1 Problem Sheet 1:
Fields and polynomials

Exercise 1.1 $\;(a)\;$ Make a list of all (monic) irreducible polynomials of degrees at most $4$ in $\mathbb{F}_2[x]$, i.e. polynomials over the field $\mathbb{F}_2$ of $2$ elements.

$\;(b)\;$ Using part $(a)$ and without finding them directly, how many irreducible polynomials of degree $5$ are there in $\mathbb{F}_2[x]$?

$\;(c)\;$ Find the polynomials from part $(b)$.

Solution

Over $\mathbb{F}_2$, coefficients are either $0$ or $1$ so there aren’t too many polynomials to test…

$(a)$ Every non-zero linear polynomial is irreducible so

in degree $1$, the irreducible polynomials are $x$ and $x+1$.

Checking for linear factors over $\mathbb{F}_2$ is really quite easy. There is a factor $x$ if $x=0$ gives a root, i.e. there’s no constant term. There is a factor $x+1$ if $x=-1=1$ is a root, i.e. if there’s an even number of non-zero coefficients.

For degrees $2$ and $3$, irreducibility is the same as having no linear factor.

in degree $2$, the only irreducible polynomial is $x^2+x+1$,
in degree $3$, the irreducible polynomials are $x^3+x+1$ and $x^3+x^2+1$.

For degree $4$, there are four polynomials without linear factors - they are of the form \[x^4+ax^3+bx^2+cx+1\] with an odd number of non-zero coefficients. Explicitly, these are \[x^4+x+1,\quad x^4+x^2+1,\quad x^4+x^3+1,\quad x^4+x^3+x^2+x+1.\] We also have to exclude products of two irreducible quadratic factors but since we know $x^2+x+1$ is the only one of these, we just exclude $x^4+x^2+1=\left(x^2+x+1\right)^2$.

in degree $4$, the three irreducible polynomials are \[x^4+x+1, \qquad x^4+x^3+1, \qquad x^4+x^3+x^2+x+1.\]

$(b)$ There are $2^5=32$ monic degree $5$ polynomials in total, but to answer this part of the question, we don’t need to write any of them down. The ones without linear factors are of the form \[x^5+ax^4+bx^3+cx^2+dx+1\] with an odd number of non-zero coefficients $a,b,c,d$ - there are $\binom{4}{1}+\binom{4}{3}=8$ of these. We also have to exclude any product of an irreducible quadratic and an irreducible cubic. Looking at the lists above, there are exactly two of these. Hence there are six irreducible polynomials of degree $5$.

$(c)$ We just have to unravel what we did in part $(b)$. Write down the eight polynomials with one or three of $a,b,c,d$ being $1$, and then eliminate the two products of an irreducible quadratic and an irreducible cubic \[(x^2+x+1)(x^3+x+1)=x^5+x^4+1 \quad\text{and}\quad (x^2+x+1)(x^3+x^2+1)=x^5+x+1. \]

in degree $5$, the irreducible polynomials are \[x^5+x^3+1,\qquad x^5+x^2+1,\qquad x^5+x^4+x^3+x^2+1,\] \[x^5+x^4+x^3+x+1,\qquad x^5+x^4+x^2+x+1,\qquad x^5+x^3+x^2+x+1.\]

The number of irreducible polynomials of degree $n$ in $\mathbb{F}_2[x]$ is quite an interesting sequence.

Exercise 1.2 $\;$Decompose the following into a product of irreducible factors in the given polynomial ring:

$\;(a)\;\;x^3+2x+3$ in $\mathbb{F}_5[x],\hspace{2em}$ $\;(b)\;\;2x^3+x^2+2x+2$ in $\mathbb{F}_5[x],$

$\;(c)\;\;x^3+2x+3$ in $\mathbb{F}_7[x],\hspace{2em}$ $\;(d)\;\;2x^3+x^2+2x+2$ in $\mathbb{F}_7[x],$

$\;(e)\;\;x^4+64$ in $\mathbb{Q}[x],\hspace{4em}$ $\;(f)\;\;x^3+2$ in $\mathbb{Q}[x],$

$\;(g)\;\;x^4+1$ in $\mathbb{R}[x],\hspace{4.5em}$ $\;(h)\;\;x^6+27$ in $\mathbb{R}[x],$

$\;(i)\;\;x^7+1$ in $\mathbb{F}_2[x],\hspace{4.35em}$ $\;(j)\;\;x^3+2$ in $\mathbb{F}_3[x],$

$\;(k)\;\;x^4+2$ in $\mathbb{F}_3[x],\hspace{4.25em}$ $\;(l)\;\;x^4+2$ in $\mathbb{F}_{13}[x].$

Solution

$(a)\;$ Nice and easy, just look for linear factors and it falls apart: \[x^3+2x+3=(x+1)^2(x+3) \quad\text{in $\mathbb{F}_5[x]$.}\]

$(b)\;$ Just check there are no roots in $\mathbb{F}_5$ so no linear factors and as it is a cubic, \[2x^3+x^2+2x+2\quad\text{is irreducible in $\mathbb{F}_5[x]$.} \]

$(c)\;$ There is only one root $-1$ in $\mathbb{F}_7$ so \[x^3+2x+3=(x+1)(x^2-x+3)\quad\text{in $\mathbb{F}_7[x]$.}\] The quadratic factor must be irreducible as otherwise we’d have found more roots.

$(d)\;$ Again, look for roots to obtain linear factors: \[2x^3+x^2+2x+2=(2x-1)(x-1)(x+2)\quad\text{in $\mathbb{F}_7[x]$.}\]

$(e)\;$ First, notice $x^4+64$ has no linear factors: there are no roots in $\mathbb{R}$ so certainly no roots in $\mathbb{Q}$. We still need to check if it is the product of two quadratic factors: by Gauss’s Lemma, we can assume these have integer coefficients. The $x^4$ coefficient tells us the $x^2$ coefficients in our quadratic factors are either both $+1$ or both $-1$. By possible multiplying each factor by $-1$, we can assume the quadratic factors are monic \[x^4+64=(x^2+ax+b)(x^2+cx+d)\quad\text{with $a,b,c,d\in\mathbb{Z}$.}\] Now

comparing $x^3$ coefficients gives $0=a+c$ so $c=-a$.
comparing $x$ coefficients gives $0=ad+bc$ so $a(d-b)=0$ and either $a=0$ or $b=d$.
If $a=0$, then the $x^2$ coefficient is $0=ac+b+d=b+d$ and the constant coefficient is $64=bd=-b^2$, not possible.
If $b=d$, then $64=bd=b^2$ is possible: take $b=d=8$ and then $0=ad+bc$ gives $a,c=\pm 4$.

We obtain the factorisation $x^4+64=(x^2+4x+8)(x^2-4x+8)$. These quadratics are automatically irreducible since we know $x^4+64$ has no linear factors.

Alternatively, we could think back to the methods we used for quartic equations in the Introduction and try to write $x^4+64$ as the difference of two squares: \[\begin{align*} x^4+64&=(x^2+8)^2-16x^2 \\ &=(x^2+8)^2-(4x)^2=(x^2-4x+8)(x^2+4x+8) \end{align*}\] This is really efficient but not always so easy to spot!

$(f)\;$ By the rational root test, the only possible roots of $x^3+2$ in $\mathbb{Q}$ are $\pm 1$, $\pm 2$ which don’t work. As the polynomial is cubic, it must be irreducible in $\mathbb{Q}[x]$. Alternatively, it follows immediately using Eisenstein’s Criterion with $p=2$.

$(g)\;$ There are no real roots, hence no linear factors. Suppose it is a product of irreducible quadratic factors. We can assume they are monic by multiplying/dividing each by a constant: \[x^4+1=(x^2+ax+b)(x^2+cx+d) \quad\text{where $a,b,c,d\in\mathbb{\mathbb{R}}$.}\]

comparing constants and $x^3$ coefficients gives $bd=1$ and $a=-c$.
comparing $x$ coefficients gives $ad+bc=a(b-d)=0$ so $a=c=0$ or $b=d$.
if $a=c=0$, the $x^2$ coefficient $b+d=0$ gives $b=-d=\pm1$ which doesn’t work. So $b=d$ and $b^2=1$.
comparing $x^2$ coefficients $0=b+d+ac$ gives $a^2=\pm2$ and we can take $a=-c=\sqrt{2}$ and $b=d=1$.

We obtain the factorisation $x^4+1=(x^2-x\sqrt{2}+1)(x^2+x\sqrt{2}+1)$ in $\mathbb{R}[x]$.

Alternatively, the quicker way is to write as the difference of two squares: \[\begin{align*} x^4+1&=(x^2+1)^2-2x^2 \\ &=(x^2+1)^2-(\sqrt2 x)^2=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1) \end{align*}\] and these factors are irreducible as there are no real roots.

$(h)\;$ A general fact is that any polynomial $f(x)\in\mathbb{R}[x]$ factors into a product of linear factors and quadratic factors in $\mathbb{R}[x]$. To see this, notice that the roots are either real (which produces linear factors with real coefficients) or complex conjugate pairs (which produce quadratic factors with real coefficients). In other words, any irreducible polynomial in $\mathbb{R}[x]$ has degree $1$ or $2$.

Now notice $x^6+27$ is a sum of two cubes. You are probably aware of a general formula for these: \[A^3+B^3=(A+B)(A^2-AB+B^2).\] (This doesn’t appear from nowhere: the left hand side vanishes when $A=-B$ so $A+B$ is a factor.) In this case, we obtain \[x^6+27=(x^2)^3+3^3=(x^2+3)(x^4-3x^2+9).\] We know the quartic must be reducible (it’s not linear or quadratic!) Using the trick of expressing as a difference of squares: \[x^4-3x^2+9=(x^2+3)^2-9x^2=(x^2+3x+3)(x^2-3x+3)\] and so we have finally found \[x^6+27=(x^2+3)(x^2+3x+3)(x^2-3x+3).\] These three quadratics are irreducible in $\mathbb{R}[x]$ as they have no real roots.

$(i)\;$ Knowing the irreducible polynomials of low degrees from Exercise 1 is useful here. First, we obviously have a root $-1=1\in\mathbb{F}_2$ so \[x^7+1=(x+1)(x^6+x^5+x^4+x^3+x^2+1).\] This degree $6$ factor has no roots in $\mathbb{F}_2$ so no more linear factors. Does it have any irreducible quadratic factors? Recall there is only one of these to check for, namely $x^2+x+1$. But applying the division algorithm gives \[x^6+x^5+x^4+x^3+x^2+1=(x^4+x)(x^2+x+1)+1\] so there are no irreducible quadratic factors.

How about irreducible cubic factors? There are only two to check, $x^3+x^2+1$ and $x^3+x+1$. It turns out \[x^6+x^5+x^4+x^3+x^2+1=(x^3+x^2+1)(x^3+x+1)\] and so we have the answer: \[x^7+1=(x+1)(x^3+x^2+1)(x^3+x+1)\quad\text{in $\mathbb{F}_2[x]$.}\]

$(j)\;$ One quickly finds that $1=-2$ is a root so factorise \[x^3+2=(x+2)(x^2+x+1).\] We then see $1=-2$ is also a root of the quadratic factor and so factorises further into linear (hence irreducible) factors \[x^3+2=(x+2)^3\in\mathbb{F}_3[x].\] Alternatively, we can do it very quickly using the Child’s Binomial Theorem in characteristic $3$: \[x^3+2=x^3-1=(x-1)^3=(x+2)^3\in\mathbb{F}_3[x].\]

$(k)\;$ Check that $1=-2$ and $2=-1$ are roots and factorise: \[x^4+2=(x+1)(x+2)(x^2+1).\] The quadratic factor is irreducible as it has no roots in $\mathbb{F}_3$.

Alternatively, we could have used the difference of two squares \[\begin{align*} x^4+2=x^4-1=(x^2-1)(x^2+1)&=(x+1)(x-1)(x^2+1) \\ &=(x+1)(x+2)(x^2+1)\in\mathbb{F}_3[x]. \end{align*}\]

$(l)$ The polynomial $x^4+2$ is irreducible in $\mathbb{F}_{13}[x]$. To see this, first check there are no roots in $\mathbb{F}_{13}$, hence no linear factors. Then check $x^4+2=(x^2+ax+b)(x^2+cx+d)$ gives a contradiction as follows: we clearly have $c=-a$ and comparing the other coefficients gives \[a^2=b+d, \quad a(d-b)=0 \quad\text{and}\quad bd=2.\] Now

if $a=0$, then $b=-d$ and $b^2=-2$. But $-2$ is not a square in $\mathbb{F}_{13}$.
if $a\neq 0$, then $b=d$ and $b^2=2$. But $2$ is not a square in $\mathbb{F}_{13}$.

Remark: Checking which elements are square in $\mathbb{F}_{13}$ is quick: just write down the squares \[0^2,\quad 1^2\equiv 12^2,\quad 2^2\equiv 11^2,\quad...\quad,6^2\equiv 7^2\bmod 13\] and see what you get! One finds they are $\{0,\pm 1,\pm 3,\pm 4\}\subset\mathbb{F}_{13}$.

Exercise 1.3 $\;$Find all prime numbers $p$ for which $x+2$ is a factor of $x^4+x^3+x^2-x+1$ in $\mathbb{F}_p[x]$.

Solution

The division algorithm is particularly simple when dividing a polynomial $f(x)$ by a linear polynomial $x-a$. Indeed, it just says there are $q(x), r(x)\in K[x]$ such that \[f(x)=q(x)(x-a)+r(x) \quad\text{where}\quad \deg r(x)<\deg(x-a)=1\] In particular, the remainder is just a constant, and furthermore, setting $x=a$ shows $r(x)=f(a)$.

For this question, dividing $f(x)=x^4+x^3+x^2-x+1$ by $x+2$ gives remainder $f(-2)=15\in\mathbb{F}_p$. Hence $x+2$ is a factor when $15\equiv 0 \bmod p$ and so only for $p=3$ or $5$.

Exercise 1.4 $\;$How many irreducible monic quadratic polynomials are there in $\mathbb{F}_p[x]$?

Solution

There are $p^2$ monic quadratics altogether and the reducible ones are of the form $(x-a)(x-b)$. There are $p$ cases with $a=b$ and $\binom{p}{2}=\frac12p(p-1)$ with $a\neq b$.

That leaves $p^2-p-\frac12p(p-1)=\frac12p(p-1)$ irreducible quadratic polynomials in $\mathbb{F}_p[x]$.

Extra Question: Can you find the number of irreducible monic cubic polynomials in $\mathbb{F}_p[x]$? Think about the ways a reducible cubic can be written as a product of irreducible polynomials of lower degree.

Extra Extra Question: How about degree 4? Or degree 6? It’s getting complicated…but next term we’ll answer questions like this using the structure of extensions of finite fields.

Exercise 1.5 $(\star)\;$ Suppose we are given $n\geq 1$ distinct integers $a_1,...,a_n\in\mathbb{Z}$. Show that the polynomial \[f(x)=(x-a_1)(x-a_2)...(x-a_n)-1\in\mathbb{Q}[x]\] is irreducible in $\mathbb{Q}[x]$. (Hint: use Gauss’s Lemma and construct a polynomial with too many roots).

Solution

Suppose that $f(x)$ is reducible and so $f(x)=g(x)h(x)$ where $g(x)$ and $h(x)$ are non-constant polynomials. By Gauss’s Lemma, we can assume that $g(x)$ and $h(x)$ are monic polynomials with integer coefficients. In particular, this crucial observation tells us that $g(a_i)$ and $h(a_i)$ are integers for each $1\leq i\leq n$.

However, $f(a_i)=-1$ for each $i$ so we have $g(a_i)h(a_i)=-1$ and hence $g(a_i)=\pm 1$ and $h(a_i)=\pm 1$.

Suppose first that $g(x)$ and $h(x)$ have distinct degrees. Without loss of generality, we can assume $\deg g(x)<\deg h(x)$ and so $\deg g(x)<n/2$. Consider the polynomial \[p(x)=g(x)^2-1.\] It has degree strictly less than $n$ but has $n$ distinct roots $a_1,...,a_n$. This is impossible and this contradiction implies $f(x)$ must have been irreducible.

We are just left to deal with the case $\deg(g)=\deg(h)=n/2$. Consider the polynomial \[q(x)=g(x)^2-h(x)^2.\] As $g(x)$ and $h(x)$ are monic with the same degree, the leading terms in $g(x)^2$ and $h(x)^2$ cancel and we have $\deg q(x)<n$. But again, $q(x)$ has $n$ distinct roots, namely $a_1,...,a_n$, and so we arrive at the same contradiction and $f(x)$ must be irreducible.

Exercise 1.6 $\;$Find generators for each of the following ideals in the given polynomial rings:

$\;(a)\;$ $I=\left\{f(x)\in\mathbb{Q}[x] \;|\; f(i)=0 \right\}\subset \mathbb{Q}[x],$

$\;(b)\;$ $I=\left\{f(x)\in\mathbb{Q}[x] \;|\; f(i\sqrt{2})=0 \right\}\subset \mathbb{Q}[x],$

$\;(c)\;$ $I=\left\{f(x)\in\mathbb{Q}[x] \;|\; f(\sqrt{2})=f(\sqrt{5})=0 \right\}\subset \mathbb{Q}[x],$

$\;(d)\;$ $I=\left\{f(x)\in\mathbb{R}[x] \;|\; f(\sqrt{2})=0 \right\}\subset \mathbb{R}[x],$

$\;(e)\;$ $I=\left\{f(x)\in\mathbb{R}[x] \;|\; f(i\sqrt{2})=0 \right\}\subset \mathbb{R}[x],$

$\;(f)\;$ $I=\left\{f(x)\in\mathbb{R}[x] \;|\; f(i)=f(\sqrt{2})=f(1)=0 \right\}\subset \mathbb{R}[x].$

Solution

Recall that ideals in polynomial rings $K[x]$ (where $K$ is a field) are always principal, so we only need a single generator in each case.

$(a)$ Note that $f(i)=0\implies f(-i)=0$ since $f(x)$ has rational (hence real) coefficients. In particular, that means $f(x)\in I\iff$ $f(x)$ is divisible by $x^2+1=(x+i)(x-i)$ and \[I=\left(x^2+1\right).\]

$(b)$ This is similar: since $f(x)$ has rational (hence real) coefficients, if $i\sqrt{2}$ is a root of $f(x)$ then so is $-i\sqrt{2}$. Hence $f(x)\in I\iff f(x)$ is divisible by $x^2+2=(x+i\sqrt2)(x-i\sqrt2)$ and \[I=\left(x^2+2\right).\] Alternatively, use the fact that $i\sqrt{2}$ is a root of the irreducible polynomial $x^2+2\in\mathbb{Q}[x]$.

$(c)$ This is similar, we just use different “conjugates”. For any polynomial $f(x)\in\mathbb{Q}[x]$, by the Division Algorithm, \[f(x)=(x^2-2)q(x)+r_0+r_1x \quad\text{where $r_0,r_1\in\mathbb{Q}$}.\] If $f(\sqrt{2})=0$, then setting $x=\sqrt{2}$ gives $0=r_0+r_1\sqrt{2}$. But $\sqrt{2}$ is irrational so $r_0=r_1=0$, i.e. $f(x)$ is divisible by $x^2-2$ and $-\sqrt{2}$ is also a root. Similarly, if $f(\sqrt{5})=0$, then $f(x)$ is divisible by $x^2-5$.

We have $f(x)\in I\iff f(\pm\sqrt2)=f(\pm\sqrt5)=0$ and \[I=\left((x^2-2)(x^2-5)\right).\]

$(d)$ Now something else happens since $\sqrt2$ is already real. The linear polynomial $x-\sqrt2\in\mathbb{R}[x]$ and \[I=\left(x-\sqrt2\right).\]

$(e)$ Again, $f(i\sqrt2)=0\implies f(-i\sqrt2)=0$ when $f(x)$ has real coefficients. So \[I=\left(x^2+2\right).\]

$(f)$ Finally, any $f(x)\in I$ has roots $1,\sqrt2$ and $\pm i$ and so \[I=\left((x-1)(x-\sqrt2)(x^2+1)\right).\]

Exercise 1.7 $\;$Use Gauss’s Lemma to show that the following polynomials are irreducible in $\mathbb{Q}[x]$:

$\;(a)\;\;x^4+8,\hspace{4em}$ $\;(b)\;\;x^4-5x^2+2,\hspace{4em}$ $\;(c)\;\;x^4-22x^2+1.$

Solution

$(a)$ By the rational root test (or Gauss’s Lemma), any rational root must be an integer dividing $8$. Testing each of $\pm 1, \pm 2, \pm 4, \pm 8$ we don’t find any rational roots. (Perhaps more simply, we could observe that if $a\in\mathbb{R}\implies a^4\neq -8$ so in fact there are no real roots). Either way, $x^4+8$ has no linear factors in $\mathbb{Q}[x]$. If $x^4+8$ has quadratic factors in $\mathbb{Q}[x]$ then (via Gauss’s Lemma) it has quadratic factors in $\mathbb{Z}[x]$. Suppose $x^4+8=(x^2+ax+b)(x^2+cx+d)$ with $a,b,c,d\in\mathbb{Z}$. Then comparing coefficients gives \[a+c=0,\quad ac+b+d=0,\quad ad+bc=0\quad\text{and}\quad bd=8. \] or equivalently $\qquad c=-a, \quad -a^2+b+d=0, \quad a(d-b)=0\quad\text{and}\quad bd=8$.

Now

if $a=0$, then $b=-d$ and $b^2=-8$ which is impossible since $b\in\mathbb{Z}$.
if $a\neq 0$, then $b=d$ and $b^2=8$ which is also impossible.

$(b)$ Similarly, checking $\pm 1, \pm 2$ we find no rational roots and hence no linear factors.

By Gauss’s Lemma, if there is a non-trivial factorisation in $\mathbb{Q}[x]$, then there is one in $\mathbb{Z}[x]$. If $x^4-5x^2+2=(x^2+ax+b)(x^2+cx+d)$ with $a,b,c,d\in\mathbb{Z}$ then we obtain \[c=-a,\quad -a^2+b+d=-5 , \quad a(d-b)=0\quad\text{and}\quad bd=2. \] Notice $bd=2$ implies $\{b,d\}=\{1,2\}$ or $\{b,d\}=\{-1,-2\}$.

if $a=0$, then $b+d=-5$ and neither of the possible pairs for $\{b, d\}$ work.
if $a\neq 0$, then $b=d$ and again, no option works.

$(c)$ Finally, by the rational root test, $x^4-22x^2+1$ has no roots in $\mathbb{Q}$ (since they would have to be $\pm 1$ and these don’t work). So there are no linear factors in $\mathbb{Q}[x]$.

If it is the product of two quadratic factors, by Gauss’s Lemma we can assume they have integer coefficients. If \[x^4-22x^2+1=(x^2+ax+b)(x^2+cx+d)\qquad \textrm{with $a,b,c,d\in\mathbb{Z}$}\] then $c=-a$, $-a^2+b+d=-22$, $a(d-b)=0$ and $bd=1$.

Now $bd=1$ implies either $b=d=1$ or $b=d=-1$.
Then $a^2=22+b+d=20$ or $24$ which is impossible for $a\in\mathbb{Z}$.

Exercise 1.8 $\;$Which of the following subsets of $\mathbb{C}$ are fields with respect to the usual addition and multiplication:

$\;(a)\;\;\mathbb{Z},\hspace{4em}$ $\;(b)\;\;\{0,\pm 1\},\hspace{4em}$ $\;(c)\;\;\{0\},$

$\;(d)\;\;\left\{a+b\sqrt{2} \;|\; a, b\in\mathbb{Q}\right\},\hspace{3em}$ $\;(e)\;\;\left\{a+b\sqrt[3]{2} \;|\; a, b\in\mathbb{Q}\right\},$

$\;(f)\;\;\left\{a+b\sqrt[4]{2} \;|\; a, b\in\mathbb{Q}\right\},\hspace{3em}$ $\;(g)\;\;\left\{a+b\sqrt{2} \;|\; a, b\in\mathbb{Z}\right\},$

$\;(h)\;\;\{z\in\mathbb{C} \;|\; |z|\leq 1\}.$

Solution

$(a)$ The integers $\mathbb{Z}$ are not a field - not all elements are invertible, e.g. $2^{-1}\not\in\mathbb{Z}$.

$(b)$ The set $\{0,\pm 1\}$ is not closed under addition (though it is closed under multiplication!)

$(c)$ The set $\{0\}$ is not a field - a field must contain at least two distinct elements $0$ and $1$.

$(d)$ This is a field - it’s straightforward to check it’s closed under addition and multiplication. For inverses, we just rationalise the denominator. If $a+b\sqrt{2}\neq 0$, then $a^2-2b^2\neq 0$ since $\sqrt{2}$ is irrational and then \[\frac{1}{a+b\sqrt{2}}=\frac{a-b\sqrt{2}}{a^2-2b^2}.\]

$(e)$ This isn’t a field. It is closed under addition but is not closed under multiplication. For instance, $\sqrt[3]{2}\cdot\sqrt[3]{2}=\sqrt[3]{4}$ is not in the set so it’s not even a ring.

But how do we prove this? In Chapter 3 of the lectures, we’ll see that it follows at once from the fact that $\sqrt[3]{2}$ is the root of an irreducible polynomial $m(x)=x^3-2\in\mathbb{Q}[x]$ of degree $3$. This $m(x)$ will be called the minimal polynomial of $\sqrt[3]{2}$ and we’ll show that every polynomial that vanishes at $\sqrt[3]{2}$ is a multiple of $m(x)$. In particular, if \[\begin{align*} \sqrt[3]{4}=a+b\sqrt[3]{2}\quad\text{where $a,b\in\mathbb{Q}$} \tag{$\star$} \end{align*}\] then $f(x)=x^2-bx-a$ vanishes at $\sqrt[3]{2}$ and so must be a multiple of $g(x)$. But this is absurd - a quadratic polynomial can’t be a multiple of a cubic polynomial.

Just for fun, let’s show that $\sqrt[3]{4}$ is not in the set directly. Suppose $(\star)$ holds and and try to find a contradiction. It’s tempting to cube both sides and rearrange. But this introduces another $\sqrt[3]{4}$ term we have to deal with and leads to a mess. Instead, multiply $(\star)$ by $\sqrt[3]{2}$ to get \[\begin{align*} 2&=a\sqrt[3]{2}+b\sqrt[3]{4} \\ &=a\sqrt[3]{2}+b\left(a+b\sqrt[3]{2}\right) \\ &= ab+(a+b^2)\sqrt[3]{2}. \tag{$\star\star$} \end{align*}\] Now, if $a+b^2\neq 0$, then $\sqrt[3]{2}=(2-ab)/(a+b^2)\in\mathbb{Q}$. But $\sqrt[3]{2}$ is irrational! Hence we must have $a+b^2=0$ and $(\star\star)$ becomes $2=ab=-b^3$ which implies $\sqrt[3]{2}=-b\in\mathbb{Q}$, contradiction.

Note to make this silly proof complete, we would also need to prove that $\sqrt[3]{2}$ is irrational. This can be done by virtually the same proof that $\sqrt{2}$ is irrational that you’ve probably seen before, but just adds to the complications of not using minimal polynomials!

$(f)$ In exactly the same way, $\sqrt[4]{2}\cdot\sqrt[4]{2}=\sqrt{2}$ doesn’t belong to this set (it has minimal polynomial $x^4-2$). Hence the set is not closed under multiplication and is not a field.

$(g)$ This is not a field either. Similarly to $(a)$, not all elements are invertible, e.g. $2^{-1}$ isn’t in the set.

$(h)$ And finally, this set is not a field. It is closed under multiplication but not under addition since e.g. $|1|\leq 1$ but $|1+1|>1$.

Exercise 1.9 $\;$Show that every subfield of $\mathbb{C}$ contains the rational numbers $\mathbb{Q}$.

Solution

Let $K\subset\mathbb{C}$ be a subfield. Then $0\in K$ and $1\in K$. We can then build a copy of $\mathbb{Q}$ in $K$ in stages:

$\overbrace{1+1+...+1}^\text{m times}=m\in K$ so $\mathbb{N}_0\subset K$.
$K$ contains additive inverses of these so $\mathbb{Z}\subset K$.
$K^\times$ contains multiplicative inverses so $n^{-1}\in K$ for each $n\in\mathbb{Z}\setminus\{0\}$.
$K$ is multiplicatively closed so $m\cdot n^{-1}\in K$ for each $m\in\mathbb{Z}\subset K$ and $n\in\mathbb{Z}\setminus\{0\}$. In other words, $m/n\in\mathbb{Q}\subset K$.

Note: This gives a proof that $\mathbb{Q}$ has no proper subfields, a fact we used in lectures.

Exercise 1.10 $\;$Give an example of an infinite field of non-zero characteristic $p$.

Solution

Take the field of fractions $\mathbb{F}_p(x)$ of the polynomial ring $\mathbb{F}_p[x]$, i.e. \[\mathbb{F}_p(x)=\left\{\frac{f(x)}{g(x)} \;\Bigg|\; f(x), g(x)\in\mathbb{F}_p[x],\; g(x)\neq 0\right\}\] This clearly has infinitely many elements, e.g. $1,x,x^2,...$ and it is a field by construction!

Exercise 1.11 $\;(a)\;$ Let $K\subset L$ be a field extension and suppose $\theta\in L$ and $\theta^2\in K$ but $\theta\not\in K$. Show that \[K(\theta)=\left\{a+b\theta \;|\; a, b \in K\right\}\] is the minimal subfield of $L$ containing $K$ and $\theta$.

$\;(b)\;$ Suppose $r, s\neq 0$ are two non-zero elements in a field $K$. Show that $K(\sqrt{r})=K(\sqrt{s})$ if and only if $rs$ is a square in $L$, i.e. $rs=t^2$ for some $t\in K$.

Solution

$(a)\;$ A field is closed under addition and multiplication so any subfield of $L$ containing $K$ and $\theta$ must contain all polynomials in $\theta$ with coefficients in $K$. Using $\theta^2\in K$, $\theta\not\in K$, we can rewrite any such polynomial uniquely in the form $a+b\theta$ for some $a,b\in K$. Hence, if $K(\theta)$ is indeed a field, then it is the minimal subfield of $L$ containing $K$ and $\theta$.

To show $K(\theta)$ is a field, we need to show non-zero elements have inverses in $K(\theta)$. If $a+b\theta\neq 0$, then $a^2-b^2\theta^2\neq 0$ (as if not then we would have $\theta\in K$). Hence, we find \[\frac{1}{a+b\theta}=\frac{a-b\theta}{(a+b\theta)(a-b\theta)}=\frac{a}{a^2-b^2\theta^2}-\frac{b}{a^2-b^2\theta^2}\theta\in K(\theta).\]

Note: in the above, $\theta$ is a root of an irreducible quadratic $f(x)=x^2-c$ for some $c\in K$. We considered a similar thing in the Introduction in the lecture notes, where $\theta$ was a root of the polynomial $f(x)=x^2+px+q\in\mathbb{Q}[x]$. We will prove a version for higher degree polynomials later in lectures.

$(b)\;$ First, suppose $rs$ is a square, so that $rs=t^2$ for some $0\neq t\in K$. Then \[\sqrt{s}=\frac{t}{r}\sqrt{r}\in K(\sqrt{r})\] and hence $K(\sqrt{s})\subset K(\sqrt{r})$. Similarly, one shows $K(\sqrt{r})\subset K(\sqrt{s})$.

In the other direction, suppose $K(\sqrt{r})=K(\sqrt{s})$. If this common field is actually just $K$, then both $r$ and $s$ are squares in $K$ and their product is also a square in $K$.

If $K(\sqrt{r})=K(\sqrt{s})$ is strictly bigger than $K$, then $\sqrt{s}=a+b\sqrt{r}$ for some $a, b\in K$. Notice we must have $b\neq 0$ as otherwise, we would have $\sqrt{s}\in K$. Squaring gives \[s=(a+b\sqrt{r})^2=a^2+b^2+2ab\sqrt{r}\in K\quad\implies\quad a=0\] and we actually have $s=b^2r$. We find $rs=b^2r^2$ is a square in $K$, as required.

Exercise 1.12 $\;$Suppose $K$ is a field of characteristic $p>0$.

$\;(a)\;$ Write the polynomial $x^p-x\in K[x]$ as a product of irreducible factors.

$\;(b)\;$ Given an arbitrary element $a\in K$, show that the polynomial $x^p-x+a\in K[x]$ has either no roots in $K$ or $p$ distinct roots in $K$.

Solution

$(a)$ Since $K$ has characteristic $p$, it contains the prime subfield $\mathbb{F}_p$. Now by Fermat’s Little Theorem, the $p$ distinct elements $b\in\mathbb{F}_p$ satisfy $b^p=b$ and so are roots of the degree $p$ polynomial $x^p-x$. As a result, we have the factorisation into linear (hence irreducible) factors
\[x^p-x=x(x-1)(x-2)...(x-p+1)\in K[x].\]

$(b)$ Suppose that $f(x)=x^p-x+a$ does have a root $\theta\in K$. Then for any $b\in\mathbb{F}_p$, by the Child’s Binomial Theorem we have \[\begin{align*} f(\theta+b)=(\theta+b)^p-(\theta+b)+a &=\theta^p+b^p-\theta-b+a \\ &=f(\theta)+b^p-b=0. \end{align*}\] In other words, $f(x)$ has $p$ distinct roots $\theta, \theta+1, \theta+2,...,\theta+p-1$ which are all elements of $K$.

Remark. It is important to remember that whilst $\alpha^p=\alpha$ is true for $\alpha\in\mathbb{F}_p$, it is not true for any other elements in an arbitrary characteristic $p$ field $K$.

It is also true that $f(x)=x^p-x+a$ has $p$ distinct roots in $K$ when $a=0$ (as in part (a)). However, it is not the case that $a\neq 0$ means there are no roots in $K$. In fact this is hardly ever the case. Here are two examples:

Let $K=\mathbb{F}_p(t)$ and $a=t-t^p$ (which is a non-zero element of $K$). Then $f(t+b)=0$ for each $b\in\mathbb{F}_p\subset K$ by the Child’s Binomial Theorem so there are indeed $p$ distinct roots in $K$.
Let $K=\mathbb{F}_4=\mathbb{F}_2(\theta)=\{0,1,\theta,1+\theta\}$ where $\theta^2+\theta+1=0$. Then $f(x)$ has two roots in $K$ when $a=0$ or $1$ but has no roots in $K$ when $a=\theta$ or $1+\theta$. (Just substitute the $4$ elements of $\mathbb{F}_4$ into $f(x)$ and see…)

Exercise 1.13 $(\star)\;$ Let $p$ be a prime and $n\geq 2$. Are the following polynomials irreducible in $\mathbb{Q}[x]$? Give reasons. $\;\;(a)\;\;x^n+p^2x+p,\hspace{2cm}$ $\;\;(b)\;\;x^n+px+p^2.$

Solution

$(a)\;$ Yes - just apply Eisenstein’s Criterion with the prime $p$.

$(b)\;$ The answer is also yes, though we can’t apply Eisenstein’s Criterion directly. Instead, we adapt the proof of the Criterion from the notes (the one using reduction mod $p$).

First, we check that $f(x)=x^n+px+p^2$ has no linear factors. Suppose there is, and hence there is an integer root $a\in\mathbb{Z}$. Then $a^n+pa+p^2=0$ implies \[a^n=p(-a-p) \quad\implies\quad p|a^n\quad\implies\quad p|a.\] Also, $a<0$ and thus $a=-kp$ for some $k\in\mathbb{N}$. Substituting into the equation and rearranging gives \[k^np^{n-2}=(-1)^n(k-1)\] which is clearly impossible (the left-hand-side is always at least $k$). That means if $f(x)$ is reducible, it must be a product of two factors of degree at least $2$.

Suppose $f(x)=g(x)h(x)$ for some $h(x), g(x)\in\mathbb{Z}[x]$ with $r=\deg(g(x))\geq 2$ and $s=\deg(h(x))\geq 2$. Writing $\bar{f}(x)=f(x)\bmod p$, we have $\overline{f}(x)=x^n\implies \bar{g}(x)\bar{h}(x)=x^n$. But $\mathbb{F}_p[x]$ is a unique factorisation domain so we must have $\bar{g}(x)=x^r$ and $\bar{h}(x)=x^s$. In particular, \[g(x)=x^r+\cdots+b_1x+b_0\quad\text{and}\quad h(x)=x^s+\cdots+c_1x+c_0\] where $p$ divides each of $b_0, b_1, c_0, c_1$. Thus the coefficient of $x$ in $f(x)=g(x)h(x)$ is divisible by $p^2$ which is false as $f(x)=x^n+px+p^2$.

Exercise 1.14 $(\star)\;$ Consider the polynomial $f(x)=x^4+1$.

$(a)\;$ Show that $f(x)$ is irreducible in $\mathbb{Q}[x]$.

$(b)\;$ Show that $f(x)$ is reducible in $\mathbb{F}_p[x]$ for every prime $p$ by expressing it as the product of two quadratic polynomials. (You may use the fact that for every $p$, at least one of $\{-1,2,-2\}$ is a square in $\mathbb{F}_p$.)

Solution

$(a)\;$ Any rational root would have to be $\pm 1$, but neither are. Hence there are no linear factors.

If $f(x)$ has quadratic factors in $\mathbb{Q}[x]$ then it has quadratic factors in $\mathbb{Z}[x]$ by Gauss’s Lemma. Suppose \[x^4+1=(x^2+ax+b)(x^2+cx+d)\quad\text{with $a,b,c,d\in\mathbb{Z}$}.\]

Comparing constant coefficients gives $bd=1$ and so $b=d=\pm 1$.
The $x^3$ coefficient gives $c=-a$.
The $x^2$ coefficient $ac+b+d=0$ then implies $a^2=\pm 2$ which is impossible.

$(b)\;$ If $-1$ is a square in $\mathbb{F}_p$, say $a^2=-1$, then \[x^4+1=x^4-a^2=(x^2+a)(x^2-a).\] If $2$ is a square in $\mathbb{F}_p$, say $b^2=2$, then \[\begin{align*} x^4+1&=(x^2+1)^2-2x^2 \\ &=(x^2+1)^2-(bx)^2=(x^2+bx+1)(x^2-bx+1). \end{align*}\]

If $-2$ is a square in $\mathbb{F}_p$, say $c^2=-2$, then \[\begin{align*} x^4+1&=(x^2-1)^2+2x^2 \\ &=(x^2-1)^2-(cx)^2=(x^2+cx-1)(x^2-cx-1). \end{align*}\]

Remark: Those who did Elementary Number Theory II last year will know that if $-1$ and $2$ are not squares in $\mathbb{F}_p$, then $-2$ is a square - the product of two quadratic non-residues is a quadratic residue.

We can also prove this fact using Algebra II methods. Notice in $\mathbb{F}_2$, we have $-1=+1=1^2$ and so if $-1$ is not a square, then $p$ must be odd. When $p$ is odd, the multiplicative group $G=\mathbb{F}_p^\times$ has order $p-1$, which is even. Now, we have $a^2\equiv (p-a)^2\bmod p$ for $a,b\in\mathbb{Z}$ and so the set of squares is \[H=\left\{1^2, 2^2, ..., \left(\frac{p-1}{2}\right)^2\right\}\subset G.\] These are distinct, since for $1\leq a, b\leq (p-1)/2$ we have $a+b\not\equiv 0$ so \[a^2\equiv b^2\bmod p\quad\implies\quad (a-b)(a+b)\equiv 0\bmod p \quad\implies a-b\equiv 0\bmod p.\]

It’s easy to check the set of squares forms a (normal) subgroup $H\subset G$ with $|H|=|G|/2$. Hence the quotient $G/H$ is a group of order $2$ consisting of $2$ cosets: the subset of squares $H$ and the subset of non-squares $gH$ for some $g$. Finally $(gH)(gH)=H$ so the product of two non-squares is a square.

1 Problem Sheet 1: Fields and polynomials

1 Problem Sheet 1:
Fields and polynomials