# Properties of composition of fields extensions in Galois Theory. [duplicate]

This question already has an answer here:

Problem. Let $K$ and $L$ be subfields of a common field, both which contain a field $F$. Prove the following statements.

(a) If $K = F(X)$ for some set $X \subset K$, then $KL = L(X)$.

(b) $[KL:F] \leq [K:F][L:F]$.

(c) If $K$ and $L$ are algebraic over $F$, then $KL$ is algebric over $F$.

(d) Prove that the previous statement remains true when $$algebraic$"$ is replaced by $$normal$"$, $$separable$"$, $$purely inseparable$"$, or $$Galois$"$.

For (a). If $K = F(X)$, since $F \subset L$ and $KL = K(L) = L(K)$ we should have $KL = L(X)$.

For (b). I tried two different ways:

1. Take $\lbrace \alpha_{i} \rbrace$ a basis for $K$ and $\lbrace \beta_{j} \rbrace$ a basis for $L$. How to find a basis for $KL$ from these? The product generates $KL$? I still get a little confused with the $$appearance$"$ of elements of $KL$ (maybe, this is a problem in (c)).

2. Write $[KL:F] = [KL:L][L:F]$ and try to show that $[KL:L] \leq [K:F]$. Intuitively this seems right, but I couldn't prove

For (c). Take $\alpha \in KL$, so exists $k_{1},...k_{n}; k'_{1},...,k'_{m};l_{1},...,l_{n};l'_{1},...,l'_{m}$ such that $$\alpha = \frac{k_{1}l_{1} + ... + k_{n}l_{n}}{k'_{1}l'_{1} + ... + k'_{m}l'_{m}}.$$ Since $k_{i}, k'_{j} \in K$ and $l_{i}, l'_{j} \in K$, they are algebraic, then $\alpha$ is algebraic, because the set of algebraics is a field. This make sense?

For (d). I have not tried it yet. Since I have not resolved the previous ones, I do not think it makes sense to try to.

I think my difficulty in this problem is due to the fact that I have not understood very well how to explicitly say who the elements of $KL$.

I would like any hint! Thanks for the advance.

## marked as duplicate by Brahadeesh, Jyrki Lahtonen abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 22 '18 at 12:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Hint for part (b): Prove that the $F$-span of product $\alpha_i\beta_j$, $i,j$ independent, is also closed under multiplication. Call that span $M$. Show that it is an integral domain, and apply this. You also need to show that $M$ is the smallest field containing both $K$ and $L$. – Jyrki Lahtonen May 20 '18 at 5:32