Smallest Field containing Subfield and Complex Number/Examples/Field Containing Rationals, Root 2 and Root 3

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Example of Smallest Field containing Subfield and Complex Number

The smallest field containing $\Q$, $\sqrt 2$ and $\sqrt 3$ is:

$\set {a + b \sqrt 2 + c \sqrt 3 + d \sqrt 6: a, b, c, d \in \Q}$

This forms a vector space of dimension $4$ which has basis $\set {1, \sqrt 2, \sqrt 3, \sqrt 6}$.


Proof

Let $K := \Q \sqbrk {\sqrt 2}$ denote the set:

$K := \set {a + b \sqrt 2: a, b \in \Q}$

that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rational numbers.

It is immediately apparent that $\sqrt 3 \notin K$.


Let $K'$ be a field containing $K$ and $\sqrt 3$.

Then $K'$ must contain all real numbers of the form:

$\paren {a + b \sqrt 2} \paren {c + d \sqrt 2} \sqrt 3$

where $a, b, c, d \in \Q$.

This can be written:

$\set {a + b \sqrt 2 + c \sqrt 3 + d \sqrt 6}$

and real numbers of this form do actually form a field.


This needs considerable tedious hard slog to complete it.
In particular: Demonstrate that they form a field. Proof of inverses for multiplication is tricky.
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Let $L := \Q \sqbrk {\sqrt 3}$ denote the set:

$L := \set {a + b \sqrt 3: a, b \in \Q}$

that is, all numbers of the form $a + b \sqrt 3$ where $a$ and $b$ are rational numbers.

Let $L'$ be a field containing $L$ and $\sqrt 2$.

Then $L'$ must contain all real numbers of the form:

$\paren {a + b \sqrt 3} \paren {c + d \sqrt 3} \sqrt 2$

where $a, b, c, d \in \Q$.

This can be written:

$\set {a + b \sqrt 3 + c \sqrt 2 + d \sqrt 6}$

which is the same set as before.

That is:

$K \sqbrk {\sqrt 2} = L \sqbrk {\sqrt 3}$

But a field containing $\sqrt 2$ and $\sqrt 3$ must also contain $\sqrt 6$.

Thus such a field contains all numbers of the form $a + b \sqrt 2 + c \sqrt 3 + d \sqrt 6$.

As these form a field, this is the smallest field containing $\Q$, $\sqrt 2$ and $\sqrt 3$.

Let this be denoted $\Q \sqbrk {\sqrt 2, \sqrt 3}$.

So:

$\Q \sqbrk {\sqrt 2, \sqrt 3} = K \sqbrk {\sqrt 3}$

We have that:

$\index {K \sqrt 3} K = 2$

and from Smallest Field containing Subfield and Complex Number: Numbers of Type $a + b \sqrt 2: a, b \in \Q$:

$\index K Q = 2$

So by Degree of Field Extensions is Multiplicative:

$\index {\Q \sqbrk {\sqrt 2, \sqrt 3} } \Q = 4$

and it is seen that $\set {1, \sqrt 2, \sqrt 3, \sqrt 6}$ forms a basis for $\Q \sqbrk {\sqrt 2, \sqrt 3}$.

By definition of basis and dimension, it is seen that $\map \dim {\Q \sqbrk {\sqrt 2, \sqrt 3} } = 4$.

$\blacksquare$


Sources

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