# Group 12

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## Initial observations of $f(x)$

We have the polynomial $f(x)=x^{4}-14x^{2}+9$ with root $\alpha=\sqrt{2}+\sqrt{5}$. Also observe $-\alpha=-\sqrt{2}-\sqrt{5}$ is such that :$f(-\alpha)=(-\alpha)^{4}-14(-\alpha)^{2}+9=\alpha^{4}-14\alpha^{2}+9=f(\alpha)=0$ Thus $-\alpha=-\sqrt{2}-\sqrt{5}$ is also a root of $f(x)$

## Factorization in Standard Fields

### Factorization over ${\mathbb Q}$

$f(x)$ is irreducible over ${\mathbb Q}[x]$. Indeed, suppose, towards a contradiction, that $f(x)$ is reducible, that is, that $f(x)=g(x)h(x)$ where $g(x),h(x)\in{\mathbb Q}[x]$. As $f(x)$ is a monic quartic, it follows that :$f(x)=(x^{2}+ax+b)(x^{2}+cx+d)=x^{4}+(a+c)x^{3}+(ac+b+d)x^{2}+(ad+bc)x+bd$ where $a,b,c,d\in {\mathbb Q}$. By equating coefficients we obtain the following system of simultaneous equations:

$a+c=0$
$ac+b+d=-14$
$ad+bc=0$
$bd=9$

From this we can immediately conclude that $a=-c$ and $d=9/b$ , which can be substituted into equation (3) to obtain: $\frac{9a}{b}-ab=0 \implies \frac{9a}{b}=ab \stackrel{if \ a\neq0}{\implies} b^{2}=9 \implies b=\pm\sqrt{9}=\pm3$ We will consider these two cases separately.

Case #1 $b=+3$

As $b=3 \implies d=\frac{9}{b} \implies d=3$, and $-c=a$, we have that $ac+b+d=-a^{2}+6=-14 \implies -a^{2}=-20 \implies a=\pm\sqrt{20} \notin {\mathbb Q}$

This contradicts the fact that $a,b,c,d\in{\mathbb Q}$

Case #2 $b=-3$

As $b=-3 \implies d=\frac{9}{b} \implies d=-3$, and $-c=a$, it follows that $ac+b+d=-a^{2}-6=-14 \implies -a^{2}=-12 \implies a=\pm\sqrt{12} \notin {\mathbb Q}$

Again, this contradicts the fact that $a,b,c,d\in{\mathbb Q}$

Note that in the above argument we have assumed that $a \neq 0$. Therefore we need to consider the additional case where $a=0$.

Suppose that $a=0$. Then $ac+b+d=-14 \implies b+d=-14 \implies b=-d-14 \implies bd=(-14-d)d=9 \implies d^2+14d+9=0$. We can use the quadratic formula to obtain solutions for $d$ of the form: $d=\frac{14\pm \sqrt{164}}{2}= -7\pm \sqrt{42} \notin {\mathbb Q}$

As we arrive at a contradiction by assuming that $f(x)$ is reducible in ${\mathbb Q}[x]$, we can conclude that $f(x)$ cannot be factorized further, that is, $f(x)$ is an irreducible polynomial in ${\mathbb Q}[x]$.

### Factorization over ${\mathbb R}$

Note that our root $\alpha=\sqrt{2}+\sqrt{5} \in {\mathbb R}$ thus by the factor theorem we have that $x-\alpha$ is a factor of $f(x)$, that is, $f(x)=(x-\alpha)g(x)$, i.e. $(x-\alpha)$ divides $f(x)$ with zero remainder. We can calculate the polynomial $g(x)$ by performing polynomial long division as follows:

$\begin{array}{r} x^3+(\sqrt{2}+\sqrt{5})x^2+(-7+2\sqrt{10})x-(3\sqrt{5}-3\sqrt{2})\\ x-(\sqrt{2}+\sqrt{5})\overline{\Big) x^4+0x^3-14x^2+0x+9}\\ \underline{x^4+(-\sqrt{2}-\sqrt{5})x^3}\\ (\sqrt{2}+\sqrt{5})x^3-14x^2+0x+9\\ \underline{(\sqrt{2}+\sqrt{5})x^3-(7+2\sqrt{10})x^2}\\ (-7+2\sqrt{10})x^2 +0x+9\\ \underline{(-7+2\sqrt{10})x^2+(3\sqrt{5}-3\sqrt{2})x}\\ -(3\sqrt{5}-3\sqrt{2})x +9\\ \underline{-(3\sqrt{5}-3\sqrt{2})x+9}\\ 0\\ \end{array}$

Thus we see that $g(x)=x^{3}+(\sqrt{2}+\sqrt{5})x^{2}+(2\sqrt{10}-7)x+(3\sqrt{2}-3\sqrt{5})$

Observe that $g(-\alpha)=g(-\sqrt{2}-\sqrt{5})=0\implies-\alpha$ is a root of $g(x)$. Thus we can reapply the factor theorem to $g(x)$ to obtain $g(x)=(x+\alpha)h(x)$, that is, $(x+\alpha)$ divides $g(x)$ exactly. Hence we can perform polynomial long division once more to obtain $h(x)$ as follows:

$\begin{array}{r} x^2+(-7+2\sqrt{10})\\ x+(\sqrt{2}+\sqrt{5})\overline{\Big) x^3 + (\sqrt{2}+\sqrt{5})x^2 + (-7+2\sqrt{10})x +(3\sqrt{2}-3\sqrt{5})}\\ \underline{x^3+(\sqrt{2}+\sqrt{5})x^2}\\ 0 + (-7+2\sqrt{10})x +(3\sqrt{2}-3\sqrt{5}) \\ \underline{(-7+2\sqrt{10})x +(3\sqrt{2}-3\sqrt{5})}\\ 0\\ \end{array}$

Thus we see that $h(x)=x^2+(2\sqrt{10}-7)$. As $2\sqrt{10}-7\lt 0$, we see that $h(x)$ has roots in ${\mathbb R}$. We can find these exclusively by observing that $deg( h(x))=2\implies h(x)=(x+a)(x+b)=x^2+(a+b)x+ab$ where $a,b\in {\mathbb R}$.

By equating coefficients we obtain the following simultaneous equations :$ab=-7+2\sqrt{10}$

$a+b=0$

Which has solution:: $a=-b\implies -a^2=-7+2\sqrt{10}\implies a^2=7-2\sqrt{10}\implies a=\pm\sqrt{7-2\sqrt{10}}\implies b=\mp\sqrt{7-2\sqrt{10}}$ As $\sqrt{7-2\sqrt{10}}=(-\sqrt{2}+\sqrt{5})\in{\mathbb R}$ it follows that $h(x)=(x+(-\sqrt{2}+\sqrt{5}))(x+(-\sqrt{2}+\sqrt{5}))$

As $f(x)=(x-\alpha)g(x)=(x-\alpha)(x+\alpha)h(x)$, it follows that the factorization of $f(x)\in {\mathbb R}[x]$ is: :$f(x)=(x-\sqrt{2}-\sqrt{5})(x+\sqrt{2}+\sqrt{5})(x-\sqrt{2}+\sqrt{5})(x+\sqrt{2}-\sqrt{5})$

Note that as $f(x)$ decomposes into linear factors, which are necessarily irreducible, this factorization is a product of irreducible polynomials.

### Factorization over ${\mathbb C}$

As our root $\alpha = \sqrt{2}+\sqrt{5} \in {\mathbb R} \subset {\mathbb C}$, it follows that the above factorization of $f(x)$ over ${\mathbb R}$ is also the factorization of $f(x)$ over ${\mathbb C}$.

## Reducibility over ${\mathbb Z}_{p}$ for prime p

The aim of this section is to find two prime numbers $p,q$ such that $f(x)$ is reducible in $p$ and irreducible in $q$. We will consider $f(x)=x^4-14x+9$ over ${\mathbb Z}_{p}$ for $p=2,3,5,7,11,13,...$ until we achieve the desired result. To consider $f(x)$ over a given ${\mathbb Z}_{p}$, we will need to evaluate its reduction modulo p, which is given by the following definition.

Definition 3.0.1: Consider $f(x)=a_{0} + a_{1}x +...+a_{n}x^{n} \in {\mathbb Z}[x]$. Then the reduction of $f(x)$ modulo $p$ is given by: $\overline{f(x)} = b_{0}+b_{1}x+...+b_{n}x^{n}$ where $a_{i}\equiv b_{i}mod(p)$ for each $i=0,2,...,n$. [1]

### Reducibility of $f(x)$ in ${\mathbb Z}_{p}[x]$ for $p=2,3,5,7,11,13,17$

This section displays the factorizations of $f(x)=x^4-14x^2+9$ in ${\mathbb Z}_{2}[x], {\mathbb Z}_{3}[x], {\mathbb Z}_{5}[x],{\mathbb Z}_{7}[x], {\mathbb Z}_{11}[x], {\mathbb Z}_{13}[x]$ and ${\mathbb Z}_{17}[x]$. The following table displays the relevant information.

p ${\mathbb Z}_p$ $\overline{f(x)}$ Factorization of $f(x)$ Verification of factorization
$2$ $\{0,1\}$ $x^4+1$ $(x+1)^4$ $(x+1)^4=(x^2+2x+1)(x^2+2x+1)=(x^2+1)(x^2+1)=x^4+2x^2+1=x^4+1=\overline{f(x)}$
$3$ $\{0,1,2\}$ $x^4+x^2$ $x^2(x^2+1)$ $x^2(x^2+1)=x^4+x^2=\overline{f(x)}$
$5$ $\{0,1,2,3,4\}$ $x^4+x^2+4$ $(x^2+3)^2$ $(x^2+3)^2=x^4+6x^2+9=x^4+x^2+4=\overline{f(x)}$
$7$ $\{0,1,2,3,4,5,6\}$ $x^4+2$ $(x^2+x+4)(x^2+x+6)$ $(x^2+x+4)(x^2+x+6)=x^4+7x^3+14x^2+28x+16=x^4+2=\overline{f(x)}$
$11$ $\{0,1,...10\}$ $x^4+8x^2+9$ $(x^2+3x+3)(x^2+8x+3)$ $(x^2+3x+3)(x^2+8x+3)=x^4+11x^3+30x^2+33x+9=x^4+8x^2+9=\overline{f(x)}$
$13$ $\{0,1,...,12\}$ $x^4+12x^2+9$ $(x^2+5)(x^2+7)$ $(x^2+5)(x^2+7)=x^4+12x^2+35=x^4+12x^2+9=\overline{f(x)}$
$17$ $\{0,1,...,16\}$ $x^4+3x^2+9$ $(x^2+5x+14)(x^2+12x+14)$ $(x^2+5x+14)(x^2+12x+14)=x^4+17x^3+88x^2+238x+196=x^4+3x^2+9=\overline{f(x)}$

As you can see, $f(x)$ appears to be reducible in ${\mathbb Z}_{p}[x]$ for any prime $p$ that we try. We could keep checking this for larger and larger primes, but the larger the value of $p$, the more difficult it is to obtain a factorization of $f(x)$ in ${\mathbb Z}_{p}[x]$. At this point a more prudent suggestion would be the following: does there even exist a prime $p$ such that ${f(x)}$ is irreducible in ${\mathbb Z}_{p}[x]$? This leads us to the next section.

### $f(x)$ is reducible in ${\mathbb Z}_{p}[x]$ for all primes $p$

In this section we will prove the following result:

Lemma 3.2.1 For all primes $p$ the polynomial $f(x)=x^4-14x^2+9$ is reducible in ${\mathbb Z}_{p}[x]$.

To prove this lemma we will use results obtained by Driver, Leonard and Williams (DLW) in their paper "Irreducible quartic polynomials with factorization modulo p"[2]. This paper explores the reducibility criteria of certain polynomials of the form $f(x)=x^4+rx^2+s$ with $r,s\in{\mathbb Z}$ (which are referred to as biquadratic), by considering the discriminant of $f(x^{1/2})$, which is $r^2-4s$. The crux of the paper's argument revolves around the case where $r^2-4s$ is a perfect square, that is, where $r^2-4s=t^2$ for some $t \in {\mathbb Z}$. Multiple results are proved throughout the paper, but for our purposes we only need to consider the following, which is listed at theorem 4, and found on page 880:

Theorem 3.2.2: (DLW) Let $r,s$ be integers such that $r^2 - 4s$ is not a perfect square. Then the polynomial $f(x) = x^4 + rx^2 + s$ is reducible modulo $p$ for every prime $p$ if and only if $s = t^2$ for some integer $t$.

The proof of this theorem involves use of previous arguments in DLW, coupled with some results in number theory, namely Legendre notation, the law of quadratic reciprocity and Dirichlet's theorem on arithmetic progressions.

The above theorem means that Lemma 3.2.1 follows as a corollary, by observing that in the polynomial $f(x)=x^4-14x^2+9$, which has $s=9=(\pm3)^2$.

## Construction of field ${\mathbb E}={\mathbb Q}[x]/I$

### Constructing ${\mathbb E}$

We consider constructing a larger field ${\mathbb E}$, such that $f(x)$ has a root in ${\mathbb E}$. We do this by letting $I = (f(x))$. We have seen that $f(x)$ is irreducible, therefore $I$ is a maximal ideal. This implies the factor ring generated by the cosets of $I$ is in fact a field. Elements of ${\mathbb E}$ are of the form $g(x) + I$, where $g(x)\in {\mathbb Q}$. By letting $\theta = x + I$ we can show that :$f(\theta) = (x + I)^4 – 14(x + I)^2 + 9$ :$f(\theta) = x^4 -14x^2 +9 + I$ :$f(\theta) = 0 + I = 0_{E}$ Consequently we can state that $\theta$ is a root of $f(x)$ in ${\mathbb E}$ and thus by the factor theorem we have that $x-\theta$ is a factor of $f(x)$. We can then go on to factorise$f(x)$ in ${\mathbb E}$ by performing polynomial long division as follows:

$\begin{array}{r} x^3+\theta x^2+ (\theta^2-14)x+(\theta^3-14\theta)\\ x- \theta \overline{\Big) x^4+0x^3-14x^2+0x+9}\\ \underline{x^4- \theta x^3)}\\ \theta x^3-14x^2+0x+9\\ \underline{\theta x^3-\theta^2x^2}\\ (-14+ \theta^2)x^2 +0x+9\\ \underline{(-14 + \theta^2)x^2+(14 \theta- \theta^3)x}\\ (\theta^3-14 \theta)x +9\\ \underline{(\theta^3-14 \theta)x+(- \theta^4+14 \theta^2)}\\ \theta^4-14 \theta^2+9 = 0_{E}\\ \end{array}$

Therefore we have constructed a field in which $f(x)$ is reducible

$f(x) = (x-\theta)(x^3+\theta x^2+(\theta^2-14)x+(\theta^3-14 \theta))$ .

### Finding a basis for ${\mathbb E}$

Elements of ${\mathbb E}$ are of the form $g(x) + I$

Proposition 4.2.1 All elements of ${\mathbb E }$ can be written as $h(\theta)$ where $deg (h(x)) \lt deg (m(x))$ where $m(x)$ is the minimal polynomial, which in our case is $f(x)$.

Proof By the Euclidean Division Algorithm

$g(x) = m(x) q(x) + h(x)$

for some $q(x)$,$h(x)\in {\mathbb Q}$ and $deg (h(x)) \lt deg (m(x))$.

$g(\theta) = m(\theta) q(\theta)+h(\theta)$

We know $m(\theta) = f(\theta) = 0$ as $\theta$ is a root.

$g(\theta) = h(\theta)$

Therefore we have shown that each element in ${\mathbb E}$ is of the form $h(\theta) = a + a_{1}\theta + a_{2} \theta^2 + a_{3} \theta^3$ , where $h(\theta)$ is of degree 3 or less. $\square$

Proposition 4.2.2 This expression of an element in ${\mathbb E}$ is unique.

Proof Assume $g(\theta) = h(\theta)$ and $g(\theta) = h'(\theta)$ . Let $h(x)-h'(x) = e(x)$, then

$e(\theta) = h(\theta) - h'(\theta) = g(\theta) - g(\theta)$
$e(\theta) = 0$

However the degree of $e(x)$ is less than the degree of $f(x)$, contradicting $f(x)$ being the minimal polynomial. Therefore $e(x) = 0$. So $e(x)=h(x)-h'(x)=0 \; \Rightarrow \; h(x)=h'(x)$. $\square$

We have shown that $h(x)$ is unique for each element of ${\mathbb E}$, which is a necessary and sufficient condition to form a basis. Consequently, $\{1,\theta,\theta^2,\theta^3\}$ forms a basis for ${\mathbb E}$ as a vector space over ${\mathbb Q}$ .

## Construction of finite field ${\mathbb E}={\mathbb F}[x]/I$

In Allenby's "Rings, Fields and Groups"[3] the following result is listed as theorem 4.5.8 part (i).

Theorem 5.0.1 Every finite field is of the form $\frac{{\mathbb Z}_{p}[x]}{[f]}$, where $f$ is a polynomial irreducible in ${\mathbb Z}_{p}[x]$.

Note that here $[f(x)]$ denotes the principal ideal generated by the polynomial $f(x)$.

For our polynomial $f(x)=x^4-14x^2+9$, we concluded in section 3.2 that $f(x)$ is reducible in ${\mathbb Z}_{p}[x]$ for all primes $p$, that is, there does not exist a prime $p$ such that $f(x)$ is irreducible in ${\mathbb Z}_{p}[x]$.

Using this result, we can apply theorem 5.0.1 to conclude that we cannot construct a finite field using $f(x)$.

## Algebraic elements and their connection to $f(x)$

We will now focus on constructing a field ${\mathbb F}$ containing $\alpha = \sqrt2+\sqrt5$, and so that $f(x)$ is the minimal polynomial of $\alpha$ over ${\mathbb F}$

### The Evaluation Homomorphism

Define the evaluation homomorphism $\phi_\alpha: {\mathbb Q}[x] \to {\mathbb Q}( \alpha )$ such that $f(x) \mapsto f(\alpha)$.

Then $\phi_\alpha$ maps functions over ${\mathbb Q} [x]$ to their evaluation at $\alpha$.

Now by considering the image and kernel of $\phi_\alpha$, we will attempt to employ the First Isomorphism Theorem in search for a field in which $\alpha$ has minimal polynomial $f(x)$

### Kernel of the Evaluation Homomorphism

$ker(\phi_\alpha)= \left \{f(x) : f(\alpha)=0 \right \}$

Proposition 6.2.1 $ker(\phi_\alpha)=(f(x))$

Proof We will start by showing the inclusion of the ideal generated by $f(x)$ inside the kernel.

Assume that $g(x) \in (f(x))$.

By the definition of the ideal $g(x) = f(x)q(x)$, for some $q(x) \in {\mathbb Q}[x]$. Then $g(\alpha) = f(\alpha) q(\alpha) = 0$ as $f(\alpha) = 0$ because $\alpha$ is a root of $f(x)$.

$g(\alpha) = 0,$ so $\phi_\alpha(g(x)) = 0$

$g(x) \in ker(\phi_\alpha)$

So we have shown that the arbitrary polynomial in the ideal $f(x)$ is also in the kernel of $\phi_\alpha$.

Now we seek to show the inclusion of the kernel inside the ideal generated by $f(x)$.

Assume $g(x) \in ker(\phi_\alpha)$, by the Euclidean algorithm we can find $q(x)$ and $r(x)$, where $degree(r(x))\lt degree(f(x))$ such that

$g(x) = f(x)q(x) + r(x)$.

As $g(x)$ is in the kernel of $\phi_\alpha$,

$g(\alpha) = f(\alpha) q(\alpha) + r(\alpha) = 0$.

Remembering that $f(\alpha) = 0$, we deduce that $r(\alpha)=0$.

Hence $\alpha$ appears to be a root of $r(x)$, which has a lower degree than $f(x)$, a contradiction.

Therefore $r(x)=0$ and $g(x) = f(x)q(x)$.

Consequently $g(x) \in (f(x))$, so we have shown that the arbitrary element of the kernel is also an element of the ideal generated by $f(x)$ and thus our claim is proved. $\square$

### Image of the Evaluation Homomorphism

We will now take on the simpler task of examining the image of $\phi_\alpha$.

$Im(\phi_\alpha) = \left \{\phi_\alpha (f(x)) | f(x) \in {\mathbb Q} [x] \right \} \\$

$Im(\phi_\alpha) = \left \{f(\alpha) | f(x) \in {\mathbb Q} [x] \right \} = {\mathbb Q}(\alpha)$

### Constructing ${\mathbb Q}[\alpha]$

We can now employ the first isomorphism theorem.

$\frac{{\mathbb Q}[x]}{ker(\phi_\alpha)} \cong im(\phi_\alpha)$

so :$\frac{{\mathbb Q}[x]}{(f(x))} \cong {\mathbb Q}[\alpha]$

As $(f(x))$ is a maximal ideal, ${\mathbb Q}[\alpha]$ is a field.

Therefore we have constructed a field ${\mathbb Q}[\alpha]$ so that $f(x)$ is the minimal polynomial of $\alpha$ over ${\mathbb Q}[\alpha]$.

We can also find a basis for ${\mathbb Q}[\alpha]$ over ${\mathbb Q}$ in the same manner we found the basis for ${\mathbb E}$.

Using the same argument we used in section 4.2 we find that $\left \{ 1, \alpha, \alpha^2, \alpha^3 \right \}$ forms the desired basis, however a more interesting observation is that the two fields we have constructed, ${\mathbb Q}[\alpha]$ and ${\mathbb E}$ are in fact isomorphic. Furthermore they are related by the isomorphism constructed by the composition of the canonical homomorphism and the evaluation homomorphism.

If we define the canonical map as $\varphi:{\mathbb E} \rightarrow{\mathbb Q}[x], \: \varphi(f(x)+I)=f(x)$ Then :${\mathbb E}\overset{\varphi }{\rightarrow}{\mathbb Q}[x]\overset{\phi_\alpha}{\rightarrow} {\mathbb Q}[\alpha]$.

It should not come as a surprise that this result is by no means a coincidence. If we look at the theorem cited in I Stewart's book "Galois Theory" [4], we see that we have stumbled on a particular example of of the more general case.

Theorem 6.4.1 Suppose $K(a):K$ and $K(b):K$ are simple algebraic extensions, such that $a$ and $b$ have the same minimum polynomial m over $K$. Then the two extensions are isomorphic, and the isomorphism of the large fields can be taken to map $a \mapsto b$.

Indeed ${\mathbb E}={\mathbb Q}(\theta)$, so both our larger fields are simple algebraic extensions with $f(x)$ the minimal polynomial of $\alpha$ and $\theta$. And lastly for completeness:

$\phi_\alpha \circ \varphi (\theta)= \phi_\alpha \circ \varphi (x+I)$

$\phi_\alpha \circ \varphi (\theta)= \phi_\alpha (x)$
$\phi_\alpha \circ \varphi (\theta)= \alpha$

## Galois Group over the field ${\mathbb Q}$

Galois Theory provides a connection between field theory and group theory. Below we have investigated the Galois Group for our polynomial over ${\mathbb Q}$.

$f(x) = x^4 -14x^2+9 = (x^2-7)^2-40$

The polynomial has four roots

$A = \sqrt2 +\sqrt5$

$B = \sqrt2 -\sqrt5$

$C = -\sqrt2 +\sqrt5$

$D = -\sqrt2 -\sqrt5$

The roots are connected by algebraic equations which have rational coefficients.

$AB = -3$

$AC = 3$

$A+D = 0$

We then consider the permutations of the roots so that these equations are satisfied before and after.

$(A B C D) \to (A B C D)$

$(A B C D) \overset{f}{\rightarrow}(C D A B)$

$(A B C D) \overset{g}{\rightarrow} (B A D C)$

$(A B C D) \overset{fg}{\rightarrow} (D C B A)$

The permutations form a permutation group which contains four elements, this is the Galois Group over ${\mathbb Q}$.

This group for our polynomial is isomorphic to the Klein-4 Group.

According to the Fundamental Theorem of Galois Theory the Galois Group for our polynomial has 5 subgroups. Each of the 5 subgroups is related by the Galois correspondance to a subfield of ${\mathbb Q}(\sqrt2,\sqrt5)$.

• The trivial subgroup (containing only the identity element) corresponds to all of ${\mathbb Q}(\sqrt2,\sqrt5)$, as the identity element is the only permutation which leaves ${\mathbb Q}(\sqrt2,\sqrt5)$ unchanged.
• The entire group corresponds to the base field ${\mathbb Q}$ as the elements of the galois group permute expressions that reside outside of the rationals, and so ${\mathbb Q}$ is fixed by the whole group.
• The two-element subgroup {$1, f$} corresponds to the subfield ${\mathbb Q}(\sqrt5)$, since $f$ fixes $\sqrt5$.
• The two-element subgroup {$1, g$} corresponds to the subfield ${\mathbb Q}(\sqrt2)$, again since $g$ fixes $\sqrt2$.
• The two-element subgroup {$1, fg$} corresponds to the subfield ${\mathbb Q} (\sqrt10)$, since $fg$ fixes $\sqrt10$.

## References

[1] Introduction to Abstract Algebra, Keith Nicholson, Wiley & Sons Publishing, page 213

[2] Irreducible quartic polynomials with factorization modulo p Eric Driver, Philip A. Leonard and Kenneth S. Williams, The American Mathematical Monthly Vol. 112, No. 10 (Dec., 2005), pp. 876-890

[3] Rings, Fields and Groups 2nd edition, R.B.J.T Allenby, Arnold Publishing, page 168

[4] Galois Theory 2nd edition, I Stewart, Chapman & Hall/CRC, page 39, Theorem 3.8