1 Introduction
1.1 Solving equations by radicals
The subject of Galois Theory deals with properties of the solutions of polynomial equations \[\begin{equation} x^n+a_{n-1}x^{n-1}+...+a_1x+a_0=0 \tag{1.1} \end{equation}\] where \(n\in\mathbb{N}\) is the degree and the \(a_i\) are some “known” coefficients. We’d like to perform the usual arithmetic operations on these coefficients, so it’s reasonable to assume they belong to some given (basic) field \(K\). For instance, we could consider \(K\) to be \(\mathbb{C}\), \(\mathbb{R}\), \(\mathbb{Q}\), \(\mathbb{F}_p=\mathbb{Z}/p\) for some prime number \(p\), or maybe something more exotic.
A first question one might ask is: can we solve the equation (1.1)?
Let’s first restrict to coefficients living in \(K=\mathbb{C}\). The Fundamental Theorem of Algebra tells us that complex solutions certainly exist so the question is really: find formulas for solutions that use (a finite number of) basic arithmetic operations (addition, subtraction, multiplication, division) on the coefficients, as well as taking radicals \(\sqrt[m]{\phantom{2}}\) for \(m\in\mathbb{N}\). This is called a solution by radicals.
The \(n=1\) case is really easy: there’s a single solution \(x=-a_0\).
We’re also very familiar with the \(n=2\) case \[x^2+a_1x+a_0=0.\] The cheap and easy trick with quadratics is always to complete the square, essentially “shifting” \(x\) to eliminate the linear term. We rewrite the equation to get the two (possibly equal) solutions as follows: \[\begin{align*} \left(x+a_1/2\right)^2-a_1^2/4+a_0=0 \implies x=-a_1/2\pm\sqrt{a_1^2/4-a_0}. \end{align*}\]
For \(n=3\) and \(n=4\), there are similar (but more complicated) formulas that involve square roots and cube roots. We’ll see more about these shortly.
For \(n\geq 5\), there is no similar formula for an equation with general coefficients! In fact, there’s not even a solution by radicals for the equation \(x^5-x-1=0\).
Galois’ primary motivation was to show that there is no solution by radicals for general polynomial equations in degree at least \(5\). This had actually been proved slightly earlier by Abel (of abelian group fame), but Galois’ methods are more elegant and go much further. In particular, using Galois Theory, one is able to
find out whether or not a given equation (1.1) can be solved in radicals,
if such solutions exist, it gives an algorithm for finding them explicitly (though applying this in practice can be extremely complicated).
We should note that, whilst there’s no general solution, there are particular equations in degree \(n\geq 5\) which can be solved by radicals. For instance, it’s easy for the equation \(x^n=2\). For a less obvious example, the equation \(x^5+x+1=0\) can be solved by radicals since \[x^5+x+1=(x^2+x+1)(x^3-x^2+1)\] and so we can handle the quadratic and cubic factors separately.
For an even less obvious example (lifted from Stewart’s book), the equation \(x^5+15x+12=0\) has a solution \[x=\sqrt[5]{\frac{-75+21\sqrt{10}}{125}}+\sqrt[5]{\frac{-75-21\sqrt{10}}{125}}+\sqrt[5]{\frac{225+72\sqrt{10}}{125}}+\sqrt[5]{\frac{225-72\sqrt{10}}{125}}\] with similar expressions for the other four roots!
Remark. We can consider solving equations by radicals in other fields than \(\mathbb{C}\) and the above formula for quadratics will work in any field where we are allowed to divide by \(2\). However, the proof fails e.g. in \(\mathbb{F}_2\) because \(2=0\) and we can’t divide by zero! In fact, the polynomial \[f(x)=x^2+x+1\neq (x+\alpha)^2+\beta\quad\text{for any $\alpha, \beta\in\mathbb{F}_2=\{0,1\}$}\] (just check each possible \(\alpha,\beta\)) so one can’t complete the square with it. Clearly \(f(x)\) has no roots in \(\mathbb{F}_2\) but later on, we’ll see how it does have roots living “somewhere above” \(\mathbb{F}_2\) in a similar way to how \(x^2+1\) has no roots in \(\mathbb{R}\) but does “somewhere above”, namely in \(\mathbb{C}\).
1.2 Cubic equations over \(\mathbb{C}\)
Attempts to find solutions of cubic equations had an important role in the history of mathematics, not least because it forced people to realise the usefulness of complex numbers. Let’s see if we can discover a solution by radicals of the general cubic equation \[\begin{equation} \tag{1.2} x^3+a_2x^2+a_1x+a_0=0. \end{equation}\] There’s a similar trick to the quadratic case called a Tschirnhaus transformation which eliminates the quadratic term. Introduce a new variable \(t=x+a_2/3\) so that by expanding \[\begin{align*} x^3+a_2x^2+a_1x+a_0 &= (t-a_2/3)^3+a_2(t-a_2/3)^2+a_1(t-a_2/3)+a_0 \\ &=(t^3-a_2t^2+...)+a_2(t^2-...)+a_1(t-a_2/3)+a_0 \end{align*}\] we obtain \[\begin{equation} \tag{1.3} t^3+pt+q=0. \end{equation}\] where \(p=(-a_2^2+3a_1)/3\) and \(q=(2a_2^3-9a_1a_2+27a_0)/27\). This equation (1.3) is called a reduced or depressed cubic equation.
Now there’s no obvious way to eliminate more terms and we need a substantially new idea to continue. The next clever idea arises from the identity: \((u+v)^3=3uv(u+v)+(u^3+v^3)\). In particular, one can see that \(u+v\) is always a solution of \[t^3-(3uv)t-(u^3+v^3)=0\] which is in the reduced cubic equation form with \[p=-3uv\quad\text{and}\quad q=-(u^3+v^3).\] So now we try to find \(u\) and \(v\) in terms of \(p\) and \(q\) that make these two equations work. Notice they tell us the sum and product of the two numbers \(u^3\) and \(v^3\). Consequently, they are the solutions to the quadratic equation: \[\begin{align*} (y-u^3)(y-v^3)&=y^2-(u^3+v^3)y+u^3v^3 \\ &=y^2+qy-\frac{p^3}{27}=0. \end{align*}\] and so \(u^3, v^3=-\dfrac{q}{2}\pm\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}\). We’ve now found a solution to \(t^3+pt+q=0\), namely \[t=u+v=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}.\] Once we have this, we can substitute to find a solution of (1.2) in terms of \(a_0\), \(a_1\) and \(a_2\).
Following this magnificent bit of algebra, we should think about the other solutions since a cubic equation ought to have three. Let \(\omega=e^{2\pi i/3}=(-1+\sqrt{-3})/2\) so that \(1\), \(\omega\) and \(\omega^2=(-1-\sqrt{-3})/2\) are the three cube roots of unity. In the above derivation, when we took the cube root of \(u^3\), there are actually three options \(u\), \(\omega u\) and \(\omega^2u\). It would seem there are similarly three options for the cube root of \(v^3\), leading to nine possible roots of our cubic equation. However, remembering that \(uv=-p/3\), once we’ve picked the cube root of \(u^3\), the cube root of \(v^3\) is fixed. Namely, the three roots to the reduced cubic are \[ u+v,\quad \omega u+\omega^2v,\quad \omega^2u+\omega v.\]
Example 1.1 \((a)\) Find the solutions of \(t^3-18t-35=0\). Here, \[u^3, v^3=35/2\pm\sqrt{35^2/4-18^3/27}=35/2\pm 19/2=8, 27\] and we find the roots are \(5\) and \((-5\pm\sqrt{-3})/2\). Note there is one real root and a complex conjugate pair since the polynomial has real coefficients.
\((b)\) Find the solutions of \(t^3-15t+22=0\). This time we get \(u^3, v^3=-11\pm 2i\) so the roots are \[\sqrt[3]{-11+2i}+\sqrt[3]{-11-2i},\] \[\omega\sqrt[3]{-11+2i}+\omega^2\sqrt[3]{-11-2i},\quad\omega^2\sqrt[3]{-11+2i}+\omega\sqrt[3]{-11-2i}.\] Now we might spot that \(t=2\) is a solution - that means one of these three numbers actually equals \(2\).
This makes sense once we (somehow) notice that \(-11\pm 2i=(1\mp 2i)^3\). Using this, we can rewrite the three roots much more simply as \(2, -1\pm 2\sqrt3\). In general, we should not expect to be so fortunate!
It’s also interesting that this polynomial has three real roots even though the intermediate calculation involved complex numbers. It turns out this is sometimes unavoidable! This is the casus irreducibilis.
Remark. The same formulas will give solutions in radicals in other situations such as over \(\mathbb{F}_p\) for \(p\geq 5\). However, they won’t work over \(\mathbb{F}_2\) or \(\mathbb{F}_3\). Notice the formulas involve division by \(2\) and \(3\) and so the above general solution by radicals of cubic equations doesn’t help over these fields.
1.3 Quartic equations over \(\mathbb{C}\)
Starting from the general form \[x^4+a_3x^3+a_2x^2+a_1x+a_0=0,\] the transformation \(t=x+a_3/4\) eliminates the cube term leaving us with the reduced quartic equation \[\begin{equation} \tag{1.4} t^4+pt^2+qt+r=0. \end{equation}\] To solve this by radicals, the trick is to manipulate the polynomial so that it is the sum or difference of two squares and then factorise over \(\mathbb{C}\) using either \(a^2+b^2=(a+ib)(a-ib)\) or \(a^2-b^2=(a+b)(a-b)\). This will express the quartic it as a product of two quadratic polynomials, as illustrated by the following two examples:
Example 1.2 \((a)\) Suppose we want to solve \(t^4+t^2+2t+1=0\).
Notice the quadratic part \(t^2+2t+1=(t+1)^2\) is already a perfect square so the equation can be rewritten \[\begin{align*} t^4+t^2+2t+1&=(t^2)^2+(t+1)^2 \\ &=(t^2+it+i)(t^2-it-i)=0. \end{align*}\] Finding the roots now amounts to solving two quadratic equations \(t^2\pm i(t+1)=0\).
\((b)\) Suppose we want to solve \(t^4+t^2-4t-3=0\).
This time the quadratic part \(t^2-4t-3\) is not a perfect square (since it has non-zero discriminant). We can get around this by “shifting” \((t^2)^2\) to \((t^2+1)^2\): \[\begin{align*} t^4+t^2-4t-3&=(t^2+1)^2-t^2-4t-4 \\ &=(t^2+1)^2-(t+2)^2 \\ &=(t^2+t+3)(t^2-t-1)=0 \end{align*}\] and again we are now just left with solving two quadratic equations.
The question remains: how did we know to shift \(t^2\) by \(1\) in the second example? For the general case, introduce a new term \(w\) and shift \(t^2\) by \(w\). The equation (1.4) becomes \[\begin{equation} (t^2+w)^2+\underbrace{(p-2w)t^2+qt+(r-w^2)}_\text{quadratic part}=0. \tag{1.5} \end{equation}\] The quadratic part of this is a perfect square if and only if it has zero discriminant, i.e. \[D=q^2-4(p-2w)(r-w^2)=0\] \[\iff w^3-\frac{1}{2}pw^2-rw+\frac{1}{8}(4pr-q^2)=0.\] This cubic equation in \(w\) (called a cubic resolvent) can now be solved by radicals as in the last section. Taking one of its solutions (any of them) and substituting into (1.5) makes the quadratic part a perfect square and this in turn converts the original reduced equation into a product of two quadratic polynomials.
Example 1.3 Solve the quartic equation \(t^4+2t^2-16t+17=0\).
Here, the cubic resolvent is \[w^3-w^2-17w-15=0.\] We could apply the general method to solve this, but it’s sometimes worth trying the rational root test: any rational root of this polynomial must be of the form \(a/b\) with \(a, b\in\mathbb{Z}\) having \(a\) dividing \(15\) and \(b\) dividing \(1\). Trying the possibilities, we find a root \(w=-1\) (the others being \(-3\) and \(5\)). We can now rewrite our equation as \[\begin{align*} t^4+2t^2-16t+17&=(t^2-1)^2+(2t-4)^2=0 \\ \iff & (t^2-1)\pm 2i(t-2)=0 \end{align*}\] Solving these two quadratics and using the fact that \((1\pm i)^2=\pm 2i\), we find the four solutions \[\sqrt{2}+(\sqrt{2}-1)i,\;\; -\sqrt{2}-(\sqrt{2}+1)i,\;\; \sqrt{2}-(\sqrt{2}-1)i,\;\; -\sqrt{2}+(\sqrt{2}+1)i.\] Notice they are two complex conjugate pairs, as we’d expect for a polynomial with real coefficients.
1.4 Galois Theory for quadratic equations
At its heart, Galois Theory studies the roots of a polynomial by considering “symmetries”. A familiar example is complex conjugation which gives a reflection symmetry of \(\mathbb{C}\). We flip the two roots of \(f(x)=x^2+1\) to give a symmetry \(a+ib\longleftrightarrow a-ib\) for \(a,b\in\mathbb{R}\) which works very nicely with addition and multiplication. A similar example is with the two roots of \(f(x)=x^2-2\) where there’s another useful type of “conjugation” given by \(a+b\sqrt{2}\longleftrightarrow a-b\sqrt{2}\) for \(a,b\in\mathbb{Q}\).
Let’s take a step back and examine the roots of a general monic quadratic polynomial \[f(x)=x^2+px+q\] from this point of view, forgetting that we’ve ever learnt how to solve quadratic equations.
For simplicity, we assume that \(p,q\in\mathbb{Q}\) and that \(f(x)\) is irreducible over \(\mathbb{Q}\), meaning that \(f(x)\) can’t be factored as a product of two linear polynomials with rational coefficients. In particular, \(f(x)\) has two (distinct) irrational roots \(\theta, \theta'\in\mathbb{C}\) and \[f(x)=x^2+px+q=(x-\theta)(x-\theta').\] Starting with the (unknown) root \(\theta\), define a subset \(\mathbb{Q}(\theta)\) of \(\mathbb{C}\) by \[\mathbb{Q}(\theta)=\left\{a+b\theta \;|\; a,b\in\mathbb{Q}\right\}.\] This is actually the minimal subring of \(\mathbb{C}\) containing all rational numbers \(\mathbb{Q}\) as well as the root \(\theta\). To see this, notice that such a minimal subring must be closed under addition and multiplication so consists of all polynomial expressions in \(\theta\) \[a_0+a_1\theta+\cdots+a_n\theta^n\quad\text{where $n\geq 0,\; a_0,...,a_n\in\mathbb{Q}$.}\] But we know that \(\theta^2=-p\theta-q\) so \(\theta^n=-p\theta^{n-1}-q\theta^{n-2}\) for each \(n\geq 2\). Repeatedly using this, we can simplify a polynomial in \(\theta\) to the form \(a+b\theta\) for suitable \(a, b\in\mathbb{Q}\).
Warning: It’s important that \(f(x)\) is quadratic for this, as otherwise we wouldn’t be able to write e.g. \(\theta^2\) in terms of lower powers.
We can say a lot more about \(\mathbb{Q}(\theta)\). First of all, its elements are uniquely represented as \(a+b\theta\) for some \(a,b\in\mathbb{Q}\). Indeed, suppose we write an element in two ways: \[a_1+b_1\theta=a_2+b_2\theta\quad\text{where $a_1,b_1,a_2,b_2\in\mathbb{Q}$.}\] If \(b_1\neq b_2\), then \(\theta=(a_1-a_2)/(b_2-b_1)\in\mathbb{Q}\) which contradicts the fact that \(\theta\not\in\mathbb{Q}\). Therefore \(b_1=b_2\) and substituting this back into the equality shows \(a_1=a_2\) as well. Thinking back to Linear Algebra I, you should now be able to see that as well as begin a ring, we can consider \(\mathbb{Q}(\theta)\) as a 2-dimensional vector space over \(\mathbb{Q}\) with basis \(\{1,\theta\}\).
Another important fact is that we can perform all ring operations on elements of \(\mathbb{Q}(\theta)\) purely using the fact that \(\theta\) is a root of \(f(x)\): \[\begin{align*} (a_1+b_1\theta)\pm(a_2+b_2\theta)&=(a_1\pm a_2)+(b_1\pm b_2)\theta \quad\text{and}\\[8pt] (a_1+b_1\theta)(a_2+b_2\theta) &=a_1a_2+(a_1b_2+a_2b_1)\theta+b_1b_2\theta^2 \\ &=(a_1a_2-qb_1b_2)+(a_1b_2+a_2b_1-pb_1b_2)\theta \end{align*}\] where we have used \(\theta^2=-p\theta-q\) in the last step.
Yet another fact is that \(\mathbb{Q}(\theta)\) is actually a field (and is called a field extension of \(\mathbb{Q}\)). For this, we need non-zero elements to have inverses in \(\mathbb{Q}(\theta)\), which we can show explicitly as follows. Notice that \[f(x)=x^2+px+q=(x-\theta)(x-\theta')\quad\implies\quad\begin{cases} \theta+\theta'\!\!\!&=-p \\ \;\;\;\theta\theta'\!&=\;q \end{cases}\] Then, given \(a+b\theta\neq 0\), we must have \(a+b\theta'\neq 0\) as well (since if \(b=0\) then \(a\neq 0\) and if \(b\neq 0\) then \(\theta'=-a/b\) contradicting \(\theta'\not\in\mathbb{Q}\)). Hence \[(a+b\theta)(a+b\theta')=a^2+ab(\theta+\theta')+b^2\theta\theta'=a^2-abp+b^2q\neq 0\] and so using \(\theta'=-p-\theta\) we get a formula \[\frac{1}{a+b\theta}=\frac{a+b\theta'}{a^2-abp+b^2q}=\frac{a-bp-b\theta}{a^2-abp+b^2q}\in\mathbb{Q}(\theta).\]
Now consider all of the above but with \(\theta\) swapped with \(\theta'\) - everything works in exactly the same way! We have a field \[\mathbb{Q}(\theta')=\left\{a+b\theta' \;|\; a, b\in\mathbb{Q}\right\}\] and all of the operations can be performed purely using that \(\theta'\) is a root of \(f(x)\). In fact, \(\mathbb{Q}(\theta')\) coincides with \(\mathbb{Q}(\theta)\) as subsets in \(\mathbb{C}\) - it’s easy to convert a linear combination of \(\{1,\theta'\}\) into a linear combination of \(\{1,\theta\}\) using the relation \(\theta'=-p-\theta\) and vice versa. This leads us to the promised symmetry. We have a map \[\begin{align*} \sigma\;:\; \mathbb{Q}(\theta) &\longrightarrow \mathbb{Q}(\theta) \\ \;\; a+b\theta &\longmapsto a+b\theta' \end{align*}\] which is compatible with the operations in \(\mathbb{Q}(\theta)=\mathbb{Q}(\theta')\) and is also a bijection. This is called a field automorphism, that is, an isomorphism of the field \(\mathbb{Q}(\theta)\) onto itself. Here are some facts about this map:
It swaps the two roots \(\theta\) and \(\theta'\). Indeed, we have \(\sigma(\theta)=\theta'\) from the definition and conversely, \[\sigma(\theta')=\sigma(-p-\theta)=-p-\theta'=\theta.\]
From the definition, rational numbers are \(\sigma\)-invariant, i.e. not changed by \(\sigma\). In fact, these are the only \(\sigma\)-invariant elements of \(\mathbb{Q}(\theta)\): \[\sigma(a+b\theta)=a+b\theta \implies (a-pb)+(-b)\theta=a+b\theta \implies b=0.\]
Applying \(\sigma\) twice is actually the identity map \(\sigma^2=\id\) on \(\mathbb{Q}(\theta)\): \[\sigma^2(a+b\theta)=\sigma(a+b\theta')=\sigma(a)+\sigma(b)\sigma(\theta')=a+b\theta.\] That means we have an order \(2\) cyclic group \(G=\{\id,\sigma\}\) of automorphisms acting on \(\mathbb{Q}(\theta)\) which fix \(\mathbb{Q}\). This is the Galois group of the field extension \(\mathbb{Q}(\theta)\) of \(\mathbb{Q}\).
Finally, let’s “discover” the formula for the roots using this symmetry. We know the sum of the two roots is given by \(\theta+\theta'=-p\). Let \(\delta=\theta-\theta'\) be the difference, so that \[\theta=\frac{-p+\delta}{2}\quad\text{and}\quad \theta'=\frac{-p-\delta}{2}.\] This is already starting to look like the quadratic formula! Before finding \(\delta\), let’s see what \(\sigma\) does to it: \[\sigma(\delta)=\sigma(\theta)-\sigma(\theta')=\theta'-\theta=-\delta\] This implies that \(\sigma(\delta^2)=\sigma(\delta)^2=(-\delta)^2=\delta^2\) so \(\delta^2\) is \(\sigma\)-invariant. But we said earlier that the only elements invariant under \(\sigma\) are the rational numbers and hence we’ve found \(\delta^2\in\mathbb{Q}\) and \(\delta\) is the square root of a rational number. In fact, we can easily find it explicitly using \[\delta^2=(\theta-\theta')^2=(\theta+\theta')^2-4\theta\theta'=p^2-4q.\] (An interesting thing here is that we’ve written \(\delta^2\) in terms of “symmetric polynomials” \(\theta+\theta'\) and \(\theta\theta'\).)
This \(\delta^2\) is the familiar discriminant of the quadratic \(f(x)=x^2+px+q\) and we’ve found the roots are \[\theta, \theta'=\dfrac{-p\pm\sqrt{p^2-4q}}{2}.\]
1.5 Final remarks
Our objective in this course is to attempt a generalisation of the above approach to the case of any irreducible polynomial \(f(x)\) of degree \(n\in\mathbb{N}\) with coefficients in an arbitrary field \(K\). This will lead us to study the following problems:
How do we construct a (minimal) field extension \(K(\theta)\) of \(K\) containing a root \(\theta\) of \(f(x)\), and how does this extension depend on the choice of \(\theta\)?
When do other roots of \(f(x)\) belong to this field extension \(K(\theta)\)?
What are the symmetries, i.e. field automorphisms of \(K(\theta)\) which fix \(K\)?
How can we use these symmetries to find the roots of \(f(x)\)?
Along the way, we’ll develop the necessary algebra to understand the elegant interplay between field extensions and their Galois groups, leading up to proving the insolubilty by radicals of general polynomials of degree \(n\geq 5\).