Let $F$ be an extension field of $K$ and $u\in F$. How do we know that adjoining an element of F to K, makes $K(u)$ a field?
I know that $Q(\sqrt2)=\{a+b\sqrt2|a,b\in Q\}$ is a field, but in the general case are we guaranteed that $K(u)$ is a field?
How can this be proved
Thanks :D
Update: Following @GitGud
Let $S= \{\cap B_i|\forall i , K\cup\{u\}\subseteq B_i\}$ be the intersection of all subfields of F that contain $K∪\{u\}$.
Proof:
Since $0,1\in B_i \ \ \forall i\rightarrow 0,1\in S$
If $a \in S\rightarrow a\in B_i \ \ \forall i\rightarrow a^{-1},-a\in B_i \ \ \forall i \rightarrow a^{-1},-a\in S$
If $a,b \in S$then $a*b^{-1}\in B_i \ \ \forall i$ and $a-b\in B_i \forall i \rightarrow$ S is closed under sums, and multiplication. Thus $S$ is a subfield of $F$ and $S=K(u)$
Unfortunately I do not have enough reputation to comment, but I would say that $K(u)$ is typically defined as the smallest field containing $K \cup \{u\}$, so taking the intersection of all subfields of $F$ with this property is really the definition of "smallest" field with that property. Then all you have to do is show the intersection of subfields is a subfield, which is straightforward.
As for your question as to "What if there isn't a bigger field?" well, it depends on your context. For one, you can always consider the algebraic closure of a field to be a "bigger" field if yours isn't algebraically closed, which exists for any field by Zorn's lemma. If your field is algebraically closed, like $\mathbb{C}$, then $\mathbb{C}(u)$ can be thought of as a field extension where $u$ is some symbol defined to be transcendental over $\mathbb{C}$. Then this would be all rational functions in $u$ over $\mathbb{C}$. I would suggest reading about function fields if this interests you.