“Give me extension and motion and I will construct the universe.”
―René Descartes
One motivation for studying field extensions is that we want to build fields in which certain polynomials have roots. Here is a classical example going back to Gauss: while over \(\mathbb{R}\) the polynomial \(f=x^{2}+1 \in \mathbb{R}[x]\) has no roots, if we want a field in which \(f\) does have a root we need to consider \(\mathbb{C}=\mathbb{R}(i)=\{a+b i \mid a, b \in \mathbb{R}\}\text{.}\)
Starting from a smaller field \(F\) and an irreducible polynomial \(f \in F[x]\text{,}\) we want to build a larger field \(L\text{.}\) One way to do this is to take a root \(a\) of \(f\) and adjoin it to \(F\) obtaining the field \(L=F(a)\text{,}\) which is the collection of all expressions that one can build using addition, subtraction, multiplication and division starting from the of elements of \(F \cup\{a\}\text{.}\) Another way to build a larger field \(L\) from a smaller field \(F\) and an irreducible polynomial \(f \in F[x]\) is to let \(L=F[x] /(f(x))\text{.}\) We will show below that these two ways of creating larger fields are one and the same.
Definition16.15.Field Extension.
A field extension is an inclusion of one field \(F\) into a larger field \(L\text{,}\) making \(F\) into a subfield of \(L\text{.}\) We will write either \(F \subseteq L\) or \(L/F\) to signify that \(L\) is a field extension of \(F\text{.}\)
Remark16.16.
Recall that if \(F\) and \(E\) are fields, then every ring homomorphism \(\phi: F \to E\) necessarily injective. [provisional cross-reference: empty] (Proof: since \(\phi(1) = 1\text{,}\)\(\ker(\phi)\) is a proper ideal of \(F\text{,}\) and since \(F\) is a field, the only proper ideal of it is \(0\text{.}\)) Thus \(\phi\) maps \(F\) isomorphically onto its image \(\phi(F)\) and \(\phi(F) \subseteq E\) is a field extension. By abuse of notation we will typically think of \(\phi\) as being a field extension, even though it is technically just an injective homomorphism of fields.
Example16.17.Field Extensions.
\(\Q \subseteq \R\) and \(\R \subseteq \C\) are basic examples of field extensions.
Recall [provisional cross-reference: empty] CITEX that \(\Q[\sqrt{2}] = \{a + b \sqrt{2} \mid a, b \in \Q\}\) is a field. So \(\Q \subseteq \Q[\sqrt{2}]\) is another example of a field extension.
Remark16.18.
The latter is a typical sort of example for us: Starting with \(\Q\text{,}\) we would like to “adjoin” a root of the irreducible (in \(\Q[x]\)) polynomial \(x^2 + 1\text{.}\) Doing so yields \(\Q[\sqrt{2}]\text{.}\)
The previous example was a quadratic extension, which is misleadingly simple.
Example16.19.Yet More Field Extensions.
Consider \(x^3 - 2 \in \Q[x]\text{.}\) It is irreducible (e.g., by Eisenstein’s Criterion) and has roots \(\a = \sqrt[3]{2}\text{,}\)\(\beta = e^{2 \pi i/3} \sqrt[3]{2}\) and \(\g = e^{4 \pi i/3} \sqrt[3]{2}\) in \(\C\text{.}\) So there are three ways in which we could “adjoin a root" to \(\Q\text{:}\) First we could form the field
\begin{equation*}
\Q[\a] = \{a + b \a + c \a^2 \mid a,b,c \in \Q\}.
\end{equation*}
It is not completely obvious this is a field, but we’ll prove it later. The reason that we don’t need third or higher powers is that, e.g., \(\a^3 = 2 \in \Q\text{.}\) Or we could instead form the field \(\Q[\beta]\) or the field \(\Q[\gamma]\text{.}\) There are not all equal since, for example, \(\Q[\a] \subseteq \R\) but \(\beta\) is not in \(\R\text{.}\) However, they are {} field extensions as we will prove below.
Let us formalize some of these definitions:
Definition16.20.\(F(\a)\).
Given a field extension \(F \subseteq L\) and an element \(\a \in L\text{,}\) set
Then \(F[\a]\) is a subring of \(L\) and in fact it is the smallest subring of \(L\) that contains \(F\) and \(\a\text{.}\) We define \(F(\a)\) to be the smallest subfield of \(L\) that contains \(F\) and \(\a\text{.}\)
Exercise16.21.Intersection of Subfields.
Show that \(F(\a)\) exists by proving that the intersection of any two subfields is again a subfield.
In some cases we have \(F(\a)=F[\a]\text{.}\)
Exercise16.22.When \(F(\a)=F[\a]\).
Prove \(F(\a)=F[\a]\) whenever \(\a\) is the root of some polynomial with coefficients in \(F\text{.}\)
Proposition16.23.
If \(F \subseteq L\) is a field extension and \(\alpha \in L\text{,}\) the field \(F(\alpha)\) is the fraction field of \(F[\alpha]=\{f(\alpha) \mid f \in F[x]\}\) : more precisely,
Soon we will give an even better description for \(F(\alpha)\) in the case where \(\alpha\) is the root of a polynomial \(p \in F[x]\text{.}\)
Example16.24.\(\Q[\sqrt 2]\) is a Field.
Take \(F = \Q\) and \(\a = \sqrt{2}\text{.}\) Then any expression of the form \(\sum_i r_i \a^i\) with \(r_i \in \Q\) is equal to one of the form \(a + b \sqrt{2}\) for \(a, b \in \Q\text{.}\) If \(c + d \sqrt{2} \ne 0\text{,}\) then
\begin{equation*}
\frac{1}{c + d \sqrt{2}} = \frac{c-d \sqrt{2}}{c^2 - 2 d^2}=c' + d' \sqrt{2}
\end{equation*}
with \(c' = \frac{c}{c^2 - 2d^2}\) and \(d' = \frac{-d}{c^2 - 2d^2}\) both in \(\Q\text{.}\) This proves that \(\Q[\sqrt2] = \Q(\sqrt2) = \{a + b \sqrt2 \mid a,b \in \Q\}\) and in particular that \(\Q[\sqrt{2}]\) is a field.
Example16.25.\(\Q[\pi]\) is not a Field.
\(\Q[\pi]\) is not a field, and so in particular it is not equal to \(\Q(\pi)\) since, for example, \(\frac{1}{\pi} \notin \Q[\pi]\text{.}\) (If it were, then we would have \(\frac{1}{\pi} = \sum_{i=0}^n q_i \pi^i\) for some \(q_i \in \Q\text{,}\) and hence \(\sum_i q_i \pi^{i+1} - 1 = 0\text{,}\) which would imply \(\pi\) is the root of a polynomial with rational coefficients. This is known to be not true.)
SubsectionDegree of a Field Extension
“I may not have a degree, but I certainly got an education.”
―Jodi Picoult
Definition16.26.Degree of a Field Extension.
The degree of a field extension \(F \subseteq L\) is
We have that \([\C: \R] = 2\) and \([\R:\Q] = \infty\text{.}\) (We could in fact say \([\R:\Q]\) is the cardinality of \(\R\text{,}\) but in general we lump all infinite field extensions together when talking about degree.) We have \([\Q[\a]: \Q] = 3\) where \(\a = \sqrt[3]{2}\text{.}\)
Theorem16.28.Properties of Extension Degrees.
Assume \(F\) is a field and \(p(x) = \sum_i a_i x^i \in F[x]\) is an irreducible polynomial. Set \(L = F[x]/(p(x))\text{,}\) and for \(f \in F[x]\text{,}\) let \(\ov{f}\) denote the coset \(f(x) + (p(x)) \in L\text{.}\) The following hold:
\(F \into L\) is a field extension via the map given by \(a \mapsto \ov{a}\) for \(a \in F\text{.}\) (This is technically an injective homomorphisms of fields.)
\([L:F]=\deg(p)\text{.}\)
\(p(x)\) has a root in \(L\text{;}\) in fact, the element \(\ov{x} \in L\) is a root of this polynomial: \(\sum_i \ov{a_i} \cdot \ov{x}^i = 0 \in L\text{.}\)
Proof.
Because \(p(x)\) is irreducible and \(F[x]\) is a PID, \((p(x))\) is a maximal ideal [provisional cross-reference: empty] CITEX. Thus \(L = F[x]/(p(x))\) is a field [provisional cross-reference: empty] CITEX. The map \(F \to L\) given by \(a \mapsto \ov{a}\) is a ring map since it is the composition of the two ring maps \(F \xra{can} F[x] \xra{can} F[x]/(p(x))\)[provisional cross-reference: empty]. Since it is a ring map between two fields, it is injective [provisional cross-reference: empty]CITEX.
The equality \([L:F]=\deg(p)\) holds since \(1, \ov{x}, \dots, \ov{x^{\deg(p)-1}}\) is a basis for \(L\) regarded as an \(F\)-vector space, as we have seen before [provisional cross-reference: empty].
The last assertion is tricky only because the notation is confusing. Say \(p(x) = a_n x^n + \cdots + a_1 x + a_0\) and just to keep things straight let’s set \(q(y) = \ov{a_n} y^n + \cdots + \ov{a_0} \in L[y]\text{.}\) We need to show \(q(\ov{x}) = 0\text{:}\) We have
The last part of the proposition is notationally confusing to prove in general but clear in examples.
Example16.29.
Say \(F = \R\) and \(p(x)= x^2 +1\text{.}\) Then \(L = \R[x]/(x^2+1)\text{.}\) The assertion is that \(\ov{x} := x + (x^2 + 1) \in L\) is a root of the polynomial \(x^2 + 1\) viewed as having coefficients in \(L\text{.}\) In other words, this element has the property that its square is \(-1\text{.}\) Let’s check: Since \(\ov{x^2 + 1} = 0\) and \(\ov{x^2 + 1} = \ov{x}^2 + 1\) we have \(\ov{x}^2 = -1\text{.}\) Indeed, there is a field isomorphism \(L \cong \C\) sending \(\ov{x}\) to \(i\) and more generally \(\ov{f}\) to \(f(i)\text{.}\)
A field extension \(L / F\) is called simple if \(L=F(\alpha)\) for some element \(\alpha\) of \(L\text{.}\) We call such an \(\alpha\) a primitive element for the extension.
Example16.31.
\(\sqrt{2}\) is a primitive element of the extension \(\Q \subseteq\Q[\sqrt{2}] = \Q(\sqrt{2})\text{.}\) So is \(1 + \sqrt{2}\) and, more generally, \(a + b \sqrt{2}\) for any \(a, b \in \Q\) with \(b \ne 0\text{.}\)
Remark16.32.
If \(L / F\) is a simple field extension, note that there might be many different elements \(\alpha \in L\) such that \(L=F(\alpha)\text{.}\) Thus primitive elements are not necessarily unique.
We can generalize this to adjoining a subset instead of a single element.
Definition16.33.Generated Subfield.
If \(F \subseteq L\) is a field extension and \(A\) is any subset of \(L\text{,}\) the subfield generated by \(A\) over \(F\text{,}\) denoted \(F(A)\text{,}\) is the smallest subfield of \(L\) that contains all of \(F\text{.}\) If \(A=\left\{a_{1}, \ldots, a_{n}\right\}\) is a finite set, we write \(F\left(a_{1}, \ldots, a_{n}\right)\) for \(F(A)\text{.}\)
Example16.34.
Regard \(\Q\) as a subfield of \(\C\) and let \(F = \Q(\sqrt{2}, \sqrt{3})\text{.}\) We may also describe \(F\) as \(F = E(\sqrt{3})\) where we set \(E =\Q(\sqrt{2})\text{.}\)
I claim that \(F\) is in fact a simple extension of \(\Q\text{.}\) For example, say \(\g = \sqrt{2} + \sqrt{3}\text{.}\) I claim that \(\Q(\sqrt{2} + \sqrt{3}) = F\text{.}\) Note that \(\g^2 = 5 + 2 \sqrt{6}\) and
So \(\Q(\g) = \Q(\sqrt{2},\sqrt{3})\text{.}\) This example shows \(\Q(\sqrt{2}, \sqrt{3})/\Q\) is simple and \(\sqrt{2} + \sqrt{3}\) is a primitive element of this field extension.
Remark16.35.
This example is an illustration of the Primitive Element Theorem (which we might or might not have time to prove this semester): Every finite extension of \(\Q\) is generated by a single element (or, in other words, is simple).
SubsectionUniqueness of Simple Extensions
“Feeling unique is no indication of uniqueness.”
―Douglas Coupland
Next we will show that if \(\alpha\) is a root of a given polynomial \(p(x) \in F[x]\text{,}\) then \(F(\alpha)\) is determined by \(p(x)\) up to isomorphism.
Theorem16.36.
Let \(L / F\) be a field extension and let \(p(x) \in F[x]\) be an irreducible polynomial. If \(p\) has a root \(\alpha \in L\text{,}\) then there is an isomorphism \(\phi\) with \(\phi|_{F}=\mathrm{id}_{F}\) and
Let \(\tilde{\phi}: F[x] \rightarrow F(\alpha)\) be the evaluation homomorphism that sends \(x \mapsto \alpha\text{;}\) more precisely, \(\tilde{\phi}(f(x):=f(\alpha)\text{,}\) and the restriction of this map to \(F\) is the identity on \(F\text{.}\) Since \(p(\alpha)=0\text{,}\) we have \((p(x)) \subseteq \operatorname{ker}(\tilde{\phi})\text{,}\) and since \((p(x))\) is a maximal ideal and \(\operatorname{ker}(\tilde{\phi}) \neq F[x]\text{,}\) we conclude that \((p(x))=\operatorname{ker}(\tilde{\phi})\text{.}\)
Now by Theorem 1.43 [provisional cross-reference: cite] we get an injective ring homomorphism
such that \(\phi(f(x)+(p(x)))=\tilde{\phi}(f(x))=f(\alpha)\text{.}\)
It remains to be shown that \(\phi\) is surjective. We will actually show more, namely that \(\operatorname{im}(\phi)=F[\alpha]=F(\alpha)\text{.}\) Note first that by the definition of \(\phi\) above, the image of \(\tilde{\phi}\) on \(F[x]\) is \(F[\alpha]\text{.}\) However, since \(\phi\) is injective the image of \(\tilde{\phi}\) is a field contained in \(F(\alpha)\text{,}\) and since the smallest field containing \(F[\alpha]\) is \(F(\alpha)\text{,}\) we must in fact have \(\operatorname{im}(\tilde{\phi})=F(\alpha)\text{.}\)
Let’s formalize the extra information we have obtained in the course of proving the theorem. First we used the following useful fact:
Exercise16.37.
If \(\phi: F \rightarrow L\) is an injective ring homomorphism and \(F\) and \(L\) are fields then the image of \(\phi\) is a subfield of \(L\text{.}\)
Corollary16.38.
Let \(L / F\) be a field extension and let \(p(x) \in F[x]\) be irreducible having a root \(\alpha \in L\text{.}\) Then \(F[\alpha]=F(\alpha)\text{.}\)
Corollary16.39.Uniqueness of \(F(\alpha)\).
Let \(p(x) \in F[x]\) be irreducible and let \(\alpha\) and \(\beta\) be two roots of \(p(x)\) in some extensions \(L\) and \(K\) of \(F\text{.}\) Then \(F(\alpha) \cong F(\beta)\text{,}\) so that the two roots are algebraically indistinguishable.
Example16.40.Complex Conjugation.
Taking \(p(x)=x^{2}+1 \in \mathbb{R}[x]\) with roots \(\alpha=i\) and \(\beta=-i\) in \(\mathbb{C}\text{,}\) we actually obtain equal fields \(\mathbb{R}(i)=\mathbb{C}=\mathbb{R}(-i)\text{.}\) But Corollary 16.39 gives that there is an interesting isomorphism \(\phi: \mathbb{C} \stackrel{\cong}{\longrightarrow} \mathbb{C}\) that sends \(i\) to \(-i\text{.}\) In general, we have \(\phi(a+b i)=a-b i\) for \(a, b \in \mathbb{R}\text{.}\)
Example16.41.
Another example illustrating Corollary 16.39 is that \(\mathbb{Q}(\sqrt{2})\) and \(\mathbb{Q}(-\sqrt{2})\) are isomorphic fields. In fact, the are equal: \(\mathbb{Q}(\sqrt{2})=\mathbb{Q}(-\sqrt{2})\text{.}\) But again Corollary 5.25 CITEX gives that there is an interesting isomorphism \(\phi: \mathbb{Q}(\sqrt{2}) \stackrel{\cong}{\longrightarrow} \mathbb{Q}(-\sqrt{2})=\mathbb{Q}(\sqrt{2})\) that sends \(\sqrt{2}\) to \(-\sqrt{2}\text{.}\) In general, we have \(\phi(a+b \sqrt{2})=a-\sqrt{2}\) for \(a, b \in \mathbb{Q}\text{.}\)
Example16.42.
Let \(\a=\sqrt[3]{2}\) (the unique real cube-root of \(2\)) and \(\beta = e^{2\pi i/3} \sqrt[3]{2}\) (one of the two imaginary cube roots of \(2\)).
Then by Corollary 16.39 (applied with \(L = L' = \C\)) there is an isomorphism \(\Q(\a) \cong \Q(\beta)\) of fields that restricts to the identity map on \(\Q\text{.}\) Note that these two fields are not equal since the former is contained in \(\R\) and the latter is not.
The two examples above preview the central idea of Galois theory.
