To define the concept of a subfield of a field and prove the stated property regarding subfields of a field.

A subfield of a field L is a subset K of L , which is also a field, with field structure inherited from L.

Let L be a field.

Definition:

A subset \(\displaystyle{K}\subset{L}\) is a subfield of L if K is a field, with the same addition and multiplcation operations as in L.

Examples:

1)\(\displaystyle\mathbb{Q},\mathbb{R}\) are subfields of \(\displaystyle\mathbb{R},\mathbb{C}\) respectively.

2)\(\displaystyle\mathbb{Q}{\left(\sqrt{{p}}\right)},{p}{p}{r}{i}{m}{e},\) is a subfield of RRZSK.

Let L be a field.

Let \(\displaystyle{\left\lbrace{K}_{{i}},{i}\in{I}\right\rbrace}\) be any collection of subfields \(\displaystyle{K}_{{i}}\) of L.

Claim: \(\displaystyle{K}=\bigcap_{{{i}\in{I}}}{K}_{{i}}\) is a field.

Proof: \(\displaystyle{K}\subset{L},{a}{s}{K}_{{i}}\subset{L}\forall{i}\in{I}.\)

\(\displaystyle{x},{y}\in{K}\Rightarrow{x}+{y},{x}-{y},{0},{1},{x}{y}\in{K}_{{i}},\forall{i}\in{I}\)

\(\displaystyle\Rightarrow{x}+{y},{x}-{y},{0},{1},{x}{y}\in\bigcap_{{{i}\in{I}}}{K}_{{i}}\)

\(\displaystyle\Rightarrow{x}+{y},{x}-{y},{0},{1},{x}{y}\in{K}\)

So, K is a commutive ring with 1

ANSWER: proved that the intersection of any collection of subfields of a field L is indeed a field, (in fact , a subfield of L)

Also, \(\displaystyle{x}\in{K},{x}\ne{0},\Rightarrow{x}\in{K}_{{i}},\forall{i}\in{I},{x}\ne{0}\)

rArr \(\displaystyle{x}^{{-{{1}}}}\in{K}_{{i}},\forall{i}\in{I}\) (as \(\displaystyle{K}_{{i}}\) are all fields)

rArr \(\displaystyle{x}^{{-{{1}}}}\in\bigcap_{{{i}\in{I}}}{K}_{{i}}={K}\)

Thus, K is a field, in fact, K is a subfield of L

A subfield of a field L is a subset K of L , which is also a field, with field structure inherited from L.

Let L be a field.

Definition:

A subset \(\displaystyle{K}\subset{L}\) is a subfield of L if K is a field, with the same addition and multiplcation operations as in L.

Examples:

1)\(\displaystyle\mathbb{Q},\mathbb{R}\) are subfields of \(\displaystyle\mathbb{R},\mathbb{C}\) respectively.

2)\(\displaystyle\mathbb{Q}{\left(\sqrt{{p}}\right)},{p}{p}{r}{i}{m}{e},\) is a subfield of RRZSK.

Let L be a field.

Let \(\displaystyle{\left\lbrace{K}_{{i}},{i}\in{I}\right\rbrace}\) be any collection of subfields \(\displaystyle{K}_{{i}}\) of L.

Claim: \(\displaystyle{K}=\bigcap_{{{i}\in{I}}}{K}_{{i}}\) is a field.

Proof: \(\displaystyle{K}\subset{L},{a}{s}{K}_{{i}}\subset{L}\forall{i}\in{I}.\)

\(\displaystyle{x},{y}\in{K}\Rightarrow{x}+{y},{x}-{y},{0},{1},{x}{y}\in{K}_{{i}},\forall{i}\in{I}\)

\(\displaystyle\Rightarrow{x}+{y},{x}-{y},{0},{1},{x}{y}\in\bigcap_{{{i}\in{I}}}{K}_{{i}}\)

\(\displaystyle\Rightarrow{x}+{y},{x}-{y},{0},{1},{x}{y}\in{K}\)

So, K is a commutive ring with 1

ANSWER: proved that the intersection of any collection of subfields of a field L is indeed a field, (in fact , a subfield of L)

Also, \(\displaystyle{x}\in{K},{x}\ne{0},\Rightarrow{x}\in{K}_{{i}},\forall{i}\in{I},{x}\ne{0}\)

rArr \(\displaystyle{x}^{{-{{1}}}}\in{K}_{{i}},\forall{i}\in{I}\) (as \(\displaystyle{K}_{{i}}\) are all fields)

rArr \(\displaystyle{x}^{{-{{1}}}}\in\bigcap_{{{i}\in{I}}}{K}_{{i}}={K}\)

Thus, K is a field, in fact, K is a subfield of L