2 Fields and polynomials

2.1 Basic properties of fields

For an overview of the properties of rings, see the Background from Algebra II section. In particular, unless stated otherwise, our rings are always commutative and contain a multiplicative identity $1$.

One of the central concepts in Algebra is a field.

Definition 2.1 A ring $R$ is a field if $0\neq 1$ and every element in $R\setminus \{0\}$ has a multiplicative inverse.

We define rings to be “objects that work a bit like $\mathbb{Z}$”. Similarly, fields are “objects that work a bit like $\mathbb{Q}$”. Familiar examples of fields are $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ and the finite fields $\mathbb{F}_p=\mathbb{Z}/p$ for each prime number $p$. We will usually denote arbitrary fields by capital letters $K$, $L$, $M$,…

Fields are especially nice rings. They have no non-trivial zero divisors and so are integral domains:

Lemma 2.1 Given $a, b$ in a field $K$, if $ab=0$ then either $a=0$ or $b=0$.

Proof. Suppose that $a\neq 0$ and $ab=0$ in $K$. Then there is a multiplicative inverse $a^{-1}\in K$ and \[b=1\cdot b=(a^{-1}a)b=a^{-1}(ab)=a^{-1}\cdot 0=0.\] Similarly, if $b\neq 0$ and $ab=0$ then $a=0$.

By a homomorphism between fields $K$ and $L$, we mean a ring homomorphism $\varphi:K\to L$. We remind the reader these are compatible with the additive and multiplicative structures, including the fact that they preserve identities, and so \[\varphi(a+b)=\varphi(a)+\varphi(b),\qquad \varphi(ab)=\varphi(a)\varphi(b)\quad\text{and}\quad\varphi(1)=1.\] These three conditions further imply that $\varphi(0)=0$, $\varphi(-a)=-\varphi(a)$ and $\varphi(a^{-1})=\varphi(a)^{-1}$ for $a\neq 0$.

Homomorphisms between fields are also especially nice:

Lemma 2.2 Let $\varphi:K\to L$ be a homomorphism between fields. Then

$\;(i)$ the image $\im(\varphi)$ is a field,

$\;(ii)$ the kernel $\ker(\varphi)=\{0\}$. As a result, every homomorphism between fields is injective.

Proof. $(i)$ In Algebra II, we saw images of ring homomorphisms are rings so all that’s left to check is that non-zero elements have multiplicative inverses in the image. Suppose $0\neq b\in\im(\varphi)$, say $\varphi(a)=b$. Then since $\varphi(0)=0$, we must have $a\neq 0$ and hence $a$ has an inverse $a^{-1}\in K$. But that’s exactly we need as $b^{-1}=\varphi(a)^{-1}=\varphi(a^{-1})\in\im(\varphi)$.

$(ii)$ Recall that the kernel $\ker(\varphi)=\left\{a\in K \;|\; \varphi(a)=0\right\}$ measures how far a homomorphism is from being injective. In particular, if $\ker(\varphi)=\{0\}$, then \[\varphi(a)=\varphi(b)\implies\varphi(a-b)=0\implies a-b\in\ker(\varphi)=\{0\}\implies a=b\] i.e. $\varphi$ is injective. It will thus be sufficient to prove that the kernel is $\ker(\varphi)=\{0\}$. Suppose on the contrary that $a\in\ker(\varphi)$ and $a\neq 0$. Then \[ 1=\varphi(1)=\varphi(a\cdot a^{-1})=\varphi(a)\cdot\varphi(a^{-1})=0\cdot\varphi(a^{-1})=0\] and so $0=1$ in $L$. But this contradicts the assumption that $L$ is a field.

A subfield $K$ of a field $L$ is a subring that is also a field. It is thus closed under the operations and has the same multiplicative identity $1$. Furthermore, the above lemma shows that the image of a homomorphism $\varphi:K\to L$ is a subfield of $L$ and since $\varphi$ is injective, we have an isomorphic copy $\im(\varphi)\cong K$ in $L$. This is sometimes called an embedding of $K$ in $L$ and $L$ is said to be a field extension of $K$, which we denote by either $K\subset L$ or $L/K$. The properties of field extensions lie at the heart of Galois Theory.

Lemma 2.3 Any intersection of subfields of a field $L$ is also a subfield of $L$.

Proof. Suppose we have subfields $K_i$ of $L$ for $i\in\mathcal{I}$ for some index set $\mathcal{I}$ and let $K=\underset{i}{\cap} K_i$

Then, for any $a, b\in K$ we have $a, b\in K_i$ and so $a\pm b, ab\in K_i$ for every $i$. As a result, $a\pm b$ and $ab$ belong to $K$. Furthermore, if $a\in K^\times$ then $a\in K_i^\times$ and $a^{-1}\in K_i^\times$ for every $i$. As a result, $a^{-1}$ belongs to $K$ as well. Finally, the multiplicative identity $1$ of $L$ is in each $K_i$ and hence $1\in K$.

The intersection of all subfields of a field $L$ is called the prime subfield of $L$. It is the minimal subfield of $L$, in the sense that any other subfield of $L$ contains it.

To determine prime subfields, we ask a silly-looking question: when is $1+1+\cdots+1$ equal to $0$? Given any field $K$, there is a ring homomorphism \[\begin{align*} \chi\;:\; \mathbb{Z} &\longrightarrow \quad K \\ m &\longmapsto \underbrace{1+1+\cdots+1}_\text{$m$ times} \end{align*}\] Its kernel is an ideal in the principal ideal domain $\mathbb{Z}$ so $\ker(\chi)=(n)$ for a unique integer $n\geq 0$. This $n$ is called the characteristic of $K$ and is denoted by $\ch(K)$. Explicitly,

Definition 2.2 The characteristic $\ch(K)$ of a field $K$ is the least integer $n\geq 1$ such that $\chi(n)=0$ if such an $n$ exists, and $\ch(K)=0$ otherwise.

Notice that $\chi(m)=\overbrace{1+1+\cdots+1}^\text{$m$ times}=0$ in $K$ if and only if $m$ is an integer multiple of $\ch(K)$.

Example 2.1 $(a)$ The fields $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ have characteristic $0$.

$(b)$ The finite field $\mathbb{F}_p=\mathbb{Z}/p$ (where $p$ is a prime) has characteristic $p$.

In these examples the characteristic of a field is either $0$ or a prime. This is always the case:

Lemma 2.4 The characteristic of a field is either $0$ or a prime number $p$.

Proof. Let $K$ be a field with $\ch(K)=n>0$. We need to prove that $n$ is a prime number. First of all, we have $1\neq 0$ in $K$ so $n\neq 1$. (Remember $1$ is not a prime!) Now suppose $n$ is composite, so $n=ab$ for some $a, b\in\mathbb{Z}$ with $1<a,b<n$. Then with $\chi:\mathbb{Z}\to K$ as above, we have \[\chi(a)\chi(b)=\chi(ab)=\chi(n)=0\] and since $K$ has no non-zero divisors of zero, we must have $\chi(a)=0$ or $\chi(b)=0$. But this contradicts the minimality of $n=\ch(K)$ and hence $n$ must be prime.

Theorem 2.1 Let $K$ be a field.

$\;(i)$ If $\ch(K)=0$ then the prime subfield of $K$ is $\mathbb{Q}$.

$\;(ii)$ If $\ch(K)=p>0$ then the prime subfield of $K$ is $\mathbb{F}_p$.

Proof. $(i)$ If $\ch(K)=0$ then by the definition of the characteristic, $\chi(n)\neq 0$ for all integers $n\neq 0$. Using this, we can extend $\chi:\mathbb{Z}\to K$ to give a ring homomorphism \[\varphi:\mathbb{Q}\to K\quad\text{where $\;\;\varphi(m/n)=\chi(m)/\chi(n)$.}\] (Check this is well-defined.) Now $\varphi$ is a homomorphism between fields so is injective and has image $\im(\varphi)\cong\mathbb{Q}$. But this image is a subfield of $K$, and since $\mathbb{Q}$ has no proper subfields, it is the prime subfield of $K$.

$(ii)$ If $\ch(K)=p>0$, the homomorphism $\chi:\mathbb{Z}\to K$ has kernel $\ker(\chi)=p\mathbb{Z}\subset\mathbb{Z}$. By the First Isomorphism Theorem for rings, we have $\im(\chi)\cong\mathbb{Z}/p=\mathbb{F}_p$. But this image is a subfield of $K$, and since $\mathbb{F}_p$ has no proper subfields, it is the prime subfield of $K$.

In fact, the ideas in this proof go a little further and we can also say the following:

Corollary 2.1 For any field $K$,

$(i)$ if $\mathbb{Q}$ is a subfield of $K$, then $\ch(K)=0$,

$(ii)$ if $\mathbb{F}_p$ is a subfield of $K$ for some prime $p$, then $\ch(K)=p$.

Remark. We can work with fields of characteristic $p>0$ in much the same way as we do with the more familiar fields of characteristic $0$. However, they do have some peculiar properties. Recall that when $p$ is prime, the binomial coefficients \[\binom{p}{i}=\frac{p!}{i!(p-i)!}\in\mathbb{Z}\quad\text{for $1\leq i\leq p-1$}\] are all multiples of $p$ (since the numerator is a multiple of $p$ but the denominator isn’t). As a consequence, their images are equal to $0$ in a field $K$ of characteristic $p$, and so we have the “Child’s Binomial Theorem” \[(a+b)^p=a^p+b^p\qquad\text{for any $a, b\in K$.}\] Along similar lines, in the polynomial ring $K[x]$, we have \[(x^p-1)=(x-1)^p\] so $1$ is the only $p\,$th root of unity in a field of characteristic $p$ (and it has multiplicity $p$).

Another important fact concerning the simplest characteristic $p$ field $\mathbb{F}_p$ is Fermat’s Little Theorem. This will be very familiar to those who did Elementary Number Theory II, but it was also hidden in the Algebra II problems sheets.

Theorem 2.2 Any element $a\in\mathbb{F}_p$ satisfies $a^p=a$.

Proof. We clearly have $0^p=0$ and assuming $a^p=a$ for some $a\in\mathbb{F}_p$, the Child’s Binomial Theorem above shows \[(a+1)^p=a^p+1=a+1 \quad\textrm{in $\mathbb{F}_p$.}\] Every element in $\mathbb{F}_p$ is of the form $1+1+\cdots+1$ and the result follows by induction.

Warning: the equivalent statement is false in other characteristic $p$ fields, since not every element is of the form $1+1+\cdots+1$.

2.2 Polynomials over fields

For a field $K$, we write $K[x]$ for the ring of polynomials over $K$ with indeterminate $x$. This is the set of polynomials \[ K[x]=\left\{ a_0+a_1x+\cdots+a_nx^n \;|\; a_i\in K\right\}\] with the usual addition and multiplication. Let’s recall some facts from Algebra II.

Every polynomial $f(x)=a_0+a_1x+\cdots+a_nx^n\in K[x]$ has a degree $\deg f(x)$:

If $a_n\neq 0$, then $\deg f(x)=n$.
In particular, degree zero polynomials are non-zero constant polynomials and these are the only invertible polynomials, i.e. the set of units is $K[x]^\times=K^\times=K\setminus\{0\}$.
If all $a_i$ are zero then $f(x)=0$ is the zero polynomial and we set $\deg(0)=-\infty$. This is the only sensible choice, in light of the following property…
For $f(x), g(x)\in K[x]$, we have $\deg(f(x)g(x))=\deg f(x)+\deg g(x)$.
For $f(x), g(x)\in K[x]$, we have $\deg(f(x)+g(x))\leq\max\left\{\deg f(x),\deg g(x)\right\}$. Note we only have an inequality since the top degree terms could cancel out in the sum.

The ring of polynomials $K[x]$ has many properties which are remarkably similar to those of the ring of integers $\mathbb{Z}$.

The ring $K[x]$ is an integral domain, i.e. $f(x)g(x)$ can only be the zero polynomial when one of $f(x)$ or $g(x)$ is the zero polynomial.
We have the division algorithm for polynomials. This states that for non-zero polynomials $f(x), g(x)\in K[x]$, there are unique $q(x), r(x)\in K[x]$ with $\deg r(x)<\deg g(x)$ such that \[ f(x)=q(x)g(x)+r(x). \]
Any $f(x), g(x)\in K[x]$ have a greatest common divisor $d(x)=\gcd(f(x),g(x))$, which is unique up to multiplication by non-zero constants. Moreover, via the Euclidean algorithm for polynomials, we can find $a(x), b(x)\in K[x]$ such that \[a(x)f(x)+b(x)g(x)=d(x).\]
The field of rational numbers $\mathbb{Q}$ is constructed by considering fractions $a/b$ of integers $a,b\in\mathbb{Z}$. In a similar way, given a polynomial ring $K[x]$ over a field $K$, we can consider rational expressions over $K$ which are ratios of two polynomials. The set of these is denoted by $K(x)$ (note the curly brackets instead of square ones) \[K(x)=\left\{\frac{f(x)}{g(x)} \;\;\Bigg|\;\; f(x),\; g(x)\in K[x],\; g(x)\neq 0 \right\}.\] Two such expressions $f_1(x)/g_1(x)$ and $f_2(x)/g_2(x)$ are regarded as being equal exactly when $f_1(x)g_2(x)=f_2(x)g_1(x)$ in $K[x]$. Rational expressions are added and multiplied in the way you’d expect and this makes $K(x)$ into a field. This is sometimes called the field of fractions of $K[x]$ and the same idea can be used to define the field of fractions $\Frac R$ of an arbitrary integral domain $R$.
Just like $\mathbb{Z}$, the ring $K[x]$ is a principal ideal domain, i.e. every ideal $I\subset K[x]$ is principal. In other words, there exists $f(x)\in K[x]$ such that \[I=(f(x))=\left\{ f(x)g(x) \;|\; g(x)\in K[x]\right\}.\]

Prime numbers can be thought of as the multiplicative building blocks of $\mathbb{Z}$. In polynomial rings, the multiplicative building blocks of polynomial rings are irreducible polynomials. These are so important to us that we’ll recall the definition from Algebra II:

Definition 2.3 A polynomial $f(x)\in K[x]$ is irreducible in $K[x]$ if

$\;(i)$ $\deg f(x)\geq 1$ and

$\;(ii)$ if $f(x)=g(x)h(x)$ for some $g(x), h(x)\in K[x]$ then either $g(x)$ or $h(x)$ is constant.

In other words, polynomials are irreducible if they can’t be factorised in a non-trivial way.

Remark. $(a)$ Any linear polynomial, i.e. $f(x)\in K[x]$ with $\deg f(x)=1$ is clearly irreducible.

$(b)$ Irreducibility depends on the field $K$. For instance, $x^2+1$ is irreducible when considered in $\mathbb{R}[x]$, but is not irreducible in $\mathbb{C}[x]$ since $x^2+1=(x+i)(x-i)$. So to be clear, we should explicitly say $f(x)$ is irreducible in $K[x]$ (or more succinctly, $f(x)$ is irreducible over $K$).

$(c)$ The ring $K[x]$ is a unique factorisation domain, i.e. it is an integral domain where any $f(x)\in K[x]$ factorises into a unique product of irreducible polynomials. Here, “unique” means up to permutation of factors and ignoring constant polynomials. As a result, we can “shuffle” factors and constant terms between factors, but that’s all we can do. For instance, \[(2x-2)(x+3),\quad (x+3)(2x-2),\quad (x-1)(2x+6),\quad (2x+6)(x-1)\] each decompose $2x^2+4x-6$ as a product of irreducible factors, but we can consider them all to be essentially the same factorisation.

$(d)$ Warning: the above definition of irreducible polynomials is only correct when $K$ is a field. Recall the general definition of irreducibility in an arbitrary integral domain $R$ from Algebra II: an element $r\in R$ is irreducible if it is not $0$ or a unit and $r=ab$ for $a,b\in R$ implies $a$ or $b$ is a unit. For instance, consider the integral domain $R=\mathbb{Z}[x]$ of polynomials with integer coefficients. Here, the set of units is $\mathbb{Z}[x]^\times=\mathbb{Z}^\times=\left\{\pm 1\right\}$ and so any prime $p\in\mathbb{Z}$ is irreducible in $\mathbb{Z}[x]$ by the definition of a prime number! In particular, constant polynomials can be irreducible in $\mathbb{Z}[x]$ and linear polynomials might not be. For example, whilst $2x+2$ is irreducible in $\mathbb{Q}[x]$, it is not irreducible in $\mathbb{Z}[x]$ since we have the non-trivial factorisation $2x+2=2(x+1)$ where neither $2$ or $x+1$ is a unit.

2.3 Tests for irreducibility

In general, deciding whether $f(x)\in K[x]$ is irreducible or not can be a very difficult question. There are efficient algorithms implemented in Computer Algebra Systems such as Mathematica/Wolfram Alpha which will factorise integer coefficient polynomials with degrees more than 1000. However, without using a computer there are a few special situations where we can do things by hand.

2.3.1 Check for linear factors

For $\deg f(x)\geq 2$, we can deduce $f(x)$ is reducible if it has a linear factor.

If $K=\mathbb{F}_p$ is a finite field, then we can just test if $f(\alpha)=0$ for each $\alpha\in K$ to find a linear factor $x-\alpha$.

If $K=\mathbb{Q}$ then this won’t work directly - we can’t test infinitely many elements! However, we can restrict to finitely many possibilities with the rational root test:

Proposition 2.1 If a degree $n$ polynomial \[f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in\mathbb{Z}[x]\] has a rational root $b/c\in\mathbb{Q}$ (expressed in lowest terms, i.e. with $\gcd(b,c)=1$), then $b|a_0$ and $c|a_n$.

(Note, if we are initially given a polynomial in $\mathbb{Q}[x]$, then we can always multiply out the denominators of the coefficients to obtain a polynomial in $\mathbb{Z}[x]$ which we then try to factorise…)

Example 2.2 For monic polynomials (i.e. with $a_n=1$), this just says a rational root must be an integer dividing $a_0$. For instance, in each of the following, the only candidate roots in $\mathbb{Q}$ are $\pm 1, \pm 2$.

$(a)$ $f(x)=x^2+3x+2$. The candidates $-1$, $-2$ are roots and $f(x)=(x+1)(x+2)$ is reducible.

$(b)$ $f(x)=x^3+3x+2$. No candidate $\pm 1,\pm 2$ is a root so $f(x)$ has no linear factors. Furthermore, a reducible cubic polynomial must have a linear factor and so $f(x)$ is irreducible in $\mathbb{Q}[x]$.

$(c)$ $f(x)=x^4+3x^2+2$. Again, no candidate $\pm 1,\pm 2$ is a root so $f(x)$ has no linear factors. However, we can not deduce from this that $f(x)$ is irreducible. It might be a product of two quadratic factors and indeed it is: $x^4+3x^2+2=(x^2+1)(x^2+2)$.

Remark. Given a monic polynomial $f(x)\in\mathbb{Z}[x]$, since rational roots must be integers, the roots of such an $f(x)$ are either integers or are irrational. In particular, using $f(x)=x^n-m\in\mathbb{Z}[x]$ we deduce that either $m$ is the $n$-th power of an integer or $\sqrt[n]{m}\not\in\mathbb{Q}$. This gives an easy way to see that numbers such as $\sqrt{3}$, $\sqrt[3]{2}$, $\sqrt[5]{27}$,… are irrational.

2.3.2 Using Gauss’s Lemma

As we just saw, non-existence of linear factors is not enough to determine irreducibility of polynomials $f(x)\in\mathbb{Z}[x]$ with $\deg f(x)\geq 4$. However, we do have a tool from Algebra II generalising Proposition 2.1 which we can use (provided the algebraic manipulation doesn’t get too nasty).

Lemma 2.5 (Gauss's Lemma) Suppose $f(x)\in\mathbb{Z}[x]$ is a product $f(x)=g(x)h(x)$ where $g(x), h(x)\in\mathbb{Q}[x]$. Then there exists $r\in\mathbb{Q}$ such that $rg(x)\in\mathbb{Z}[x]$ and $r^{-1}h(x)\in\mathbb{Z}[x]$.

In other words, if we can factorise $f(x)$ in $\mathbb{Q}[x]$, then we can actually factorise it in $\mathbb{Z}[x]$.

Example 2.3 $(a)$ Consider the polynomial $f(x)=x^4-3x^3+1$ in $\mathbb{Q}[x]$.

First check $\pm 1$ are not roots so $f(x)$ has no linear factors in $\mathbb{Q}[x]$. To check if $f(x)$ is a product of two quadratic factors, \[f(x)=x^4-3x^3+1=(ax^2+bx+c)(rx^2+sx+t),\] Gauss’s Lemma allows us to restrict to integer coefficients $a, b, c, r, s, t\in\mathbb{Z}$.

Now the $x^4$ and constant coefficients tell us $ar=ct=1\implies a=r=\pm 1$, $c=t=\pm 1$ and by possibly swapping $g(x)$, $h(x)$ with $-g(x)$, $-h(x)$ we can assume $a=r=1$.

Then the $x^3$ and $x$ coefficients give $b+s=-3$ and $c(b+s)=0$. But this is impossible since $c\neq 0$ so $f(x)$ is irreducible in $\mathbb{Q}[x]$.

$(b)$ Consider the polynomial $f(x)=x^4-3x^2+1$ in $\mathbb{Q}[x]$.

Again, $\pm 1$ are not roots so $f(x)$ has no linear factors. For a product of two quadratics as above, \[f(x)=x^4-3x^2+1=(ax^2+bx+c)(rx^2+sx+t),\] we are led to $a=r=1$ and $c=t=\pm 1$ again.

The other coefficients tell us $b+s=0$ and $bs+2c=-3$ $\implies$ $b^2=3\pm 2$ and this time there are solutions and $f(x)$ is reducible: \[x^4-3x^2+1=(x^2-x-1)(x^2+x-1).\]

Recall the reduction mod p map for a prime number $p$ is the ring homomorphism \[\begin{align*} \varphi\;:\; \mathbb{Z}[x] &\longrightarrow \;\,\mathbb{F}_p[x] \\ f(x) &\longmapsto \overline{f}(x)=f(x) \bmod p. \end{align*}\] which replaces all the coefficients in a polynomial with their residues modulo $p$. The following corollary of Gauss’s Lemma can sometimes give a quick alternative method for proving irreducibility in $\mathbb{Q}[x]$, provided we can spot a suitable prime number.

Proposition 2.2 Suppose $f(x)=a_nx^n+\cdots+a_1x+a_0\in\mathbb{Z}[x]$. If there exists a prime $p\!\not|\;a_n$ such that $\overline{f}(x)$ is irreducible in $\mathbb{F}_p[x]$, then $f(x)$ is irreducible in $\mathbb{Q}[x]$.

Example 2.4 Let’s use this to show that $f(x)=8x^3+14x-9$ is irreducible in $\mathbb{Q}[x]$.

Taking $p=7$, we have $\overline{f}(x)=x^3-2\in\mathbb{F}_7[x]$ and as this is degree $3$, we just check that $\overline{f}(x)$ has no linear factors in $\mathbb{F}_7[x]$. However, this amounts to checking none of $\alpha\in\{0,\pm 1, \pm 2, \pm 3\}$ satisfy $\alpha^3\equiv 2\bmod 7$ and we can conclude $f(x)\in\mathbb{Q}[x]$ is irreducible.

Be careful with the direction of the implication here - this result only lets us show a polynomial over $\mathbb{Q}$ is irreducible, not reducible. For instance, taking $p=3$ instead, we find $\overline{f}(x)=2x^3+2x$ is reducible in $\mathbb{F}_3[x]$ but this doesn’t imply $f(x)$ is reducible in $\mathbb{Q}[x]$.

Also, note the condition $p\!\not|\;a_n$ is important here. For instance, with $f(x)=3x^2+8x+5$ and $p=3$ we find $\overline{f}(x)=2x+2$ is irreducible in $\mathbb{F}_3[x]$ even though $f(x)=(x+1)(3x+5)$ is reducible in $\mathbb{Q}[x]$.

Remark. Gauss’s Lemma extends beyond polynomials with integer coefficients $\mathbb{Z}[x]$ inside $\mathbb{Q}[x]$. It can be stated (and proved) more generally for a unique factorisation domain $R$ inside its field of fractions $K=\Frac R$. Here’s an interesting case. Let $k$ be a field and $R=k[t]$ be the ring of polynomials over $k$ with indeterminate $t$. Then $K=k(t)$ is the field of rational expressions with coefficients in $k$ and we can think of $k[t]$ inside $k(t)$ in a very similar way to how we think of $\mathbb{Z}$ inside $\mathbb{Q}$.

Consider the polynomial ring $K[x]$, which is a unique factorisation domain since $K$ is a field. Its elements are polynomials with indeterminate $x$ where each of the coefficients are themselves rational expressions in the indeterminate $t$ \[K[x]=\left\{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 \;|\; a_i\in k(t)\right\}.\] The analogue of Gauss’s Lemma here concerns factorisations of elements of $R[x]=k[t][x]$ inside the bigger ring $K[x]=k(t)[x]$. It says that if a given polynomial \[f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 \qquad\text{with $a_i\in k[t]$} \] factorises as $f(x)=g(x)h(x)$ in $K[x]$, then we can assume that $g(x)$ and $h(x)$ have coefficients in $R=k[t]$, not just in $K=k(t)$.

As an example, let’s show that $f(x)=x^4-tx^2+1\in\mathbb{R}[t][x]$ is irreducible in $\mathbb{R}(t)[x]$. It has degree $4$ (in the indeterminate $x$) so we need to check it has no linear or quadratic factors.

Linear factors: If there is $\alpha\in \mathbb{R}[t]$ such that $x-\alpha$ is a linear factor, then Gauss’s Lemma implies $\alpha$ must divide the constant term $1$ in $f(x)$. That means $\alpha$ is a polynomial in $t$ dividing a constant so $\alpha$ must itself be constant, i.e. $\alpha\in\mathbb{R}$. But now $\alpha$ is a root of $f(x)$ so $\alpha^4-t\alpha^2+1=0$ in $\mathbb{R}[t]$. The coefficients of this polynomial in $t$ must be zero, i.e. $\alpha^4+1=0$ and $\alpha^2=0$. But this is impossible and hence $f(x)$ has no linear factors.

Quadratic factors: 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, $\deg(ac)=2\deg(a)=1$, but a polynomial having degree $1/2$ is nonsense!

Hence $f(x)$ has no linear or quadratic factors and so must be irreducible in $\mathbb{R}(t)[x]$.

2.3.3 Eisenstein’s Criterion

There is another test from Algebra II which makes use of a suitable prime number.

Proposition 2.3 (Eisenstein's Criterion) Suppose we have an integer polynomial \[f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in\mathbb{Z}[x]\qquad\text{with $a_n\neq 0$.}\] If there is a prime number $p\in\mathbb{Z}$ such that \[p\,|\,a_0,\; p\,|\,a_1,...,\;p\,|\,a_{n-1}\quad\text{but $\;p\!\not|\;a_n$ and $p^2\!\not|\;a_0$}\] then $f(x)$ is irreducible in $\mathbb{Q}[x]$.

Proof. This was proved in Algebra II by expanding products of polynomials and comparing coefficients. Instead, let’s give a slick proof here applying the reduction mod p map from above. Given $f(x)\in\mathbb{Z}[x]$ as in the Criterion, almost all coefficients disappear and $\overline{f}(x)=a_nx^n\bmod p$. Note this is not the zero polynomial since $p\!\not|\,a_n$.

Now suppose also that $f(x)$ is reducible. By Gauss’s Lemma, we know $f(x)=g(x)h(x)$ for some $g(x), h(x)\in\mathbb{Z}[x]$ with $r=\deg g(x)\geq 1$, $s=\deg h(x)\geq 1$ and $r+s=n$.

However, $\mathbb{F}_p[x]$ is a unique factorisation domain - the only way to factorise $\overline{f}(x)=\overline{g}(x)\overline{h}(x)$ is with $\overline{g}(x)=\alpha x^r$ and $\overline{h}(x)=\beta x^s$ for some non-zero $\alpha,\beta\in\mathbb{F}_p^\times$ with $\alpha\beta=a_n\!\!\mod p$.

In particular, the constant coefficients in $g(x)$ and $h(x)$ must both be divisible by $p$. But that implies the constant coefficient in $f(x)=g(x)h(x)$ is divisible by $p^2$ which contradicts the assumption $p^2\!\not|\;a_0$. Hence $f(x)$ is not reducible.

Example 2.5 $(a)$ Using $p=3$ in the Criterion shows $f(x)=x^4+3x+3\in\mathbb{Q}[x]$ is irreducible.

$(b)$ Using $p=2$ shows $f(x)=x^9+16x^3+48x^2+8x+18\in\mathbb{Q}[x]$ is irreducible.

$(c)$ We can’t immediately apply the Criterion to $f(x)=x^3-3x+1$. However, replacing $x$ with $x-1$, we get $f(x-1)=x^3-3x^2+3$ and applying Eisenstein’s criterion with $p=3$ shows $f(x-1)$ is irreducible in $\mathbb{Q}[x]$. But any factorisation of $f(x)$ gives a factorisation of $f(x-1)$ and hence $f(x)$ is irreducible as well.

$(d)$ A similar trick shows that the $p$-th cyclotomic polynomial \[f(x)=\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1\] for prime $p$ is irreducible in $\mathbb{Q}[x]$. Indeed, we have \[\begin{align*} f(x+1)&=\frac{(x+1)^p-1}{(x+1)-1} \\ &=\frac{1}{x}\left[x^p+\binom{p}{1}x^{p-1}+\binom{p}{2}x^{p-2}+\cdots+\binom{p}{p-1}x+1-1\right] \\ &=x^{p-1}+px^{p-2}+\binom{p}{2}x^{p-3}+\cdots+\binom{p}{p-2}x+p. \end{align*}\] However, the binomial coefficients $\binom{p}{i}=\frac{p!}{i!(p-i)!}$ for $1\leq i\leq p-1$ are all divisible by $p$. Furthermore, the constant term is $p$ so isn’t divisible by $p^2$. In particular, we find $f(x+1)$ is irreducible by applying Eisenstein’s Criterion with $p$, and hence $f(x)$ is irreducible as well.

This trick of making a transformation and then applying Eisenstein’s criterion is very nice. Unfortunately, finding a suitable transformation can require a bit of luck and ingenuity!

Remark. As with Gauss’s Lemma, it’s possible to extend Eisenstein’s Criterion to other situations by replacing $\mathbb{Z}$ with another integral domain $R$. For instance, let $k$ be a field and set $R=k[t]$ as before. Here’s the statement of the Criterion: suppose that \[f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\qquad\text{where $a_i\in k[t]$ and $a_n\neq 0$.}\] If there is an irreducible polynomial $p\in k[t]$ such that \[p\,|\,a_0,\; p\,|\,a_1,...,\;p\,|\,a_{n-1}\quad\text{but $p\!\not|\;a_n$ and $p^2\!\not|\;a_0$}\] then $f(x)$ is irreducible in $k(t)[x]$.

For example, we find the polynomial \[x^4-tx^2+(t+t^2)x+5t\in\mathbb{R}(t)[x]\] is irreducible by taking $p=t$ in the Criterion.