Field Automorphisms: A Nice Touch on "Ambiguity"

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I am sure that the title of the post is going to confuse many readers, but for lack of a better title please be content with it. Rest assured that the post itself is quite unambiguous. We would like to throw some light on the a special kind of ambiguity which we find in mathematical systems (here mainly in algebra).

Thus for example let us take a look at the simplest irrational number $ \sqrt{2}$. This number has a relation with the rationals in the form of the fact that it is the root of a polynomial $ x^{2} - 2$ over rationals and that's the only fundamental relation it has with the rationals. But if we now look at its cousin $ -\sqrt{2}$ we see that it shares the same relation with the rationals, namely that it is also a root of $ x^{2} - 2$.

The reader who is all too familiar with the real number system might point out that these two numbers $ \sqrt{2}$ and $ -\sqrt{2}$ can be distinguished by using the fact that one is positive and other is negative. But this statement itself can not be made only on the basis of rational number system. My point is that as far the rationals are concerned, the numbers $ \sqrt{2}$ and $ -\sqrt{2}$ are indistinguishable. Although these are two different numbers from point of view of algebra, but they bear the same relationship with the rationals in the context of algebraic operations.

This makes the symbol $ \sqrt{2}$ ambiguous. It can be used to refer to any one of the roots of $ x^{2} - 2$ and then the other could be referred as $ -\sqrt{2}$. It is this kind of ambiguity in Algebra which we wish to talk about in this post.

The above argument could be carried out for the imaginary unit $ i$ which is a root of $ x^{2} + 1$. In this case we can not distinguish between $ i$ and $ -i$ over the real numbers.

We now formalize this concept a bit in the language of Modern Algebra.

Field Automorphisms

Let us start with the example polynomial $ x^{2} - 2$ in $ \mathbb{Q}[x]$. And as explained in previous post we have a field extension $ \mathbb{Q} \subset K = \mathbb{Q}[x]/\langle x^{2} - 2 \rangle$ in which we have a member $ \alpha = [x]$ such that $ \alpha^{2} = 2$ in $ K$. This $ \alpha$ we traditionally denote by $ \sqrt{2}$, but we shall refrain from using the traditional symbolism because the symbol $ \sqrt{2}$ is generally related with many of its non-algebraic properties in the real number system.

Since the members of $ K$ are equivalence classes modulo $ x^{2} - 2$ they are of the form $ [a + bx]$ where $ a, b$ are rational. Thus every member $ k \in K$ is expressible as $ k = [a + bx] = [a] + [b][x] = a + b\alpha$. It is now clear that by changing the sign of $ b$ we can express every $ k \in K$ as $ a - b\alpha$.

Now consider the mapping $ \sigma : K \rightarrow K$ given by $ \sigma(a + b\alpha) = a - b\alpha$. This mapping is one-one and onto and moreover it preserves the field operations i.e $$ \sigma((a + b\alpha)(c + d\alpha)) = \sigma(a + b\alpha)\sigma(c + d\alpha)$$ and $$ \sigma((a + b\alpha) + (c + d\alpha)) = \sigma(a + b\alpha) + \sigma(c + d\alpha)$$ In such cases we say that $ \sigma$ is an isomorphism from $ K$ onto itself and since the isomorphism is from the given field onto itself it is more properly called an automorphism. Thus $ \sigma$ is an example of a field automorphism. The existence of a field automorphism (different from identity map) implies the existence of at least two elements in the field which are indistinguishable from each other as far as the field operations are concerned. Thus in the above example $ \alpha$ can be replaced by $ -\alpha$ without affecting the field structure of $ K$.

Since the identity map on any field is a trivial automorphism, we are almost always interested in non-trivial (i.e. different from identity map) automorphisms. However there are cases when a field might have only the trivial automorphism. For example the field $ \mathbb{Q}$ has identity map as the only automorphism. This is easy to see because an automorphism maps identity to identity and thereby leaves integers fixed and hence leaves the rationals fixed too.

The same holds true for the field of real numbers $ \mathbb{R}$. In this case we show that if $ \sigma$ is an automorphism then $ \sigma$ is a continuous function. Clearly if we have $ x < y$ in $ \mathbb{R}$ then we can write $ y - x > 0$ and so $ y - x = z^{2}$ for some $ z \in \mathbb{R}$. Then $$ \sigma(y) - \sigma(x) = \sigma(y - x) = \sigma(z^{2}) = (\sigma(z))^{2} > 0 \Rightarrow \sigma(x) < \sigma(y)$$ Thus the automorphism preserves order relations. Let $ \epsilon > 0$ be given and we can choose a rational $ \delta$ with $ 0 < \delta < \epsilon$. If $ -\delta < y - x < \delta$ then $$ \sigma(-\delta) < \sigma(y - x) < \sigma(\delta) \Rightarrow -\delta < \sigma(y) - \sigma(x) < \delta$$ and thus $ -\epsilon < \sigma(y) - \sigma(x) < \epsilon$. So the automorphism is a continuous function. Now let $ r$ be an arbitrary real number then there is a sequence of rationals $ \{q_{n}\}$ such that $ \lim_{n \to \infty}q_{n} = r$. Then we have $$ \sigma(r) = \sigma(\lim_{n \to \infty}q_{n}) = \lim_{n \to \infty}\sigma(q_{n}) = \lim_{n \to \infty}q_{n} = r$$ Thus we see that $ \sigma$ is the identity map.

Now let's take an example which is algebraic in nature. Let $ p(x) = x^{3} - 2$ and then $ p(x)$ is irreducible in $ \mathbb{Q}[x]$ and hence we have a field extension $ \mathbb{Q} \subset K = \mathbb{Q}[x]/\langle p(x) \rangle$. Let $ \alpha = [x] \in K$ then every element $ k \in K$ is expressible as $$ k = [a + bx + cx^{2}] = a + b\alpha + c\alpha^{2}$$ The reader may note that changing the sign of $ b, c$ we can also write $ k = a - b\alpha -b\alpha^{2}$, but the mapping $$ \sigma(a + b\alpha + c\alpha^{2}) = a - b\alpha - c\alpha^{2}$$ fails to be an automorphism because it does not preserve the multiplication in $ K$.

Now let $ \sigma$ be an automorphism on $ K$ and let $ \beta = \sigma(\alpha)$. Then we have $$ \beta^{3} = (\sigma(\alpha))^{3} = \sigma(\alpha^{3}) = \sigma(2) = \sigma(1 + 1) = \sigma(1) + \sigma(1) = 1 + 1 = 2$$ In the last equality we have used the simple fact that isomorphism maps identity element to identity element.

Now we shall prove that $ \beta = \alpha$. Since $ \beta \in K$ we can write $ \beta = a + b\alpha + c\alpha^{2}$ and then \begin{align} 2 &= \beta^{3} = \beta\cdot\beta^{2} = \beta(a + b\alpha + c\alpha^{2})^{2}\notag\\ &= \beta(a^{2} + b^{2}\alpha^{2} + c^{2}\alpha^{4} + 2ab\alpha + 2bc\alpha^{3} + 2ca\alpha^{2})\notag\\ &= (a + b\alpha + c\alpha^{2})(a^{2} + 4bc + (2c^{2} + 2ab)\alpha + (b^{2} + 2ca)\alpha^{2})\notag\\ &= (a^{3} + 4abc + 2b^{3} + 4abc + 4c^{3} + 4abc)\notag\\ &\,\,\,\,\,\,\,\, + (2c^{2}a + 2a^{2}b + a^{2}b + 4b^{2}c + 2b^{2}c + 4c^{2}a)\alpha\notag\\ &\,\,\,\,\,\,\,\, + (2bc^{2} + 2ab^{2} + ab^{2} + 2ca^{2} + ca^{2} + 4bc^{2})\alpha^{2}\notag\\ &= (a^{3} + 2b^{3} + 4c^{3} + 12abc) + (3a^{2}b + 6b^{2}c + 6c^{2}a)\alpha + (3ab^{2} + 6bc^{2} + 3ca^{2})\alpha^{2}\notag \end{align} Thus we get the following three equations \begin{align} a^{3} + 2b^{3} + 4c^{3} + 12abc &= 2\tag{1}\\ 3a^{2}b + 6b^{2}c + 6c^{2}a &= 0\tag{2}\\ 3ab^{2} + 6bc^{2} + 3ca^{2} &= 0\tag{3} \end{align} Clearly not all $ a, b, c$ can be zero because of the first equation. Also if $ b = 0$ then from the last two equations either $ a = 0$ or $ c = 0$ in which case from first equation either $ 4c^{3} = 2$ or $ a^{3} = 2$ and these options are clearly not possible since $ 2$ is not the cube of a rational number. Thus it follows that $ b \neq 0$.

Now using last two equations we get $$ (a^{2} + 2bc) = \dfrac{-2c^{2}a}{b},\,\,\,\,\,\,(ab + 2c^{2}) = \dfrac{-ca^{2}}{b}$$ and hence after some manipulation we get $ a^{3} - 4c^{3} = 0$ and this is possible only when $ a = c = 0$ because $ 4$ is not the cube of a rational number. It follows that we have $ a = 0, b = 1, c = 0$ and hence $ \beta = a + b\alpha + c\alpha^{2} = \alpha$. Thus we see that $ \sigma(\alpha) = \alpha$ and therefore (here $ a, b, c$ again represent arbitrary rational numbers) $$ \sigma(a + b\alpha + c\alpha^{2}) = a + b\alpha + c\alpha^{2}$$ It thus follows that $ \sigma$ is the identity map.

The above examples suggest that having non-trivial automorphisms is indeed an special property which all fields might not possess. Only in the very special cases do we have such automorphisms possible. To understand what lies behind the existence of non-trivial automorphisms we need to dig further. If we look closely at the field $ K_{1} = \mathbb{Q}[x]/\langle x^{2} - 2\rangle$ and $ K_{2} = \mathbb{Q}[x]/\langle x^{3} - 2\rangle$, we see that the field $ K_{1}$ is special in the sense that it contains all the roots (both $ [x]$ and $ [-x]$ are the roots)of $ x^{2} - 2$ whereas the field $ K_{2}$ contains only one root of $ x^{3} - 2$. This property of field extensions is called normality which we discuss next.

Normal Extensions

Let $ F$ be a field of characteristic $ 0$ and let $ F \subset L$ be a field extension. If every irreducible polynomial $ p(x) \in F[x]$ which has a root in $ L$ has all its roots in $ L$ then $ F \subset L$ is called a normal extension.

In this parlance, we say that $ \mathbb{Q}[x]/\langle x^{2} - 2 \rangle$ is a normal extension of $ \mathbb{Q}$, but $ \mathbb{Q}[x]/\langle x^{3} - 2 \rangle$ is not a normal extension of $ \mathbb{Q}$.

Since an isomorphism maps identity to identity, any automorphism of a field extension $ \mathbb{Q} \subset L$ leaves the integers and hence the rationals fixed. In case of a general field extension $ F \subset L$ where $ F$ is of characteristic $ 0$ it makes sense to study only those automorphisms which leave the base field $ F$ fixed.

Normal extensions guarantee the existence of non-trivial field automorphisms. To see this let's suppose we have an irreducible polynomial $ p(x) \in F[x]$ and let $ F \subset L$ be a field which contains all the roots of $ p(x)$. Then for any root $ \alpha$ of $ p(x)$ and any automorphism $ \sigma$ we have $ 0 = \sigma(0) = \sigma(p(\alpha)) = p(\sigma(\alpha))$ so that $ \sigma(\alpha)$ is also a root of $ p(x)$. Hence an automorphism maps one root of $ p(x)$ to another root of $ p(x)$. Since the distinct roots of $ p(x)$ all lie in $ L$, there will as many automorphisms as the number of roots of $ p(x)$.

The definitions which we have given above are not fully precise as they require lot of machinery of field extensions but our point was to discuss the important concepts of field extensions with suitable and simple examples without getting bogged down into too much formalism which is prevalent in the subject of Modern Algebra. In later posts we will discuss the theory of field extensions in general with bit more formalism and the concepts presented here will receive fuller attention.

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