1.3 Field extensions
Let \(K\) be a field containing a field \(F\). Then we call \(K\) a field extension of \(F\) (and \(F\) a subfield of \(K\)). This is denoted by \(K/F\). The degree of a field extension \(K/F\), denoted \([K:F]\) is the dimension of \(K\) as a vector space over \(F\). A field extension is said to be finite if \([K:F]\) is finite, otherwise we say its infinite.
Let \(F\) be a field and \(p(x)\) be an irreducible polynomial in \(F[x]\) of degree \(n\). Then \(K:=F[x]/(p(x))\) is a field extension of \(F\) and \([K:F]=n\).
I claim that the image of \(1,x,x^2,\dots ,x^{n-1}\) in \(K\) form a basis for \(K/F\). To prove this, first note that, if \(f(x) \in F[x]\) has degree less than \(n\) then its written in terms of this basis, so when we look at the image in \(K\) the same is true. So now assume that \(\deg (f(x))\geq n\) them we can do polynomial long division to write
with \(\deg (r(x)) {\lt} n\). So \(f(x) \equiv r(x) \mod (p(x))\). So again we see that in \(K\), \(f(x)\) can be written in terms of this basis, so this basis spans \(K\).
So we just need to prove that this basis is linearly independent. Let \(\bar{x}^i\) denote the image of \(x^i\) in \(K\). Then assume for contradiction, that \(1,\bar x, \dots ,\bar{x}^{n-1}\) is not linearly independent. Then we can find \(a_i \in F\) (not all zero) such that
this means
which means that \(p(x)\) divides \(a_0+a_1 x+\cdots ,+a_{n-1}x^{n-1} \) but this is impossible as \(p(x)\) has degree \(n\) and \(\deg ( a_0+a_1 x+\cdots ,+a_{n-1}x^{n-1} ){\lt}n\).
Let \(K/F\) be a field extension and let \(\alpha _1,\dots ,\alpha _n \in K\). Let \(F(\alpha _1,\dots ,\alpha _n)\) denote the smallest subfield of \(K\) containing \(F\) and \(\alpha _1,\dots ,\alpha _n\). We call it the field generated by \(\alpha _1,\dots ,\alpha _n\).
A field generated by a single element is called a simple extension. In other words, \(F(\alpha )\) is a simple extension of \(F\). We call \(\alpha \in K\) the primitive element for the extension.
Now, here is an important result:
Let \(F\) be a field and \(p(x)\) an irreducible polynomial in \(F[x]\). Moreover, let \(K/F\) be a field extension containing a root \(\alpha \) of \(p(x)\). Then there is an isomorphism
given by sending \(f(x)\) to \(f(\alpha )\). 1
Let me denote by \(\phi '\) the map \(F[x] \to F(\alpha )\) sending \(f(x)\) to \(f(\alpha )\). First note that \(p(x)\) is in the kernel of \(\phi '\), since \(p(\alpha )=0\) as \(\alpha \) is taken to be a root of \(p(x)\). Therefore, \(\phi \) also induces a new map \(F[x]/(p(x)) \to F(\alpha )\) which we call \(\phi \) (we say that \(\phi '\) "factors through" \(F[x]/(p(x))\)). Now we want to show \(\phi \) is an isomorphism. First note that \(\phi (x+(p(x)))=\alpha \) and if \(a \in F\) then \(\phi (a)=a\), so the image of \(\phi \) has \(F\) and \(\alpha \) in its image. Moreover, \(\phi \) is a field homomorphism, so the image is again a field. This means that \(F(\alpha )\) is in the image of \(\phi \) (since by Definition 1.3.3, \(F(\alpha )\) is defined to be the smallest such field). So we just need to check this map is injective.
Injectivity is easy, since if you recall, the kernel of any non-zero ring homomorphism to an integral domain is a prime ideal, and the only prime ideal in a field is the zero ideal. So since \(F[x]/(p(x))\) is a field and our map is not the zero map, it must have kernel being the other prime ideal, which is \((0)\) and thus is injective.
Ok, so why is this so important. Well lets consider the first example in Example 1.2.18. Here we took \(\mathbb {Q}[x]/(x^2-2)\), now the above result tells us that this field is isomorphic to \(\mathbb {Q}(\sqrt{2})\) which you may remember as the field whose elements look like \(a+b\sqrt{2}\) with \(a,b \in \mathbb {Q}\). But note that the theorem above doesn’t say anything about which root of \(x^2-2\) one should take. So we equally have
One slightly more philosophical observation, is that even to define \(F(\alpha )\) in Definition 1.3.3 we needed to assume the existence of a field \(K\) containing a root of my polynomial \(p(x)\), so by definition \(F(\alpha )\) depends on \(K\). So just for the moment let me highlight this dependence on \(K\) by writing \(F(\alpha )\) as \(F_{K}(\alpha )\).
Lets look at the example above. Here we have to assume that there is some field which contains \(\sqrt{2}\). But how do we choose such a \(K\)? for example we could have \(K\) being \(K_1:=\mathbb {Q}(\sqrt{2},\sqrt{3})\) or \(K_2:=\mathbb {Q}(2^{1/4})\) or \(\mathbb {C}\) or something else. In each case we get our own version \(\mathbb {Q}(\sqrt{2})\) which we are for the moment denoting as \(\mathbb {Q}_{K_1}(\sqrt{2}),\mathbb {Q}_{K_2}(\sqrt{2})\) and \(\mathbb {Q}_{\mathbb {C}}(\sqrt{2})\). Now, one can ask, are all these versions the same? well in each case they are isomorphic to \(\mathbb {Q}[x]/(x^2-2)\) but to construct the isomorphisms we had to make choices, particularly we had to pick a root. So all the versions are in fact isomorphic, but they aren’t "equal", since to be equal we would require the existence of a canonical isomorphisms between them, i.e. choice free isomorphisms.
What’s the point of all this? what I want to highlight is how defining \(\mathbb {Q}(\sqrt{2})\) requires some choices, but defining \(\mathbb {Q}[x]/(x^2-2)\) is choice free. In practice what we will do is just find some \(K\) in Definition 1.3.3 which works in all cases. Meaning, we fix a \(K\) which contains the roots of all polynomials in \(F[x]\). When \(F=\mathbb {Q}\), then we will just take \(K=\mathbb {C}\).