2 Problem Sheet 2:
Field extensions and minimal polynomials

Exercise 2.1 By considering degrees of field extensions, determine which of the following numbers are algebraic (over $\mathbb{Q}$).

$\;(a)\;\;1+\sqrt{3}+\sqrt{5},\hspace{2em}$ $\;(b)\;\;\sqrt{2}+\sqrt[4]{2},\hspace{2em}$ $\;(c)\;\;\sqrt{2}+\sqrt[3]{3},$

$\;(d)\;\;\sqrt{\pi}-1,\hspace{4.4em}$ $\;(e)\;\;\sqrt{\pi}+\sqrt{7},\hspace{1.9em}$ $\;(f)\;\;\sqrt[5]{1+\sqrt{e}}.$

Solution

One way to test if a number is algebraic (or transcendental) is to find a polynomial equation it satisfies (or show that no such polynomial equation exists).

It’s often easier to use degrees of extensions: $\theta$ is algebraic over $\mathbb{Q}$ if and only if $[\mathbb{Q}(\theta):\mathbb{Q}]<\infty$. In particular, if $\theta\in L=\mathbb{Q}(\alpha_1,...,\alpha_n)$ where $[L:\mathbb{Q}]<\infty$, then $\theta$ is algebraic over $\mathbb{Q}$.

The even easier (perhaps too easy!) way is to use the fact that the algebraic numbers form a field, so sums of algebraic numbers are algebraic.

$(a)$ Yes, since $\theta=1+\sqrt{3}+\sqrt{5}\in\mathbb{Q}(\sqrt{3},\sqrt{5})$ and \[[\mathbb{Q}(\sqrt{3},\sqrt{5}):\mathbb{Q}]\leq [\mathbb{Q}(\sqrt{3}):\mathbb{Q}] [\mathbb{Q}(\sqrt{5}):\mathbb{Q}] = 2\times 2<\infty.\]

Taking the more explicit route, just eliminate the irrationalities, e.g. \[\begin{align*} (\theta-1-\sqrt{3})^2=5 \;&\implies\; (\theta-1)^2+3-2\sqrt{3}(\theta-1)=5 \\ \;&\implies\; (\theta^2-2\theta-1)=2(\theta-1)\sqrt{3} \\ \;&\implies\; (\theta^2-2\theta-1)^2=12(\theta-1)^2 \\ \end{align*}\]

Though we didn’t need it, we’ve found $\theta$ is a root of $x^4-4x^3-10x^2+28x-11$ (and actually this is the minimal polynomial).

$(b)$ Yes, since $\theta=\sqrt{2}+\sqrt[4]{2}\in\mathbb{Q}(\sqrt[4]{2})$ and $[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]= 4<\infty$.

(Alternatively, show $\theta^4-4\theta^2-8\theta+2=0$.)

$(c)$ Yes, since $\theta=\sqrt{2}+\sqrt[3]{3}\in\mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ and \[[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]\leq [\mathbb{Q}(\sqrt{2}):\mathbb{Q}] [\mathbb{Q}(\sqrt[3]{3}):\mathbb{Q}]=2\times 3<\infty.\]

(The minimal polynomial is actually $x^6-6x^4-6x^3+12x^2-36x+1$.)

$(d)$ This time the answer is no. Let $\theta=\sqrt{\pi}-1$. Then $\pi=\theta^2+1\in\mathbb{Q}(\theta)$ and $\mathbb{Q}(\pi)\subset\mathbb{Q}(\theta)$. But $\pi$ is transcendental and $[\mathbb{Q}(\pi):\mathbb{Q}]=\infty$ so $[\mathbb{Q}(\theta):\mathbb{Q}]=\infty$ as well.

$(e)$ No, for similar reasons. If $\theta=\sqrt{\pi}+\sqrt{7}$, then $\pi=(\theta-\sqrt{7})^2$ and so $\mathbb{Q}(\pi)\subset\mathbb{Q}(\theta,\sqrt{7})$, hence \[ [\mathbb{Q}(\theta,\sqrt{7}):\mathbb{Q}]=\infty\quad\implies\quad [\mathbb{Q}(\theta,\sqrt{7}):\mathbb{Q}(\theta)] [\mathbb{Q}(\theta):\mathbb{Q}]=\infty\] But $[\mathbb{Q}(\theta,\sqrt{7}):\mathbb{Q}(\theta)]\leq 2$ so we must have $[\mathbb{Q}(\theta):\mathbb{Q}]=\infty$.

$(f)$ And again, no: $\theta=\sqrt[5]{1+\sqrt{e}}$ implies $e=(\theta^5-1)^2\in\mathbb{Q}(\theta)$ so $\mathbb{Q}(e)\subset\mathbb{Q}(\theta)$. But $e$ is transcendental and $[\mathbb{Q}(e):\mathbb{Q}]=\infty$ so $[\mathbb{Q}(\theta):\mathbb{Q}]=\infty$ as well.

Exercise 2.2 Find the minimal polynomial of $\theta$ over the field $K$ where:

$\;(a)\;\;\theta=\sqrt[3]{1+\sqrt{3}}$ and $K=\mathbb{Q},\hspace{2em}$ $\;(b)\;\;\theta=\sqrt{2}+\sqrt[3]{2}$ and $K=\mathbb{Q},$

$\;(c)\;\;\theta=e^{2\pi i/5}$ and $K=\mathbb{Q},\hspace{3.75em}$ $\;(d)\;\;\theta=\sqrt{2}$ and $K=\mathbb{Q}(i),$

$\;(e)\;\;\theta=\sqrt[3]{2}$ and $K=\mathbb{Q}(\sqrt{2}).$

Solution

General tip: To find the minimal polynomial $m(x)$ of some $\theta$ over $K$, it’s often necessary to find its degree $\deg(m(x))=[K(\theta):K]$.

$(a)$ With $\theta=\sqrt[3]{1+\sqrt{3}}$, we have \[\theta^3=1+\sqrt{3}\quad\implies\quad (\theta^3-1)^2=3\quad\implies\quad \theta^6-2\theta^3-2=0.\] So $\theta$ is a root of the monic polynomial $x^6-2x^3-2\in\mathbb{Q}[x]$. This polynomial is also irreducible over $\mathbb{Q}$ by using Eisenstein’s Criterion with $p=2$. Hence it is the minimal polynomial of $\theta$ over $\mathbb{Q}$.

Problems Class Discussion. We were quite lucky here, as it’s not always so obvious that a polynomial is irreducible or we might have forgotten about Eisenstein’s Criterion. How could we still answer the question? If we could show that $[\mathbb{Q}(\theta):\mathbb{Q}]=6$, we’d be done, as that means the minimal polynomial of $\theta$ has degree $6$. Consider the tower $\mathbb{Q}\subset\mathbb{Q}(\sqrt{3})\subset\mathbb{Q}(\theta)$.

Since $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]=2$, we would like to show that $[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{3})]=3$. As $\alpha^3=1+\sqrt{3}$, this is the same as showing $1+\sqrt{3}$ is not a cube in $\mathbb{Q}(\sqrt{3})$.

Suppose $(a+b\sqrt{3})^3=1+\sqrt{3}$ for some $a,b\in\mathbb{Q}$.
Then, $\;\;a^3+3a^2b\sqrt{3}+9ab^2+3b^3\sqrt{3}=1+\sqrt{3}\quad\implies\quad\begin{cases} \;\,a^3+9ab^2=1 \\ 3a^2b+3b^3=1\end{cases}$
Now try to show there are no $a,b\in\mathbb{Q}$ satisfying these two equations. If $a,b$ were integers, it would be easy, but we only know $a,b\in\mathbb{Q}$. It can be done by writing $a,b$ with a minimal common denominator, then using an “infinite descent” argument but this is quite hard! So let’s find a better way.

Here’s a really useful fact: if $\sqrt{3}$ is a root of a polynomial $g(x)\in\mathbb{Q}[x]$, then so is $-\sqrt{3}$. (Notice how similar this is to the fact that for any root of $g(x)\in\mathbb{R}[x]$, the complex conjugate is also a root.)

How to prove this remarkable fact? By the Division Algorithm, \[g(x)=(x^2-3)q(x)+r_0+r_1x \quad\text{where $r_0,r_1\in\mathbb{Q}$}\] and setting $x=\sqrt{3}$ gives $0=r_0+r_1\sqrt{3}$. But $\sqrt{3}$ is irrational so $r_0=r_1=0$, i.e. $g(x)$ is divisible by $x^2-3$. So let’s have another go at the question using this.

Suppose $(a+b\sqrt{3})^3=1+\sqrt{3}$ for some $a,b\in\mathbb{Q}$.
Then $(a-b\sqrt{3})^3=1-\sqrt{3}$ and multiplying gives $(a+b\sqrt{3})^3(a-b\sqrt{3})^3=(1+\sqrt{3})(1-\sqrt{3})$.
Thus $(a^2-3b^2)^3=-2$ and so $\sqrt[3]{-2}\in\mathbb{Q}$. This is false and so we are done! The minimal polynomial of $\theta$ over $\mathbb{Q}$ does have degree $6$ so must equal $x^6-2x^3-2$.

Note it is generally necessary to do extra work to find the degree of the minimal polynomial. Suppose we considered the similar looking number $\theta=\sqrt[3]{2+\sqrt{5}}$. Then \[\theta^3=2+\sqrt{5}\quad\implies\quad (\theta^3-2)^2=5\quad\implies\quad \theta^6-4\theta^3-1=0\] so $\theta$ is a root of the monic polynomial $x^6-4x^3-1\in\mathbb{Q}[x]$. However, this is not the minimal polynomial of $\theta$ over $\mathbb{Q}$. Indeed, notice that $\left(\dfrac{1+\sqrt{5}}{2}\right)^3=2+\sqrt{5}$ so $\theta=\dfrac{1+\sqrt{5}}{2}$ actually has a quadratic minimal polynomial $x^2-x-1$ over $\mathbb{Q}$.

$(b)$ Given $\theta=\sqrt{2}+\sqrt[3]{2}$, eliminating the irrationalities gives \[\begin{align*} (\theta-\sqrt{2})^3=2 \quad &\implies\quad \theta^3+6\theta-2=\sqrt{2}(3\theta^2+2) \\ \quad &\implies\quad (\theta^3+6\theta-2)^2=2(3\theta^2+2)^2 \\ \quad &\implies\quad \theta^6-6\theta^4-4\theta^3+12\theta^2-24\theta-4=0 \end{align*}\] Thus we find a degree $6$ polynomial \[x^6-6x^4+12x^2-24x-4\in\mathbb{Q}[x]\] with root $\theta$. But how do we know this is the minimal polynomial of $\theta$? We don’t know it is irreducible and it’s not obvious why it might be… Instead, we will show that $[\mathbb{Q}(\theta):\mathbb{Q}]=6$ directly. To do this, we could compute that $1,\theta,\theta^2,\theta^3,\theta^4,\theta^5$ are linearly independent over $\mathbb{Q}$. However, we can save a bit of effort by noticing that $\theta=(2^{1/6})^3+(2^{1/6})^2$ and so \[\mathbb{Q}\subset\mathbb{Q}(\theta)\subset\mathbb{Q}(2^{1/6}).\] Since $[\mathbb{Q}(2^{1/6}):\mathbb{Q}]=6$, the Tower Law implies that $[\mathbb{Q}(\theta):\mathbb{Q}]=1,2,3$ or $6$.

Working in the $\mathbb{Q}$-basis $1,2^{1/6},2^{2/6},2^{3/6},2^{4/6},2^{5/6}$ of $\mathbb{Q}(2^{1/6})/\mathbb{Q}$, after a bit of calculation we find

$1$ has coordinates $(1,0,0,0,0,0)$,
$\theta$ has coordinates $(0,0,1,1,0,0)$,
$\theta^2$ has coordinates $(2,0,0,0,1,2)$,
$\theta^3$ has coordinates $(2,6,6,2,0,0)$.

These vectors are clearly linearly independent (the last two entries are only non-zero for $\theta^2$.) Hence $[\mathbb{Q}(\theta):\mathbb{Q}]\geq 4$ and so must equal $6$. Hence \[x^6-6x^4+12x^2-24x-4\in\mathbb{Q}[x]\] is indeed the minimal polynomial.

$(c)$ We have $0=\theta^5-1=(\theta-1)(\theta^4+\theta^3+\theta^2+\theta+1)$ and so $\theta$ is a root of \[f(x)=x^4+x^3+x^2+x+1\in\mathbb{Q}[x].\] We have already seen in lectures that this is irreducible - it is the $p$th cyclotomic polynomial $(x^p-1)/(x-1)$ where $p=5$ is prime. Hence $f(x)$ is the minimal polynomial of $\theta$.

Recall this proof of irreducibility: applying Eisenstein’s criterion to \[f(x+1)=x^4+5x^3+10x^2+10x+5\] with $p=5$ shows $f(x+1)$ is irreducible, hence $f(x)$ is as well.

$(d)$ There are various ways to do this, all ending with the minimal polynomial $f(x)=x^2-2\in\mathbb{Q}(i)[x]$. We could work by looking for roots: if $f(x)$ was reducible in $\mathbb{Q}(i)[x]$, then it would have a root in $\mathbb{Q}(i)$. But for $a,b\in\mathbb{Q}$, \[\begin{align*} (a+bi)^2=2 \quad &\implies\quad a^2-b^2+2abi=2 \\ \quad &\implies\quad a^2-b^2=2\;\;\text{and}\;\; 2ab=0 \\ \quad &\implies\quad a=0,\; b^2=-2\;\;\text{or}\;\; b=0,\; a^2=2 \\ \end{align*}\] which is impossible as $\sqrt{2}$ is irrational.

Or we could work with the Tower Law, using $[\mathbb{Q}(i):\mathbb{Q}]=[\mathbb{Q}\sqrt{2}):\mathbb{Q}]=2$: \[\begin{align*} i\not\in\mathbb{Q}(\sqrt{2}),\; i^2\in\mathbb{Q}(\sqrt{2}) \quad &\implies\quad [\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}(\sqrt{2})]=2 \\ \quad &\implies\quad [\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}]=4 \\ \quad &\implies\quad [\mathbb{Q}(i,\sqrt{2}):\mathbb{Q}(i)]=2 \end{align*}\] so the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}(i)$ has degree $2$ and must be $f(x)$.

$(e)$ The minimal polynomial of $\sqrt[3]{2}$ over $K=\mathbb{Q}(\sqrt{2})$, has degree $[K(\sqrt[3]{2}):K]$.

Clearly $\sqrt{2},\sqrt[3]{2}\in\mathbb{Q}(\sqrt[6]{2})$ so $K(\sqrt[3]{2})\subset\mathbb{Q}(\sqrt[6]{2})$.
Also, $\sqrt[6]{2}=\sqrt{2}/\sqrt[3]{2}\in K(\sqrt[3]{2})$ and so $K(\sqrt[3]{2})=\mathbb{Q}(\sqrt[6]{2})$, which has degree $6$ over $\mathbb{Q}$.
Hence using $[K(\sqrt[3]{2}):\mathbb{Q}]=[K(\sqrt[3]{2}):K] [K:\mathbb{Q}]$ we get $[K(\sqrt[3]{2}):K]=3$.
Thus the minimal polynomial has degree $3$ so must be $x^3-2$.

Exercise 2.3 For the following field extensions $L/K$, find a basis of $L$ (as a vector space over $K$).

$\;(a)\;\;L=\mathbb{Q}(\sqrt{2}\,,\,i)$ over $K=\mathbb{Q},\hspace{5.1em}$ $\;(b)\;\;L=\mathbb{Q}(\sqrt{2}\,,\,\sqrt{5})$ over $K=\mathbb{Q},$

$\;(c)\;\;L=\mathbb{Q}(\sqrt[3]{5}\,,\,i)$ over $K=\mathbb{Q},\hspace{5.1em}$ $\;(d)\;\;L=\mathbb{Q}(\sqrt[3]{2}-2\sqrt[3]{4})$ over $K=\mathbb{Q},$

$\;(e)\;\;L=\mathbb{Q}(\sqrt{2}\,,\,\sqrt{3}\,,\,\sqrt{7})$ over $K=\mathbb{Q},\hspace{2em}$ $\;(f)\;\;L=\mathbb{Q}(\sqrt[3]{1+\sqrt{3}})$ over $K=\mathbb{Q}(\sqrt{3}),$

$\;(g)\;\;L=\mathbb{R}(x)$ over $K=\mathbb{R}(x+x^{-1}),\hspace{3em}$ $\;(h)\;\;L=\mathbb{R}(x)$ over $K=\mathbb{R}(x^2+x^{-2}),$

$\;(i)\;\;L=\mathbb{F}_2(\alpha)$ over $K=\mathbb{F}_2$ where $\alpha^4+\alpha+1=0,$

$\;(j)\;\;L=\mathbb{F}_3(\alpha)$ over $K=\mathbb{F}_3$ where $\alpha^3+\alpha^2+2=0.$

Solution

Recall the key to proving the Tower Law: given $K\subset M\subset L$, if $\{\alpha_i \mid i\in I \}$ is a basis of $M/K$ and $\{\beta_j\mid j\in J\}$ is a basis of $L/M$, then $\{\alpha_i\beta_j\; \mid i\in I, j\in J \}$ is a basis of $L/K$.

$(a)$ Apply this to the tower $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt{2},i)$:

We have $i\not\in\mathbb{Q}(\sqrt{2})$ so $\mathbb{Q}(\sqrt{2},i)$ has a basis $\{1,i\}$ over $\mathbb{Q}(\sqrt{2})$.
Also $\mathbb{Q}(\sqrt{2})$ has a basis $\{1,\sqrt{2}\}$ over $\mathbb{Q}$.
So $\mathbb{Q}(\sqrt{2},i)$ has a basis $\{1,\sqrt{2},i,i\sqrt{2}\}$ over $\mathbb{Q}$.

$(b)$ Using the tower $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt{2},\sqrt{5})$:

Show that $\sqrt{5}\not\in\mathbb{Q}(\sqrt{2})$ so that $\mathbb{Q}(\sqrt{2},\sqrt{5})$ has a basis $\{1,\sqrt{5}\}$ over $\mathbb{Q}(\sqrt{2})$.
Also $\mathbb{Q}(\sqrt{2})$ has a basis $\{1,\sqrt{2}\}$ over $\mathbb{Q}$.
So $\mathbb{Q}(\sqrt{2},\sqrt{5})$ has a basis $\{1,\sqrt{2},\sqrt{5},\sqrt{10}\}$ over $\mathbb{Q}$.

$(c)$ Similarly, using $\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{5})\subset\mathbb{Q}(\sqrt[3]{5},i)$

We have $i\not\in\mathbb{Q}(\sqrt[3]{5})$ so $\mathbb{Q}(\sqrt[3]{5},i)$ has a basis $\{1,i\}$ over $\mathbb{Q}(\sqrt[3]{5})$.
Also $\mathbb{Q}(\sqrt[3]{5})$ has a basis $\{1,\sqrt[3]{5},\sqrt[3]{25}\}$ over $\mathbb{Q}$.
So $\mathbb{Q}(\sqrt[3]{5},i)$ has a basis $\{1,\sqrt[3]{5},\sqrt[3]{25},i,i\sqrt[3]{5},i\sqrt[3]{25}\}$ over $\mathbb{Q}$.

$(d)$ Notice that $\mathbb{Q}\subset L\subset\mathbb{Q}(\sqrt[3]{2})$, $L\neq\mathbb{Q}$ and $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ has prime degree $3$. This implies $L=\mathbb{Q}(\sqrt[3]{2})$, and so a basis is $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$.

$(e)$ Using $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2},\sqrt{3})\subset\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{7})$:

In the same way as above, show that $\sqrt{3}\not\in\mathbb{Q}(\sqrt{2})$ and find $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has a basis $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ over $\mathbb{Q}$.
Now show that $\sqrt{7}\not\in\mathbb{Q}(\sqrt{2},\sqrt{3})$ e.g. as follows: if $\sqrt{7}=\alpha+\beta\sqrt{3}$ for some $\alpha,\beta\in\mathbb{Q}(\sqrt{2})$, then \[\begin{align*} (\alpha+\beta\sqrt{3})^2=7 \quad &\implies\quad (\alpha^2+3\beta^2)+2\alpha\beta\sqrt{3}=7 \\ \quad &\implies\quad \alpha^2+3\beta^2=7\;\;\text{and}\;\; 2\alpha\beta=0 \\ \quad &\implies\quad \alpha=0,\; 3\beta^2=7\;\;\text{or}\;\; \beta=0,\; \alpha^2=7 \\ \quad &\implies\quad 3\sqrt{7/3}=\sqrt{21}\in\mathbb{Q}(\sqrt{2})\;\;\text{or}\;\; \sqrt{7}\in\mathbb{Q}(\sqrt{2}) \\ \end{align*}\] Both cases are impossible by a similar argument to the proof that $\sqrt{3}\not\in\mathbb{Q}(\sqrt{2})$. Hence $L$ has basis $\{1,\sqrt{7}\}$ over $\mathbb{Q}(\sqrt{2},\sqrt{3})$.
Finally $[L:\mathbb{Q}]=8$ and we have a basis $\{1,\sqrt{2},\sqrt{3},\sqrt{6},\sqrt{7},\sqrt{14},\sqrt{21},\sqrt{42}\}$

More generally, you might guess that for distinct primes $p_1,...,p_n$, one has \[[\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}):\mathbb{Q}]=2^n\] and you’d be right!

$(f)$ The polynomial $x^3-(1+\sqrt{3})$ is irreducible in $K[x]=\mathbb{Q}(\sqrt{3})[x]$. To prove this,

suppose $(a+b\sqrt{3})^3=1+\sqrt{3}$ for some $a,b\in\mathbb{Q}$.
Taking conjugates gives $(a-b\sqrt{3})^3=1-\sqrt{3}$.
Then $(a^2-3b^2)^3=(a+b\sqrt{3})^3(a-b\sqrt{3})^3=(1+\sqrt{3})(1-\sqrt{3})=-2$ which is impossible since $\sqrt[3]{2}\not\in\mathbb{Q}$.
Thus the cubic polynomial $x^3-(1+\sqrt{3})$ has no linear factors in $K[x]$ so must be irreducible.

This implies a $K$-basis of $L$ is $\{1,\theta,\theta^2\}$ where $\theta=\sqrt[3]{1+\sqrt{3}}$.

$(g)$ We are considering the extension $\mathbb{R}(t)\subset\mathbb{R}(x)$ where $t=x+x^{-1}$. Notice then that $x^2-tx+1=0$, i.e. $x$ is a root of the quadratic polynomial \[f(X)=X^2-tX+1\in K[X].\] We claim that $f(X)$ is irreducible in $K[X]$. By the analogue of Gauss’s Lemma from lectures, if $f(X)$ has a root in $K=\mathbb{R}(t)$, then it has a root $\alpha\in\mathbb{R}[t]$. Furthermore, $\alpha$ divides the constant term $1$ in $f(X)$ and so $\alpha\in\mathbb{R}$. But now $\alpha^2-t\alpha+1=0$ in $\mathbb{R}[t]$ and a polynomial is zero only when its coefficients are all zero, i.e. $\alpha^2+1=0$ and $\alpha=0$. This is clearly impossible for any $\alpha\in\mathbb{R}$.

Thus $f(X)$ is the minimal polynomial of $x$ over $K$ so $[L:K]=\deg(f(x))=2$ and a basis is given by $\{1,x\}$. Explicitly, \[L=\mathbb{R}(x)=\left\{a+bx \mid a,b\in K=\mathbb{R}(x+x^{-1})\right\}.\]

$(h)$ Let $t=x^2+x^{-2}$ so that $K=\mathbb{R}(t)$ and $L=K(x)$. Then $x^4-tx^2+1=0$, i.e. $x$ is a root of \[f(X)=X^4-tX^2+1\in K[X].\] At the end of Section 2.3.2 of the notes, we actually showed this polynomial is irreducible in $K[X]$. Here’s a quick reminder:

It has degree $4$ (in the indeterminate $x$) so we need to check it has no linear or quadratic factors.
By Gauss’s Lemma, if $f(X)$ has a root in $K=\mathbb{R}(t)$, then it has a root $\alpha\in\mathbb{R}[t]$. But $\alpha$ must divide the constant term $1$ in $f(X)$ and so $\alpha$ must itself be constant, i.e. $\alpha\in\mathbb{R}$. We have $\alpha^4-t\alpha^2+1=0$ in $\mathbb{R}[t]$ and the coefficients of this polynomial in $t$ must be zero, i.e. $\alpha^4+1=0$ and $\alpha^2=0$. This is impossible and hence $f(X)$ has no linear factors.
Again by Gauss’s Lemma, if $f(X)$ is a product of two quadratic polynomials in $\mathbb{R}(t)[X]$ then \[f(X)=X^4-tX^2+1=(X^2+aX+b)(X^2+cX+d)\] for some polynomials $a, b, c, d\in\mathbb{R}[t]$. The constant coefficient gives $bd=1$ which implies that $b, d\in\mathbb{R}^\times$, the $x^3$ coefficient gives $a+c=0$ which implies that $\deg(a)=\deg(c)$ and the $x^2$ coefficient gives $b+d+ac=-t$ which implies that $\deg(ac)=1$. As a result, $\deg(ac)=2\deg(a)=1\implies \deg(a)=1/2$ which is impossible.

Thus $f(X)$ is the minimal polynomial of $x$ over $K$ so $[L:K]=\deg(f(X))=4$ and a $K$-basis for $L$ is $\{1,x,x^2,x^3\}$.

$(i)$ The polynomial $x^4+x+1\in\mathbb{F}_2[x]$ is irreducible, as we saw in Exercise 1.1. (It has no roots in $\mathbb{F}_2$ so no linear factors, and no quadratic factors since the only irreducible quadratic is $x^2+x+1$).

We have a simple extension with $[\mathbb{F}_2(\alpha):\mathbb{F}_2]=4$ and a basis is $\{1,\alpha,\alpha^2,\alpha^3\}$.

$(j)$ The polynomial $x^3+x^2+2\in\mathbb{F}_3[x]$ has degree $3$ but no roots in $\mathbb{F}_3$. That means it is irreducible, $[\mathbb{F}_3(\alpha):\mathbb{F}_3]=3$ and a basis is $\{1,\alpha,\alpha^2\}$.

Exercise 2.4 Write each of the following in the form $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ with $a, b, c\in\mathbb{Q}$:

$\;(a)\;\;\dfrac{1}{\sqrt[3]{2}},\hspace{2em}$ $\;(b)\;\;\dfrac{1}{1+\sqrt[3]{2}},\hspace{2em}$ $\;(c)\;\;\dfrac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}},$

$\;(d)\;(\star)\;\dfrac{1}{r+s\sqrt[3]{2}+t\sqrt[3]{4}}\;\;$ for arbitrary $r,s,t\in\mathbb{Q}$ with $r+s\sqrt[3]{2}+t\sqrt[3]{4}\neq 0$.

Solution

Recall that $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ has basis $\{1,\sqrt[3]{2},\sqrt{4}\}$ so this question reduces to linear algebra.

$(a)$ We have \[\begin{align*} \dfrac{1}{\sqrt[3]{2}}=a+b\sqrt[3]{2}+c\sqrt[3]{4} \quad &\iff\quad a\sqrt[3]{2}+b\sqrt[3]{4}+2c=1 \\ \quad &\iff\quad a=0,\;\;b=0, \;\; c=1/2 \\ &\,\qquad\qquad\text{so}\;\; \dfrac{1}{\sqrt[3]{2}}=\dfrac{\sqrt[3]{4}}{2} \end{align*}\]

$(b)$ We have \[\begin{align*} \dfrac{1}{1+\sqrt[3]{2}}=a+b\sqrt[3]{2}+c\sqrt[3]{4} \quad &\iff\quad (1+\sqrt[3]{2})(a+b\sqrt[3]{2}+2c\sqrt[3]{4})=1 \\ \quad &\iff\quad (a+2c)+(a+b)\sqrt[3]{2}+(b+c)\sqrt[3]{4}=1 \\ \quad &\iff\quad a+2c=1,\;\; a+b=0 \;\;\text{and}\;\; b+c=0 \\ \quad &\iff\quad a=1/3,\;\; b=-1/3 \;\;\text{and}\;\; c=1/3 \\ &\,\qquad\qquad\text{so}\;\; \dfrac{1}{1+\sqrt[3]{2}}=\dfrac{1-\sqrt[3]{2}+\sqrt[3]{4}}{3} \end{align*}\]

Notice we could have done this more quickly, using $A^3+B^3=(A+B)(A^2-AB+B^2)$ with $A=1$ and $B=\sqrt[3]{2}$.

$(c)$ We have \[\dfrac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}=a+b\sqrt[3]{2}+c\sqrt[3]{4}\] \[\begin{align*} \quad &\iff\quad (1+\sqrt[3]{2}+\sqrt[3]{4})(a+b\sqrt[3]{2}+c\sqrt[3]{4})=1+\sqrt[3]{2} \\ \quad &\iff\quad (a+2b+2c)+(a+b+2c)\sqrt[3]{2}+(a+b+c)\sqrt[3]{4}=1 \\ \quad &\iff\quad a+2b+2c=1,\;\; a+b+2c=1 \;\;\text{and}\;\; a+b+c=0 \\ \quad &\iff\quad a=-1,\;\; b=0 \;\;\text{and}\;\; c=1 \\ &\,\qquad\qquad\text{so}\;\; \dfrac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}=-1+\sqrt[3]{4} \end{align*}\] The simple form of this answer suggests there might have been a cheeky shortcut and indeed there is: \[\dfrac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\dfrac{(1+\sqrt[3]{2})(1-\sqrt[3]{2})}{(1+\sqrt[3]{2}+\sqrt[3]{4})(1-\sqrt[3]{2})} =\dfrac{1-\sqrt[3]{4}}{1-2}=-1+\sqrt[3]{4}\]

$(d)$ This can be done by brute force in the same way. Multiplying out \[(r+s\sqrt[3]{2}+t\sqrt[3]{4})(a+b\sqrt[3]{2}+c\sqrt[3]{4})=1\] leads to three linear equations. They can be expressed in matrix form as \[\begin{pmatrix} r & 2t & 2s \\ s & r & 2t \\ t & s & r \end{pmatrix}\begin{pmatrix} a \\ b \\ c \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix}\] and solving for $a,b,c$ (using a computer helps!) gives \[\frac{1}{r+s\sqrt[3]{2}+t\sqrt[3]{4}}=\frac{(r^2-2st)+(2t^2-rs)\sqrt[3]{2}+(s^2-rt)\sqrt[3]{4}}{D}\] where $D=r^3+2s^3+4t^3-6rst$ is the determinant of the matrix. Of course, we are left with the problem that $D$ might equal zero but I’ll let you worry about that…(it only happens if $r=s=t=0$).

There is a more sophisticated way to think about this. Let $\omega=e^{2\pi i/3}$, then we can think about “conjugates” of $\sqrt[3]{2}$ - they are the other roots of the polynomial $x^2-3$, i.e. $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$.

And multiplying “conjugates” does magic: \[(r+s\sqrt[3]{2}+t\sqrt[3]{4})(r+s\omega\sqrt[3]{2}+t\omega^2\sqrt[3]{4})(r+s\omega^2\sqrt[3]{2}+t\omega\sqrt[3]{4})=D\] Try multiplying the second two factors, using $\omega^2+\omega+1=0$ - you’ll find all $\omega$’s disappear and we get the formula for the inverse given above. For Number Theory III people, the rational number $D$ is the called the norm of $r+s\sqrt[3]{2}+t\sqrt[3]{4}$ in the cubic number field $\mathbb{Q}(\sqrt[3]{2})$.

Exercise 2.5 Write each of the following in the form $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ with $a, b, c, d\in\mathbb{Q}$:

$\;(a)\;\;\dfrac{1}{\sqrt{2}+\sqrt{3}},\hspace{2em}$ $\;(b)\;\;\dfrac{1}{1+\sqrt{2}+\sqrt{3}},\hspace{2em}$ $\;(c)\;\;\dfrac{\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}+\sqrt{6}}.$

Solution

Similarly to the last question, $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$ has basis $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ so we could always do some linear algebra. However, we perhaps might be able to spot a shortcut…

$(a)$ The difference of two squares is useful - we have $(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=1$ so \[\frac{1}{\sqrt{2}+\sqrt{3}}=-\sqrt{2}+\sqrt{3}.\]

$(b)$ $(1+\sqrt{2}+\sqrt{3})(1+\sqrt{2}-\sqrt{3})=(1+\sqrt{2})^2-3=2\sqrt{2}$ so \[\frac{1}{1+\sqrt{2}+\sqrt{3}}=\frac{1+\sqrt{2}-\sqrt{3}}{2\sqrt{2}}=\frac{2+\sqrt{2}-\sqrt{6}}{4}.\]

$(c)$ This time, spot that the denominator factorises: \[\begin{align*} \frac{\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}+\sqrt{6}}=\frac{\sqrt{2}+\sqrt{3}}{(1+\sqrt{2})(1+\sqrt{3})} &=\frac{(\sqrt{2}+\sqrt{3})(1-\sqrt{2})(1-\sqrt{3})}{(1+\sqrt{2})(1-\sqrt{2})(1+\sqrt{3})(1-\sqrt{3})} \\ &=\frac{(\sqrt{2}+\sqrt{3})(1-\sqrt{2}-\sqrt{3}+\sqrt{6})}{2} \\ &=\frac{-5+4\sqrt{2}+3\sqrt{3}-2\sqrt{6}}{2} \end{align*}\]

Something to notice - multiplying by “conjugates”, i.e. swapping $\sqrt{2}\leftrightarrow-\sqrt{2}$ and/or $\sqrt{3}\leftrightarrow-\sqrt{3}$ appears to be very useful. This is not an accident…

Exercise 2.6 Let $L=\mathbb{F}_2(\theta)$ where $\theta$ is a root of $f(x)=x^4+x+1\in\mathbb{F}_2[x]$. Write each of the following in the form $a+b\theta+c\theta^2+d\theta^3$ with $a, b, c, d\in\mathbb{F}_2$:

$\;(a)\;\;\theta^{-1},\hspace{3em}$ $\;(b)\;\;\theta^5,\hspace{3em}$ $\;(c)\;\;\theta^{15},\hspace{3em}$ $\;(d)\;\;(1+\theta+\theta^2)^{-1}.$

Solution

Working in $L=\mathbb{F}_2(\theta)$, we can repeatedly use $2=0$ and $\theta^4=-1-\theta=1+\theta$. This makes calculations quite easy…

$(a)$ Notice $\theta^4+\theta+1=0$ implies $\theta(\theta^3+1)=1$ so $\theta^{-1}=1+\theta^3$.

$(b)$ This time, $\theta^5=\theta\theta^4=\theta(1+\theta)=\theta+\theta^2$.

$(c)$ We have \[\begin{align*} \theta^{15}=(\theta^5)^3=(\theta+\theta^2)^3 &=\theta^3+\theta^4+\theta^5+\theta^6 \\ &=\theta^3+(1+\theta)+(\theta+\theta^2)+(\theta^2+\theta^3)=1. \end{align*}\] A more efficient way would be to remember squaring is really nice in characteristic $2$: \[\theta^{16}=(\theta^4)^4=(1+\theta)^4=(1+\theta^2)^2=1+\theta^4=\theta\] and so $\theta^{15}=1$. More generally, we will prove next term for any finite field $\mathbb{F}_q$, we have $\alpha^q=\alpha$ for all $\alpha\in\mathbb{F}_q$, so this is the special case $\theta^{16}=\theta$ in $\mathbb{F}_{16}=\mathbb{F}_2(\theta)$.

$(d)$ If you can’t spot any tricks, the linear algebra route will always work: multiplying $(1+\theta+\theta^2)(a+b\theta+c\theta^2+d\theta^3)=1$ gives \[\begin{align*} &\quad a+(a+b)\theta+(a+b+c)\theta^2+(b+c+d)\theta^3+(c+d)\theta^4+d\theta^5=1 \\ \quad &\iff\quad (a+c+d)+(a+b+c)\theta+(a+b+c+d)\theta^2+(b+c+d)\theta^3=1 \\ \quad &\iff\quad a+c+d=1,\;\; a+b+c=0,\;\; a+b+c+d=0,\;\; b+c+d=0 \\ \quad &\iff\quad a=0,\;\; b=1,\;\; c=1,\;\; d=0 \end{align*}\] and so $(1+\theta+\theta^2)^{-1}=\theta+\theta^2$.

There are quicker ways, however. For instance,

recall a difference of two cubes is $(A-B)(A^2+AB+B^2)=A^3-B^3$.
In our question, this means $(1+\theta)(1+\theta+\theta^2)=1+\theta^3$.
Now multiply both sides by $\theta$ to get $(\theta+\theta^2)(1+\theta+\theta^2)=\theta+\theta^4=1$.

Exercise 2.7 Let $L=\mathbb{F}_2(\theta)$ where $\theta$ is a root of $f(x)=x^3+x^2+1\in\mathbb{F}_2[x]$. Show that $f(x)$ can be written as a product of three linear factors with coefficients in $L$.

Solution

To factorise $f(x)$ into linear factors, we need to find three roots. The field $L=\mathbb{F}_2(\theta)$ only has $8$ elements \[\mathbb{F}_2(\theta)=\left\{0,\, 1,\, \theta,\, 1+\theta,\, \theta^2,\, 1+\theta^2,\, \theta+\theta^2,\, 1+\theta+\theta^2 \right\}\] so we could just test which are roots. By definition, $\theta$ is one of them and $\theta^3=-\theta^2-1=\theta^2+1$. Trying $\theta^2$, for instance, we get \[f(\theta^2)=\theta^6+\theta^4+1=(\theta^2+1)^2+\theta^4+1=\theta^4+1+\theta^4+1=0\] so $\theta^2$ is also a root. We could continue through the list in the same way to find the third root is the last element in the list $1+\theta+\theta^2$. This results in the factorisation \[f(x)=x^3+x^2+1=(x-\theta)(x-\theta^2)(x-\theta^2-\theta-1).\]

Thankfully, there is a more efficient way: remember that in characteristic $2$, we have \[A^2+B^2+C^2+...=(A+B+C+...)^2\] In particular, $f(\theta^2)=f(\theta)^2=0$ so $\theta^2$ is a root. Similarly, $f(\theta^4)=f(\theta^2)^2=0$ so we also get $\theta^4$ is a root. Continuing, we find $\theta,\theta^2,\theta^4,\theta^8,\theta^{16},...$ are all roots!

However, the set of non-zero elements $\mathbb{F}_2(\theta)$ is a group under multiplication of order $7$ so $\theta^7=1$, and we obtain exactly three roots $\theta, \theta^2, \theta^4$. Notice $\theta^4=\theta^3+\theta=\theta^2+\theta+1$ is the root from earlier.

The same trick will work for any polynomial $f(x)\in\mathbb{F}_p[x]$ for other primes $p$, using $(A+B)^p=A^p+B^p$. If $\theta$ is a root, then so are $\theta^p,\theta^{p^2},\theta^{p^3},...$

Exercise 2.8 Find the minimal polynomial of $\alpha=\sqrt{7}+i\in\mathbb{C}$ over the field $K$ in the following cases:

$\;(a)\;\;K=\mathbb{R},\hspace{3em}$ $\;(b)\;\;K=\mathbb{Q}(i),\hspace{3em}$ $\;(c)\;\;K=\mathbb{Q}.$

Solution

$(a)$ Since $\sqrt{7}\in\mathbb{R}$, we have $\mathbb{R}(\alpha)=\mathbb{R}(i)=\mathbb{C}$ and this has degree $2$ over $\mathbb{R}$. Also, \[(\alpha-\sqrt{7})^2=-1\quad\implies\quad \alpha^2-2\sqrt{7}\alpha+8=0\] so the minimal polynomial of $\alpha$ over $\mathbb{R}$ is $x^2-2\sqrt{7}x+8\in\mathbb{R}[x]$.

$(b)$ This time, $\mathbb{Q}(i)(\alpha)=\mathbb{Q}(i)(\sqrt{7})$ which has degree $2$ over $\mathbb{Q}(i)$. Also, \[(\alpha-i)^2=7\quad\implies\quad \alpha^2-2i\alpha-8=0\] so the minimal polynomial of $\alpha$ over $\mathbb{Q}(i)$ is $x^2-2ix-8\in\mathbb{Q}(i)[x]$.

$(c)$ The extension $\mathbb{Q}(i,\sqrt{7})/\mathbb{Q}$ has basis $\{1,i,\sqrt{7},i\sqrt{7}\}$ (in a similar way to other examples we’ve seen such as $\mathbb{Q}(\sqrt{2},\sqrt{7})/\mathbb{Q}$). We have $\mathbb{Q}\subset\mathbb{Q}(\alpha)\subset\mathbb{Q}(i,\sqrt{7})$ so $[\mathbb{Q}(\alpha):\mathbb{Q}]$ divides $[\mathbb{Q}(i,\sqrt{7}):\mathbb{Q}]=4$.

However, the three elements $1$, $\alpha=\sqrt{7}+i$ and $\alpha^2=6+2i\sqrt{7}$ are linearly independent over $\mathbb{Q}$ and so $[\mathbb{Q}(\alpha):\mathbb{Q}]\geq 3$. Hence we must have $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$ and the minimal polynomial of $\alpha$ over $\mathbb{Q}$ has degree $4$. To find it, eliminate the irrationalities: \[\begin{align*} (\alpha-\sqrt{7})^2=-1 \quad &\implies\quad \alpha^2-2\sqrt{7}\alpha+8=0 \\ \quad &\implies\quad (\alpha^2+8)^2=(2\sqrt{7}\alpha)^2=28\alpha^2 \\ \quad &\implies\quad \alpha^4-12\alpha^2+64=28 \end{align*}\] so the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $x^4-12x^2+64\in\mathbb{Q}[x]$.

An alternative, perhaps quicker way to show $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$ is to show directly that $\mathbb{Q}(\alpha)=\mathbb{Q}(i,\sqrt{7})$. Obviously, $\alpha$ can be written in terms of $\sqrt{7}$ and $i$. We just have to write $\sqrt{7}$ and $i$ in terms of $\alpha$.

Notice $(\sqrt{7}+i)(\sqrt{7}-i)=8$ so
\[ \begin{cases} \sqrt{7}+i=\alpha \\\sqrt{7}-i=8\alpha^{-1} \end{cases}\implies \begin{cases} \sqrt{7}=(\alpha+8\alpha^{-1})/2\in\mathbb{Q}(\alpha) \\ \;\;\;\,i=(\alpha-8\alpha^{-1})/2\in\mathbb{Q}(\alpha)\end{cases}\]

Exercise 2.9 Find the minimal polynomials over $\mathbb{Q}$ of the following algebraic numbers:

$\;(a)\;\;\sqrt{1+\sqrt{2}},\hspace{3em}$ $\;(b)\;\;\sqrt{\sqrt{2}-i}.$

Solution

Tip: it is often easier to show a polynomial is a minimal polynomial by first determining its degree, rather than using a Gauss’s Lemma argument to show it is irreducible.

$(a)$ Let $\alpha=\sqrt{1+\sqrt{2}}$ so that $\alpha^2=1+\sqrt{2}\in\mathbb{Q}(\sqrt{2})$ and we have a tower \[\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})=\mathbb{Q}(\alpha^2)\subset\mathbb{Q}(\alpha).\] We might guess that $[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{2})]=2$ so that the minimal polynomial of $\alpha$ over $\mathbb{Q}$ has degree $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$, but we need to prove it. We want to show $\alpha\not\in\mathbb{Q}(\sqrt{2})$.

Suppose that $1+\sqrt{2}$ is a square in $\mathbb{Q}(\sqrt{2}$), i.e. $(a+b\sqrt{2})^2=1+\sqrt{2}$ for $a,b\in\mathbb{Q}$.
Take conjugates to get $(a-b\sqrt{2})^2=1-\sqrt{2}$.
Then $0\leq(a^2-2b^2)^2=(a+b\sqrt{2})^2(a-b\sqrt{2})^2=(1+\sqrt{2})(1-\sqrt{2})=-1$ which is false.

Hence $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$ and rearranging $(\alpha^2-1)^2=2$ shows the minimal polynomial is $x^4-2x^2-1$.

$(b)$ Let $\alpha=\sqrt{\sqrt{2}-i}$ and $\beta=\alpha^2=\sqrt{2}-i$. We clearly have $\mathbb{Q}\subset\mathbb{Q}(\beta)\subset\mathbb{Q}(\sqrt{2},i)$.

Also, $(\sqrt{2}-i)(\sqrt{2}+i)=3$ and \[\begin{cases} \sqrt{2}-i=\beta \\ \sqrt{2}+i=3\beta^{-1} \end{cases}\implies \begin{cases} \sqrt{2}=(\beta+3\beta^{-1})/2\in\mathbb{Q}(\beta) \\ \;\;\;\,i=(3\beta^{-1}-\beta)/2\in\mathbb{Q}(\beta)\end{cases}\] Hence $\mathbb{Q}(\beta)=\mathbb{Q}(\sqrt{2},i)$ and by similar arguments as before, $[\mathbb{Q}(\sqrt{2},i):\mathbb{Q}]=4$.

Consider the tower \[\mathbb{Q}\subset\mathbb{Q}(\beta)=\mathbb{Q}(\sqrt{2},i)\subset\mathbb{Q}(\alpha).\] We will show that $\sqrt{2}-i$ is not a square in $\mathbb{Q}(\sqrt{2},i)$, as then $[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{2},i)]=2$. By the Tower Law, we will then know the minimal polynomial of $\alpha$ over $\mathbb{Q}$ has degree $[\mathbb{Q}(\alpha):\mathbb{Q}]=8$.

Suppose that $(a+bi)^2=\sqrt{2}-i$ where $a,b\in\mathbb{Q}(\sqrt{2})$
Taking (complex) conjugates gives $(a-bi)^2=\sqrt{2}+i$
Multiplying gives $(a^2+b^2)^2=3$ and so $\sqrt{3}\in\mathbb{Q}(\sqrt{2})$. This is false.

Hence the minimal polynomial of $\alpha$ over $\mathbb{Q}$ does indeed have degree $8$. We just need to find it: \[\begin{align*} &\quad \alpha^2=\sqrt{2}-i \quad \implies\quad (\alpha^2+i)^2=2 \quad\implies\quad \alpha^4-3=-2i\alpha^2 \\ \quad &\implies\quad (\alpha^4-3)^2=-4\alpha^4 \quad\implies\quad \alpha^8-2\alpha^4+9=0 \end{align*}\] and the required minimal polynomial is $x^8-2x^4+9$.

Exercise 2.10 Find the degree $[L:K]$ of the following field extensions and give a basis for $L$ over $K$:

$\;(a)\;\;L=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ over $K=\mathbb{Q},\hspace{3.75em}$ $\;(b)\;\;L=\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$ over $K=\mathbb{Q},$

$\;(c)(\star)\;\;L=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{6},\sqrt[3]{24})$ over $K=\mathbb{Q},\hspace{2em}$ $\;(d)\;\;L=\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $K=\mathbb{Q}(\sqrt{2}+\sqrt{3}),$

$\;(e)\;\;L=\mathbb{Q}(\sqrt{2},\sqrt{6}+\sqrt{10})$ over $K=\mathbb{Q}(\sqrt{3}+\sqrt{5}).$

Solution

$(a)$ We have $[L:K]=8$ and a basis is $\{1,\sqrt{2},\sqrt{3},\sqrt{6},\sqrt{5},\sqrt{10},\sqrt{15},\sqrt{30}\}$.

As we’ve seen before, $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$ has degree $4$ and basis $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$.
The key point is to show that $\sqrt{5}\not\in\mathbb{Q}(\sqrt{2},\sqrt{3})$. Suppose on the contrary that $\sqrt{5}=\alpha+\beta\sqrt{3}$ for some $\alpha,\beta\in\mathbb{Q}(\sqrt{2})$, then \[\begin{align*} (\alpha+\beta\sqrt{3})^2=5 \quad &\implies\quad (\alpha^2+3\beta^2)+2\alpha\beta\sqrt{3}=5 \\ \quad &\implies\quad \alpha^2+3\beta^2=5\;\;\text{and}\;\; 2\alpha\beta=0 \\ \quad &\implies\quad \alpha=0,\; 3\beta^2=5\;\;\text{or}\;\; \beta=0,\; \alpha^2=5 \\ \quad &\implies\quad 3\sqrt{5/3}=\sqrt{15}\in\mathbb{Q}(\sqrt{2})\;\;\text{or}\;\; \sqrt{5}\in\mathbb{Q}(\sqrt{2}) \\ \end{align*}\]
Both cases are impossible and $L$ has basis $\{1,\sqrt{5}$ over $\mathbb{Q}(\sqrt{2},\sqrt{3})$.
Using the Tower Law (and its proof) then gives the degree and a basis of $L/K$.

$(b)$ We have $[L:K]=6$ and a basis is $\{1,2^{1/6},2^{2/6},2^{3/6},2^{4/6},2^{5/6}\}$.

Notice that $\sqrt{2},\sqrt[3]{2}\in\mathbb{Q}(2^{1/6})$ and $2^{1/6}=\sqrt{2}/\sqrt[3]{2}\in\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$.
Hence $L=\mathbb{Q}(\sqrt{2},\sqrt[3]{2})=\mathbb{Q}(2^{1/6})$ is a simple extension and $2^{1/6}$ is algebraic of degree $6$.

$(c)$ We have $[L:K]=9$ and a basis is $\{1,\sqrt[3]{2},\sqrt[3]{4},\sqrt[3]{3},\sqrt[3]{6},\sqrt[3]{12},\sqrt[3]{9},\sqrt[3]{18},\sqrt[3]{36}\}$.

First notice $24=2^2\times 6$ so $L=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{6})$. Also $6=2\times 3$ so we can write $L$ in the simpler form $L=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$ and can consider the tower \[K=\mathbb{Q}\;\;\subset\;\; M=\mathbb{Q}(\sqrt[3]{2})\;\;\subset\;\;L=\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})\]
We know $M/K$ has degree $3$ and basis $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$.
For $L/M$, suppose that $\sqrt[3]{3}\in \mathbb{Q}(\sqrt[3]{2})$, i.e. $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^3=3$ for some $a,b,c\in\mathbb{Q}$. Then \[(a^3+2b^3+4c^3+12abc)+3(a^2b+2b^2c+2ac^2)\sqrt[3]{2}+3(a^2c+2bc^2+ab^2)\sqrt[3]{4}=3\] \[\implies \begin{cases} a^3+2b^3+4c^3+12abc=3 \\ a^2b+2b^2c+2ac^2=0 \\ a^2c+2bc^2+ab^2=0 \end{cases}\] Multiply the 2nd equation by $c$ and subtract it from the 3rd equation multiplied by $b$. This give $a(b^3-2c^3)=0$. If $a\neq 0$, then $b^3=2c^3$ implies $\sqrt[3]{2}\in\mathbb{Q}$ which is false. So $a=0$ and the 2nd equation implies $b$ or $c$ is zero. But then the first equation reduces to $4c^3=3$ or $2b^3=3$, both of which are impossible.
Hence $[L:M]\neq 1$ and so must equal $3$ (since we are adjoining a cube root). A basis for $L/M$ is $\{1,\sqrt[3]{3},\sqrt[3]{9}\}$ and combining with the basis for $M/K$ we get the basis for $L/K$ given above.

$(d)$ We have $[L:K]=1$ and a basis is just $\{1\}$. (!)

$\alpha=\sqrt{2}+\sqrt{3}$ is in $\mathbb{Q}(\sqrt{2},\sqrt{3})$
$(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})=-1$ so $\alpha^{-1}=\sqrt{3}-\sqrt{2}$ and $\begin{cases} \sqrt{2}=(\alpha-\alpha^{-1})/2\in\mathbb{Q}(\sqrt{2}+\sqrt{3}) \\ \sqrt{3}=(\alpha+\alpha^{-1})/2\in\mathbb{Q}(\sqrt{2}+\sqrt{3})\end{cases}$
So $L=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})=K$ is a degree $1$ extension.

$(e)$ We have $[L:K]=2$ and a basis is $\{1,\sqrt{2}\}$.

To see this, notice $L=\mathbb{Q}(\sqrt{2},\sqrt{6}+\sqrt{10})=\mathbb{Q}(\sqrt{3}+\sqrt{5})(\sqrt{2})=K(\sqrt{2})$.
Then show $2$ is not a square in $K=\mathbb{Q}(\sqrt{3}+\sqrt{5})$.

Exercise 2.11 Prove that $f(x)=x^2-5$ is irreducible over $\mathbb{Q}(\sqrt[3]{7})$.

Solution

Suppose on the contrary that $x^2-5$ is not irreducible over $\mathbb{Q}(\sqrt[3]{7})$ and so $\sqrt{5}\in\mathbb{Q}(\sqrt[3]{7})$. Then we have a tower of fields \[\mathbb{Q}\subset\mathbb{Q}(\sqrt{5})\subset\mathbb{Q}(\sqrt[3]{7})\] and the Tower Law implies $[\mathbb{Q}(\sqrt{5}):\mathbb{Q}]=2$ divides $[\mathbb{Q}(\sqrt[3]{7}):\mathbb{Q}]=3$, contradiction.

Notice how considering degrees of extensions can show polynomials are irreducible really quickly!

Exercise 2.12 $(\star)\;$ Let $p>2$ be an odd prime. Show that there is exactly one field (up to isomorphism) with $p^2$ elements. Hint: try completing the square.

Solution

Any field $K$ with $p^2$ elements contains a prime subfield and since $p$ is the only prime dividing $p^2$, the only possible choice is $\mathbb{F}_p$. Now $K$ is a quadratic extension of $\mathbb{F}_p$ so is of the form $\mathbb{F}_p(\theta)$ where $\theta$ is a root of an irreducible monic quadratic polynomial $f(x)=x^2+ax+b\in\mathbb{F}_p[x]$. Since $p$ is odd, we can divide by $2$ in $\mathbb{F}_p$ so can complete the square to find $\theta$ via the usual quadratic formula \[\theta^2+a\theta+b=0\implies \theta=-a/2\pm\sqrt{r}\quad\text{where $r=a^2/4-b$.}\] Notice $\sqrt{r}\not\in\mathbb{F}_p$ as otherwise $\theta\in\mathbb{F}_p$. So $K=\mathbb{F}_p(\theta)=\mathbb{F}_p(\sqrt{r})$ where $r\in\mathbb{F}_p$ is not a square, i.e. every quadratic extension of $\mathbb{F}_p$ is formed by adjoining a square root.

Now $p$ is odd, so the multiplicative group of non-zero elements $\mathbb{F}_p^\times$ has order $p-1$, which is even. Furthermore, the set of non-zero squares, denoted by ${\mathbb{F}_p^\times}^2$, is a subgroup of index $2$ in $\mathbb{F}_p^\times$. It’s easy to check it is a subgroup. To see that it has index $2$, notice $a^2=(p-a)^2$ for $a\in\mathbb{F}_p$ so \[{\mathbb{F}_p^\times}^2=\left\{1^2, 2^2, ..., \left(\frac{p-1}{2}\right)^2\right\}\subset \mathbb{F}_p^\times.\] These are distinct, since for $a, b\in\{1,2,...,(p-1)/2\}\in\mathbb{F}_p$, we have $a+b\neq 0$ so \[a^2=b^2\;\;\implies\;\; (a-b)(a+b)=0\;\;\implies\;\; a-b=0\;\;\implies\;\; a=b.\] Hence $\mathbb{F}_p^\times$ is the disjoint union of ${\mathbb{F}_p^\times}^2$ (which are the squares) and a coset $r{\mathbb{F}_p^\times}^2$ (which are the non-squares) for any given non-square $r\in\mathbb{F}_p^\times$. In particular, if $s$ is any other non-square in $\mathbb{F}_p^\times$, then $s=rt^2$ for some $t\in\mathbb{F}_p$ and $\mathbb{F}_p(\sqrt{r})=\mathbb{F}_p(\sqrt{s})$.

Next term, we will see an alternative, more general method which will show there is a unique (up to isomorphism) field with $p^m$ elements for any $m\geq 1$.

2 Problem Sheet 2: Field extensions and minimal polynomials

2 Problem Sheet 2:
Field extensions and minimal polynomials