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Consider the field extension $L:K$, with finite degree, and consider $E$, $F$ two intermediate field.

Define, $$EF=\left\{ \sum_{1\leq i\leq n} x_iy_i; n\in \mathbb{N}^*, x_i\in E, y_i\in F, \ \forall i(1\leq i\leq n) \right\}.$$

Show that $EF$ is a subfield of $L$. (It's enough to show that it is a domain of integrity, there's a proposition that takes from there.) I am having trouble to show the properties of the multiplication, there's a easy way to do this?

Check that $EF$ is the sub field of $L$ generated by $E\cup F$.

121212
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  • Try to multiply $\left(\sum_{1\le i\le n}x_iy_i\right) \cdot\left(\sum_{1\le j\le m}z_jw_j\right)$. What do you get explicitly? The use the fact that both $E$ and $F$ are contained in the field $L$, which has associative and commutative multiplication. – Lios Oct 28 '20 at 16:48
  • But I can't associate and commute elements of $E$ with elements of $F$ can I?
    And to show that $EF$ is closed under subtraction I took two elements of $EF$ but I can't write the subtraction of the elements as an element of $EF$.     – 121212 Oct 29 '20 at 18:07
  • You can commute elements of $E$ with elements of $F$ since both fields are contained in $L$, which is (among the other things) a commutative ring. For the sottraction, notice that sottracting is equal to adding the opposite, therefore... – Lios Oct 29 '20 at 19:28
  • Yes, but I want to show that the addition or subtraction are closed, so I can't use that if I did not prove that is closed first – 121212 Oct 29 '20 at 19:45
  • Ok first moment I have tomorrow I will write an explicit answer – Lios Oct 29 '20 at 19:54
  • Ok, what I trying to show is that $EF$ is closed under adition and multiplication, then $EF$ will be a ring. Thanks :) – 121212 Oct 29 '20 at 22:21

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The key fact on which this exercise is based is that the set $EF$ is contained in $L$, a field, therefore we can nicely operate on its elements. Indeed, an element of the form $\sum_i x_iy_i=x_1y_1+\cdots+x_ny_n$ is well defined since for every $i$, $x_i\in E\subset L$, $y_i\in F\subset L$, and in $L$ we can sum and multiply.

Step 1: $EF$ is an abelian group. The addition is the following: $$(x_1y_1+\cdots+x_ny_n)+(z_1w_1+\cdots+z_mw_m)=x_1y_1+\cdots+x_ny_n+z_1w_1+\cdots+z_mw_m.$$ All this is happening in $L$. Clerly we get an element again in $EF$, and this operation is trivially associative. The neutral element is $0\in L$, it belongs to $EF$ since for example it can be written as $0_E\cdot 0_F$, and the opposite of $\sum_i x_iy_i=x_1y_1+\cdots+x_ny_n$ is $\sum_i -x_iy_i=-x_1y_1-\cdots-x_ny_n\in EF$. Indeed: $$x_1y_1+\cdots+x_ny_n+(-x_1y_1-\cdots-x_ny_n)=x_1y_1-x_1y_1+\cdots+x_ny_n-x_ny_n=0,$$ where again this is happening in $L$, which is commutative as additive group. We conclude that $EF$ has a natural structure of abelian group.

Step 2: EF is a ring. The product is defined using the product of $L$ and the distributive property. To lighten the notation, I make a simple example: $$(x_1y_1+x_2y_2)\cdot(z_1w_1+z_2w_2+z_3w_3)=x_1y_1z_1w_1+x_1y_1z_2w_2+x_1y_1z_3w_3+x_2y_2z_1w_1+x_2y_2z_2w_2+x_2y_2z_3w_3.$$ Again, this is happening in $L$, a commutative ring, so we can commute elements of the form $x_iy_iz_jw_j$, writing $$x_iy_iz_jw_j=x_iz_jy_iw_j$$ and deducing that $EF$ is closed under this product. The identity element is $1\in L$, we can write it as $1_E\cdot 1_F$, and as for the sum, associativity (and also distributivity) follow straightforwardly (try to do it yourself if you are not convinced).

Step 3: now $EF$ is clearly an integral domain, being a subring of a field. Can you conclude (with your proposition) that $EF$ is actually a subfield? (here you surely need that the two extensions are finite and therefore algebraic over $K$, otherwise it is not always true that $EF$ is a field. See for example Is the composite of two subfields just their product?).

Step 4: $EF$ is the field generated by $E\cup F$, that is, it is the smallest subfield of $L$ containing $E\cup F$. Indeed, it is a field, it contains both, so it contains the union, and if $\Omega$ is a field containing $E\cup F$, then it contains all the possibile sums $x_1y_1+\cdots+x_ny_n$, so it has to contain $EF$.

I hope this is clear.

Lios
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