Smallest Field containing Subfield and Complex Number/Examples/Field Containing Rationals, Root 2 and Root 3
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. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
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
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Field Extensions: $\S 36$. The Degree of a Field Extension: Example $73$