This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Let us examine the overarching theme of Galois theory that we will study, through a very simple example. For instance, consider the degree \(4\) extension \(\mathbb{Q}(\sqrt{2}, \sqrt{3})\) of \(\mathbb{Q}\). The newly added elements \(\sqrt{2}\) and \(\sqrt{3}\) arise from minimal polynomials with rational coefficients
\[\x^2-2,\qquad \x^2-3\]However, looking at each of these two polynomials, they each have two roots \(\pm \sqrt{2}\), \(\pm\sqrt{3}\), and there is no algebraic way to distinguish them in \(\mathbb{Q}\). Therefore, considering the action of swapping these roots (or the \(\mathbb{Q}\)-automorphisms of \(\mathbb{Q}(\sqrt{2},\sqrt{3})\)), that is, considering the permutation group \(S_2\times S_2\), this is a subgroup of \(S_4\).
In this way, whenever a polynomial is given, we can define an appropriate Galois group, and the philosophy of Galois theory is that looking at these groups allows us to classify extensions of \(\mathbb{Q}\).
However, thinking on the basis of this philosophy, if for instance the minimal polynomial has a multiple root, defining a permutation action becomes quite awkward. This is an unfounded concern in \(\mathbb{Q}\), but in some cases such a thing can actually happen.
Remark All fields appearing in this post have characteristic exponent \(p\).
\(p\)-Radical Extensions
Definition 1 For a field extension \(\mathbb{L}/\mathbb{K}\), an element \(x\in \mathbb{L}\) is said to be \(p\)-radical if there exists some \(m\geq 0\) such that \(x^{p^m}\in \mathbb{K}\). The smallest such \(m\) is called the height of \(x\).
If \(p=1\), the above definition has little meaning, and the same is true for the rest of the content in this post. In other words, essentially all content in this post can be regarded as being about fields of characteristic \(p\).
Proposition 2 Fix a field extension \(\mathbb{L}/\mathbb{K}\) and a \(p\)-radical element \(x\in \mathbb{K}\) of height \(e\). Then for \(a=x^{p^e}\in \mathbb{K}\), the minimal polynomial of \(x\) is given by
\[\x^{p^e}-a\in \mathbb{K}[\x]\]Therefore \([\mathbb{K}(x):\mathbb{K}]=p^e\).
We write the image of the Frobenius endomorphism \(\Frob_p:\mathbb{K}\rightarrow \mathbb{K}\) as \(\mathbb{K}^p\). By the minimality of \(e\), we have \(a\not\in \mathbb{K}^p\), so the claim follows from the next lemma.
Lemma 3 Assume that an element \(a\) of a field \(\mathbb{K}\) satisfies \(a\not\in \mathbb{K}^p\). Then for any \(e\geq 0\), the polynomial \(f(\x)=\x^{p^e}-a\) is an irreducible polynomial in \(\mathbb{K}[\x]\).
Proof
The following definition would have been natural even right after Definition 1.
Definition 4 A field extension \(\mathbb{L}/\mathbb{K}\) is said to be \(p\)-radical if every element of \(\mathbb{L}\) is \(p\)-radical. If there exists an integer \(e\) such that \(x^{p^e}\in \mathbb{K}\) for all elements \(x\) of \(\mathbb{L}\), then the smallest such \(e\) satisfying this property is called the height of \(\mathbb{L}\).
In other words, the height of \(\mathbb{L}/\mathbb{K}\) can be thought of as (if defined) the maximum of the heights of the elements of \(\mathbb{L}\). Also, by Proposition 2, any \(p\)-radical extension is naturally an algebraic extension.
If the Frobenius endomorphism \(\Frob_p:A\rightarrow A\) is a bijection, we called \(A\) a perfect ring. Therefore, if \(\mathbb{K}\) were a perfect field, then \(\mathbb{K}^p=\mathbb{K}\), so any \(p\)-radical extension of a perfect field must be itself. Moreover, it is obvious from the definition that the compositum of \(p\)-radical extensions is \(p\)-radical. The following proposition concerns the existence of a (relative) \(p\)-radical closure.
Proposition 5 Fix a field extension \(\mathbb{L}/\mathbb{K}\) and for each \(n\geq 0\) define
\[\mathbb{L}_n=\{x\in \mathbb{L}\mid\text{$x$ is $p$-radical of height $\leq n$}\}\]Then the union \(\mathbb{L}_\infty\) of the increasing sequence \(\mathbb{L}_n\) is the largest \(p\)-radical subextension of \(\mathbb{L}\) containing \(\mathbb{K}\).
The proof of this is essentially obvious.
In the previous post, we saw that any field \(\mathbb{K}\) has an algebraic closure \(\overline{\mathbb{K}}\). Therefore, in Proposition 5 we may set \(\mathbb{L}=\overline{\mathbb{K}}\). Then \(\overline{\mathbb{K}}\) is a perfect field, and moreover, we know that for each \(n\), \(\overline{\mathbb{K}}_n\) is exactly \(\mathbb{K}^{p^{-n}}\). Let us write the (relative) \(p\)-radical closure in this situation as \(\mathbb{K}^{p^{-\infty}}\). This is exactly the same as §Fields, ⁋Theorem 15. If \(\mathbb{K}\) is imperfect, that is, if \(\mathbb{K}\neq \mathbb{K}^p\), then the above ascending sequence is strictly increasing, and therefore \(\mathbb{K}^{p^{-\infty}}/\mathbb{K}\) becomes an extension of infinite degree.
Meanwhile, the following holds.
Proposition 6 Suppose that a field extension \(\mathbb{L}/\mathbb{K}\) is a \(p\)-radical extension and a homomorphism \(u\) from \(\mathbb{K}\) to some perfect field \(\mathbb{F}\) is given. Then there exists a unique homomorphism \(v:\mathbb{L} \rightarrow \mathbb{F}\) extending \(u\).
Therefore, the following holds.
Corollary 7 A field extension \(\mathbb{L}/\mathbb{K}\) is the perfect closure of \(\mathbb{K}\) if and only if \(\mathbb{L}\) is a \(p\)-radical extension of \(\mathbb{K}\) and \(\mathbb{L}\) is a perfect field.
Proposition 8
We conclude this post by introducing the counterexample mentioned in the introduction.
Example 9 Consider the field \(\mathbb{K}=\mathbb{F}_p(t)\). Then consider the polynomial \(u(\x)=\x^p-t\in \mathbb{K}[\x]\), and through this we can consider the \(p\)-radical extension \(\mathbb{L}=\mathbb{K}[\x]/(\x^p-t)\). Then the minimal polynomial of a root \(\alpha\) of \(u(\x)=0\) in \(\mathbb{L}\) must be \(u(\x)\) (Proposition 2), and differentiating this gives \(Du=p\x^{p-1}=0\), so by [Ring Theory] §Polynomial Rings, ⁋Proposition 11 we know that \(\alpha\) is a multiple root of \(u\). In fact, by §Fields, ⁋Theorem 10 we have \((\x-\alpha)^p=\x^p-\alpha^p=\x^p-t\), so \(\alpha\) has multiplicity \(p\).
Now, in the next post and the one after that, we examine how to exclude such cases from the discussion.
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