K is a field. Are the following properties always true?

  1. For every field extension L|K with $[L:K]\in\{1,\infty\}$, K is algebraically closed

  2. K algebraically closed, so for every field extension L|K we have $[L:K]\in\{1,\infty\}$

I dont see how to solve that and why it should be true at all. Can anyone help me there?

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    $\begingroup$ As stated, these are false, because they are asserting that every field is algebraically closed... $\endgroup$ – Eric Wofsey Feb 2 '18 at 6:30
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    $\begingroup$ Neither statement is stated very clearly. $\endgroup$ – carmichael561 Feb 2 '18 at 6:31
  • $\begingroup$ @EricWofsey what do you mean? can you elaborate why they are false? Or do you mean i phrased it poorly $\endgroup$ – SquareJoe Feb 2 '18 at 6:36

Restated, it appears your question is how to prove the following statement:

A field K is algebraically closed if and only if the only field $L$ which is a finite extension of $K$ is $L=K$.

The proof is very easy . . .

The forward direction:

Assume $K$ is algebraically closed, and $L$ is a finite extension of $K$.

But a finite extension of a field is an algebraic extension, hence $L$ is algebraic over $K$.

Since $K$ is algebraically closed, $L \subseteq K$.

But $L$ is an extension of $K$, so $K \subseteq L$.

Therefore $L=K$.

The reverse direction:

Let $K$ be a field, and suppose the only field $L$ which is a finite extension of $K$ is $L=K$.

Let $\bar{K}$ be an algebraically closed field containing $K$.

Of course $K \subseteq \bar{K}$.

For the reverse inclusion, let $a \in \bar{K}$.

Then $a$ is algebraic over $K$, so $a$ is a root of some polynomial $p \in K[x]$, of degree $n$, say.

Letting $L=K(a)$, it follows that $[L:K] \le n$, so $L$ is a finite extension of $K$, hence $L = K$, so $a \in K$.

Thus, $\bar{K} \subseteq K$, and hence $K = \bar{K}$.

Therefore $K$ is algebraically closed.

  • $\begingroup$ Im a bit confused. Are you sure this proves the original 2 points? What confuses me the most is that you dont talk about $[L:K]=\infty$ ($[L:K]=\{1,\infty\}$) $\endgroup$ – SquareJoe Feb 2 '18 at 7:36
  • $\begingroup$ @SquareJoe: $[L:K]=\infty$ is equivalent to saying "$L$ is an extension of $K$, but not a finite extension of $K$", and $[L:K]=1$ is equivalent to saying "$L=K$". Thus, saying $[L:K]\in \{1,\infty\}$ is the same as saying "If $L$ is a finite extension of $K$, then $L=K$". $\endgroup$ – quasi Feb 2 '18 at 7:44
  • $\begingroup$ and what if $[L:K]=\infty$? is it still algebraic then? $\endgroup$ – SquareJoe Feb 2 '18 at 8:01
  • $\begingroup$ Oh my godness. I completely misunderstood. This explains why i struggled so much finding a solution. Thank you a lot! $\endgroup$ – SquareJoe Feb 2 '18 at 8:12
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    $\begingroup$ Statement 1 (recast): If for all extensions $L$ of $K$, we have $[L:K]\in\{1,\infty\}$, then $K$ is algebraically closed. $$$$ Statement 2 (recast): If $K$ is algebraically closed, then for all extensions $L$ of $K$, we have $[L:K]\in\{1,\infty\}$. $\endgroup$ – quasi Feb 2 '18 at 8:17

For (2), by contradiction, if there is some field $L$ such that $[L:K]$ is finite (but not $1$), then all elements in $L$ satisfy a polynomial with finite degree. But if $K$ is algebraically closed then the polynomial be solved in $K$ and hence $L=K$ and $[L:K]=1$.

For (1), if we have a polynomial $f(x)$, and its roots are not all in $K$, then there are some element $\alpha$ not in $K$ such that it satisfies this polynomial so that we can write a new extension $K[\alpha]/K$ which has finite degree (but not $1$ since $\alpha\notin K$.

The proofs above are not very rigorous but I guess it is a good draft for how to solve them. For comments above that stating that those statements are false, I would like to know what additional assumption I have made in proofs above. Thank you very much.


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