p(x)) $ has dimension $1$ as $K$- vector space so necessarily the degree of $p(x)$ must be $1$ (I'm considering the homomorphism of valutation: $\psi : K[x]\to L $ such that $\psi (p)= p(\alpha )\space \forall \space p \in K[x]$ with $ \alpha \in L $).
Therefore $K/(p(x))=K$ and being $K/(p(x))\cong L$, $K\cong L$.
(We can say are equal).

$\Leftarrow)$ Obvious.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.