Question

Let $\left\{E_i|i\in l\right\}$ be a family of subfields of K. Prove that$\underset{\mathbf{i}\mathbf{\in }\mathbf{I}}{\mathbf{\cup }{\mathbf{E}}_{\mathbf{i}}}$ is a subfield of K.

Short answer

It is proved that $\underset{i\in l}{{\cup E}_i}$is a subfield of K.

Step-by-step solution

Definition of subfield

A subfield is a subset of a mathematical field that is itself a field. It is a subdivision of a field.

Prove that ∪Eii∈lis a subfield of K

Let $\mu =\left\{E_i|i\in I\right\}$be a family of subfield of K .

Claim that intersection of family of subfield is a subfield of K .

That means$\underset{i\in l}{{\cup E}_i}$ is subfield of K .

Let $P=\underset{i\in l}{{\cup E}_i}$.

Since$0€E$ for all$i_l\in l$ . So,role="math" localid="1659157104382" $0\in \underset{i\in l}{{\cup E}_i}=P$ .

Therefore, P is non-empty.

Let$a,b\in P$ , then$a,b\in E$ for all$E\in \mu$ .

This implies that their difference also belongs to E as follows.

$a-b\in E$for all$E\in \mu$

So,

$\underset{}{a-b\in \underset{i\in l}{\cup }{E}_i}$

Thus,

$a-b=P$

And,

$a,b\in E,b\ne 0$

This implies,

role="math" localid="1659156256742" $a{b}^{-1}\in E$for all$E\in \mu$

So,

$a{b}^{-1}\underset{i\in l}{\in {\cup E}_i}$

Thus,

$a{b}^{-1}=P$

Therefore, the set P is a subfield of K.

Hence, it is proved thatrole="math" localid="1659156595040" $\underset{i\in l}{{\cup E}_i}$ is a subfield of K.