existence of some algebraic closure

Question

I was curious somethings when I studied algebraic closure in Hungerford's algebra.

I know that every field has an algebraic closure. But I have the following question.

Let $K$ be a field and $F$ an algebraic extension of $K$. Then is there an algebraic closure of $K$ containing $F$?

I know that its answer is "yes", but I have difficulty to say its reason logically.

Roughly, I can only say that 'because every field has an algebraic closure, and any two algebraic closures of $K$ are $K$-isomorphic'. Is my answer reasonable? If not, please tell me the correct answer. Thanks.

Answer 1

Your answer is on the right track. Consider the algebraic closure $F'$ of $F$. Then $K \subset F \subset F'$, and since $F'$ is algebraic over $F$ and $F$ is algebraic over $K$, $F'$ is also algebraic over $K$. Moreover, $F'$ is algebraically closed. This shows that $F'$ is an algebraic closure for $K$ which contains $F$. Because the algebraic closure is unique up to isomorphism, we can say that $F'$ is contained in the algebraic closure of $K$. That is, given any other algebraic closure $K'$ of $K$, we can embed $F$ into $K'$ by restricting an isomorphism between $F'$ and $K'$.

Answer 2

Take an algebraic closure $\overline F$ of $F$. Then $\overline F$ is algebraic over $K$ and is algebraically closed, hence an algebraic closure of $K$