as an offering

Today, I was feeling unwell. It all started the day before yesterday when I had a terrible experience on Twitter. I was answering a question from an undergraduate student when someone else interfered and got in the way. To make matters worse, someone retweeted it, adding insult to injury. Yesterday, I made a casual comment on Twitter that unexpectedly went viral, which was exhilarating at the moment. However, upon reflection, it seemed meaningless. This is because even if I followed up with deeper replies, no one responded. It appears that people are only enjoying the momentary thrill, which absolutely prevents the systematic accumulation of knowledge. Disheartened by this realization, I decided to check out Math Stack Exchange. But I encountered problems there as well. While answering a question about algebra, I realized halfway through that my answer was flawed, which was discouraging. My clumsiness makes me think I should be more careful when answering other people's questions. I begin to doubt my ability to solve unknown problems if I can't even solve such simple ones.

However, my attitude today towards not bothering with questions that aren't worth contemplating might be considered a form of pride. Perhaps I should just treat the problems I do want to think about as part of my job as a TA and tackle them with determination.

Here, I'll leave my attempted solution to the problem as an offering. Oh, how embarrassing.

供養1

When $K/F$ is Galois, $\alpha \in F$ is equivalent to $\forall \sigma \in \text{Gal}(K/F), \sigma(\alpha)=\alpha$.

Let start with the case $K/F$ is quadratic and Galois.

$\forall{\alpha}\in \text{Gal}(K/F), \sigma(\alpha+\sigma(\alpha))=\sigma \alpha+{\sigma}^2\alpha=\sigma \alpha +\alpha$.

Thus $\alpha+\sigma \alpha \in F$ by the above equivalence.

If $K/F$ is not Galois, $\forall \sigma \in \text{Aut}(K/F), \sigma(\alpha)=\alpha$ does not always imply $\alpha \in F$.

So we take Galois closure $L/F$.

I leave the general case to you.

In response to the comment : $\sigma$ runs finite candidate. If we let call it $\sigma_1,\sigma_2,・・・, \sigma_n$, notice that $\{\sigma_1,\sigma_2,・・・,\sigma_n\}=\{\tau \sigma_1,・・・,\tau \sigma_n\}$ as a set for all $\tau \in \text{Gal}(K/F)$. Then the total sum is the same.

供養２

Let $a_1,...,a_n\in E$ be a basis of $E/F$.

Let $b_1,...,b_m \in K$ be a basis of $K/F$.

If $K=E$, there is nothing to prove.

When $E$ is a proper subspace of $K$, we want to prove $m >n $.

Take $k\in K-E\neq \emptyset$, then $a_1,...,a_n,k\in K$ are linearly independent. This is just because if they are dependent, $k$ should be written as a linear combination of $a_1,...,a_n$, and $k\in E$.

Thanks to an additional element $k\in K$,we realize that basis of $K/F$ should consists of more than $a_1,...,a_n,k$'s $n+1$ elements. Thus  $m\ge n+1$ holds and we are done.