Question

Find the basis of the given extension field of .

(a) $\mathbf{Q}\left(\sqrt{5},i\right)$

(b)role="math" localid="1657946461951" $\mathbf{Q}\mathbf{(}\sqrt{\mathbf{5}}\mathbf{,}\sqrt{\mathbf{7}}\mathbf{)}$

(c)role="math" localid="1657946632953" $\mathbf{Q}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{,}\mathbf{)}$

(d)$\mathbf{Q}\mathbf{(}{}^{\mathbf{3}}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{,}\mathbf{)}$

Short answer

(a)$\left\{1,i,\sqrt{5},\sqrt{5i}\right\}$is the basis.

(b) $\left\{1,\sqrt{5},\sqrt{7},\sqrt{35}\right\}$is the basis

(c)$\left\{1,\sqrt{2},\sqrt{3},\sqrt{6},\sqrt{5},\sqrt{10},\sqrt{15},\sqrt{30}\right\}$is the basis.

(d)$\left\{1,{}^3\sqrt{2},{\left({}^3\sqrt{2}\right)}^2,\sqrt{3},{}^3\sqrt{2}.\sqrt{3},{\left({}^3\sqrt{2}\right)}^2.\sqrt{3}\right\}$ is the basis.

Step-by-step solution

Definition of minimal polynomial.

The monic polynomial $p\left(x\right)$ over a field of the least degree such that $p\left(\alpha \right)=0$ for an algebraic element $\alpha$ over a field.

Step-2: Showing that the basis is {1,i,5,5i}  .

(a)

To find the basis of $Q\left(\sqrt{5},i\right)$over Q . As $Q\subseteq Q\left(i\right)\subseteq Q\left(\sqrt{5,}i\right)$. This implies role="math" localid="1657948787491" $\left[Q\left(\sqrt{5},i\right):Q\left(i\right)\right]=2\left[Q\left(i\right):Q\right]=2$.

First as $\left[Q\left(i\right):Q\right]=2$ and $x^2+1$ is minimal polynomial of $i\in Q\left(i\right)$ over . Second as and is minimal polynomial of $\sqrt{5}\in \left[Q\left(\sqrt{5},i\right)\right]$ over Q and also $x^2-5$ is irreducible polynomial over Q because root of this polynomial does not belong to $Q\left(i\right)$.

That is $\sqrt{5}\notin Q\left(i\right)$. Therefore basis for $Q\left(\sqrt{5},i\right)$ over $Q$ are

$\left\{1,\sqrt{5}\right\}\times \left\{1,i\right\},\sqrt{5},\sqrt{5i}$.

Hence the required basis is $\left\{1,i,\sqrt{5},\sqrt{5i}\right\}$.

Showing that the basis is {1,5,7,35}  .

(b)

To find the basis of $Q\left(\sqrt{5},\sqrt{7}\right)$over Q . As $Q\subseteq Q\left(\sqrt{5}\right)\subseteq Q\left(\sqrt{5},\sqrt{7}\right)$.

This implies role="math" localid="1657950200835" $\left[Q\left(\sqrt{5},\sqrt{7}\right):Q\left(\sqrt{5}\right)\right]=2,\left[Q\left(\sqrt{5}\right):Q\right]=2$

First as $\left[Q\left(\sqrt{5},\sqrt{7}\right):Q\left(\sqrt{5}\right)\right]=2$ and $x^2-7$is minimal polynomial of $\sqrt{7}$over Q . Second as and is minimal polynomial of over .

Therefore basis for $Q$$\left(\sqrt{5},\sqrt{7}\right)$over $Q$ are

$\left\{1,\sqrt{7}\right\}\times \left\{1,\sqrt{5}\right\}=\left\{1,\sqrt{5},\sqrt{7},\sqrt{35}\right\}$

.

Hence the required basis is $\left\{1,\sqrt{5},\sqrt{7},\sqrt{35}\right\}$.

Step-4: Showing that the basis is {1,2,3,6,5,10,15,30} .

(c)

To find the basis of $Q\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$over Q .

As $Q\subseteq Q\left(\sqrt{2}\right)\subseteq Q\left(\sqrt{2},\sqrt{3}\right)\subseteq Q\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$. This implies

localid="1657951649895" $\left[Q\left(\sqrt{2},\sqrt{3},\sqrt{5}\right):Q\left(\sqrt{2},\sqrt{3}\right)\right]=2,\left[Q\left(\sqrt{2},\sqrt{3},\right):Q\sqrt{2}\right]=2\left[Q\left(\sqrt{2}\right):Q\right]=2$

First as $\left[Q\left(\sqrt{2},\sqrt{3},\sqrt{5}\right):Q\left(\sqrt{2},\sqrt{3}\right)\right]=2$and $x^2-5$is minimal polynomial of $\sqrt{5}$over localid="1657951997668" $Q\left(\sqrt{2},\sqrt{3}\right):Q\left(\sqrt{2}\right)=2$. Second as and is minimal polynomial of $\sqrt{3}$ over $Q\left(\sqrt{2}\right)$

Third as $\left[Q\left(\sqrt{2}\right):Q\right]=2$ and $x^2-2$ is minimal polynomial of $\sqrt{2}$ over$Q$ .

Basis of role="math" localid="1657952996573" $Q\left(\sqrt{2},\sqrt{3}\right)$over $Q$ is

$\left\{1,\sqrt{2}\right\}\times \left\{1,\sqrt{3}\right\}=\left\{1,\sqrt{2},\sqrt{3},\sqrt{6}\right\}$

Therefore basis of $Q\left(\sqrt{2},\sqrt{3},\sqrt{5}\right)$ over $Q$ is

role="math" localid="1657953421519" $\left\{1,\sqrt{2},\sqrt{3},\sqrt{6}\right\}\times \left\{1,\sqrt{5}\right\}=\left\{1,\sqrt{2},\sqrt{3},\sqrt{6},\sqrt{5},\sqrt{10},\sqrt{15},\sqrt{30}\right\}$

Hence the required basis is $\left\{1,\sqrt{2},\sqrt{3},\sqrt{6},\sqrt{5},\sqrt{10},\sqrt{15},\sqrt{30}\right\}$.

Step-5: Showing that the basis is  {1,23,232,3,23.3,232.3}

(d)

To find the basis $Q\left\{\sqrt[3]{2},\sqrt{3}\right\}$over Q. As $Q\subseteq Q\left(\sqrt{3}\right)\subseteq Q\left\{\sqrt[3]{2},\sqrt{3}\right\}$$Q\subseteq Q\left(\sqrt{3}\right)\subseteq Q\left\{\sqrt[3]{2},\sqrt{3}\right\}$ .

This impliesrole="math" localid="1657954664976" $\left[Q\left\{\sqrt[3]{2},\sqrt{3}\right\}:Q\left(\sqrt{3}\right)\right]=2\left[Q\left\{\sqrt{3}\right\}:Q\right]=2$ .

First as role="math" localid="1657954912428" $\left[Q\left\{\sqrt[3]{2},\sqrt{3}\right\}:Q\left(\sqrt{3}\right)\right]=2$ and $x^2-2$ is minimal polynomial of $\sqrt[3]{2}$over $Q\left(\sqrt{3}\right)$. Second as $\left[Q\left\{\sqrt{3}\right\}:Q\right]=2$ and $x^3-3$ is minimal polynomial of $\sqrt{3}$over $Q$.

Basis of $Q\left\{\sqrt[3]{2},\sqrt{3}\right\}$ over $Q\left(\sqrt{3}\right)$ is

$\left\{1,\sqrt[3]{2},{\left(\sqrt[3]{2}\right)}^2\right\}$

Basis of $Q\left(\sqrt{3}\right)$ over Q is

$\left(1,\sqrt{3}\right)$

Therefore the basis of $Q\left\{\sqrt[3]{2},\sqrt{3}\right\}$over $Q$.

role="math" localid="1657955408287" $\left\{1,\sqrt[3]{2},{\left(\sqrt[3]{2}\right)}^2\right\}\times \left\{1,\sqrt{3}\right\}=\left\{1,\sqrt[3]{2},{\left(\sqrt[3]{2}\right)}^2,\sqrt{3},\sqrt[3]{2}.\sqrt{3},{\left(\sqrt[3]{2}\right)}^2.\sqrt{3}\right\}$

Hence the required basis is$\left\{1,\sqrt[3]{2},{\left(\sqrt[3]{2}\right)}^2,\sqrt{3},\sqrt[3]{2}.\sqrt{3},{\left(\sqrt[3]{2}\right)}^2.\sqrt{3}\right\}$ .