Algebraic extensions: The very basics (part 1)

Let \(F\) be a field. The degree of a field extension \(E/F\), denoted \([E : F]\), is the dimension of \(E\) as a vector space over \(F\). For example \([\mathbb{C} : \mathbb{R}] = 2\), where a basis is \(\{1,i\}\). Note that it is not necessarily finite, and finding an explicit basis is not always possible.

There is a particular type of field extension whose aforementioned properties are known. Consider an irreducible polynomial \(p(x) \in F[x]\). Then we can construct a field \(F'\) containing a root \(\alpha\), and an isomorphic copy of \(F\). In particular, we let \(F'\) be the quotient \(F[x]/(p(x))\), whose elements can be represented by polynomials of degree \(< \text{deg}
\ p\).

Note that the condition that \(p(x)\) is irreducible is required for this quotient to be a field. Furthermore, it forces the \(p(x)\) to be of minimal degree. (If irreducible polynomials \(a(x),b(x)\) have the same root \(\alpha\), but \(a(x)\) is of minimal degree while \(b(x)\) is not, we obtain a contradiction from dividing \(b(x)\) by \(a(x)\) and considering the remainder.)

If we impose the further condition that \(p(x)\) be monic, then we obtain the minimal polynomial \(m_{\alpha,F}(x)\) which is unique. In particular, \([F' : F] = \deg m_{\alpha,F}\) (where \(F\) is identified with its isomorphic copy). Also, the element \(\bar{x} \in F'\) is the desired root of \(p(x)\), and \(F'\) has basis \(\{1,\bar{x},\cdots,\bar{x}^{n-1}\}\), where \(n = \deg m_{\alpha,F}\). Furthermore, the image of \(F\) under the projection is precisely the isomorphic copy of \(F\).

As an example, the field \(F = \mathbb{R}[x]/(x^2+1)\) contains a root \(\alpha\) to the polynomial \(x^2+1\). The degree of \(F\) over \(\mathbb{R}\) is 2, and every element of \(F\) can be represented in the form \(a+b\alpha\), where \(a,b \in \mathbb{R}\). \(F\) contains a copy of \(\mathbb{R}\), namely the set of elements with \(b=0\). The rules of complex addition and multiplication can then be derived via usual addition and multiplication of polynomials, mod \(\alpha^2+1\).

If we know a priori some field extension \(E/F\) containing a root \(\alpha\) to \(p(x)\), then we use \(F(\alpha)\) to denote the smallest subfield containing \(F\) and \(\alpha\). One key isomorphism which arises is \(F(\alpha) \cong F[x]/(p(x))\), so it does not matter whether we construct a field containing a root, or simply take an existing one. Taking the abovementioned example, we have \(\mathbb{C} \cong \mathbb{R}(i) \cong \mathbb{R}[x]/(x^2+1)\).

Now, suppose there are 2 distinct roots \(\alpha,\beta\) of \(p(x) \in F[x]\) in some field \(F'\). Then \(F(\alpha) \cong F[x]/(p(x)) \cong F(\beta)\), i.e. the roots are algebraically indistinguishable. Furthermore, there is a canonical isomorphism \(F(\alpha) \rightarrow F(\beta)\) mapping \(a_0+a_1\alpha+\cdots+a_n\alpha^n\) to \(a_0+a_1\beta+\cdots+a_n\beta^n\). For example, the map \(a+bi \rightarrow a-bi\) is an isomorphism.

This isomorphism can be generalized to the case where the fields \(F\) and \(F'\) where \(\alpha\) and \(\beta\) are adjoined to are isomorphic. Let \(\varphi : F \rightarrow F'\) be an isomorphism. Then \(\varphi\) extends naturally to an isomorphism \(F[x] \rightarrow F'[x]\) and thus we have \(F[x]/(p(x)) \cong F'[x]/(p(x))\). So we can conclude that \(F(\alpha) \cong F'(\beta)\). This fact is useful later on in proving the uniqueness of splitting fields up to isomorphism.