# On a formula of the norm of an element of a finite extension of a field

Theorem Let $F$ be a field. Let $K$ be a finite extension of $F$. Let $[K : F]_i$ be the inseparable degree of $K/F$. Let $\bar{K}$ be an algebraic closure of $K$. Let $S$ be the set of $F$-embeddings of $K$ into $\bar{K}$. Let $\alpha \in K$. Then $N_{K/F}(\alpha) = (\prod_{\sigma \in S}\sigma(\alpha))^{[K : F]_i}$

This is proved in Theorem 60 in page 39 of the lecture note written by Pete L. Clark. I don't understand the proof. Would any one please enlighten me?

EDIT Why down votes?? Asking for help to understand a proof should be frowned upon?

EDIT I understand the prerequisites for the proof, i.e. the content of section 6 and Corollary 58.

EDIT Let $f(X)$ be the characteristic polynomial of $\alpha$. Let $g(X)$ be the minimal polynomial of $\alpha$. $By$  $Corollary$  58, $f(X)$ = $g(X)^{[K:F(\alpha)]}$. The set {$\sigma(\alpha); \sigma ∈ S$} is the set of the roots of g(X). However, it is not clear to me that the equation $N_{K/F}(\alpha) = (\prod_{\sigma \in S}\sigma(\alpha))^{[K : F]_i}$ follows immediately. Would anyone please explain why the equation follows other than just saying it is straightforward?

EDIT It's amazing that some people regard questioning a proof as an attack to the author's credibilty. Everybody makes a mistake. Even Grothendieck made a non-trivial mistake(see Misconceptions About $K_{X}$ by Kleiman). I think this attitude is harmful to healthy development of mathematics. I'm not claiming that the proof is wrong, though.

EDIT To anyone who thinks the proof is straightforward, please explain to me in detail. I am not as smart as you.

EDIT It's surprising that no one explained the proof so far. I believe every step of any proof can be reduced to (really) trivial statements. If you think it's straightforward, please reduce it to more trivial statements that anybody who has basic knowledge of abstract algebra can understand.

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@Arturo I thought it's obvious that the proof is too terse even assuming Corollary 58 etc.. Now I know some people do think otherwise. Perhaps it's straightforward for smart people, but alas I'm not that smart. –  Makoto Kato May 19 '12 at 0:58
If I may venture a piece of advice: when you are reading a text and come to a point that seems obscure to you, usually the fastest and best remedy is just to consult a different text. In this case any reasonably serious field theory text should do: for instance, see Section 8 of Patrick Morandi's Field and Galois Theory, Section 6.5 of Lang's Algebra, and so forth. –  Pete L. Clark May 19 '12 at 2:31
For my part, I'm sorry that you read my notes (which are, by the way, signalled on my webpage as being a first, rough draft) and found something not to your liking. But "it's obvious that the proof is too terse" is not constructive feedback: it doesn't give me any indication what to change. If someone else has a suggestion for how to improve this part of the exposition, please do say so. –  Pete L. Clark May 19 '12 at 2:33
@Pete L. Clark I edited my question. I believe my question is clear enough now. –  Makoto Kato May 26 '12 at 22:19

Well the theorem is a bit too hastily stated, for example if you happen to pick $\alpha \in F$, then $f(t) = (t-\alpha)^n$ and has only one distinct root.

The correct statement is : Let $K/F$ be a field extension of degree $n < \infty$ and separable degree $m$. Let $\overline{K}$ be an algebraic closure of $K$. Let $\alpha \in K$ and let $f(t)$ be the characteristic polynomial of $\alpha \bullet \in End_F(K)$. Let $\sigma_1 \ldots \sigma_m$ be the distinct $F$-algebra embeddings of $K$ into $\overline{K}$. Then the factorisation of $f(t)$ on $\overline{K}$ is $f(t) = \prod_{i=1}^m (t- \sigma_i(\alpha))^{n/m}$.

First, let $K_0 = F(\alpha)$, $n_0 = [K_0 : F]$, $m_0$ be the separable degree of $[K_0 : F]$, $\tau_1 \ldots \tau_{m_0}$ be the embeddings of $K_0$ into $\overline{K}$, and $f_0(t)$ be the minimal polynomial of $\alpha$ over $F$.
Then, by corollary 58, $f(t) = f_0(t)^{[K : K_0]} = f_0(t)^{n/n_0}$.
According to the proof of theorem 38, each $\tau_i$ is the restriction to $K_0$ of $m/m_0$ many embeddings $\sigma_j$, therefore $\prod_{i=1}^m (t- \sigma_i(\alpha))^{n/m} = \prod_{i=1}^{m_0} (t- \tau_i(\alpha))^{n/m_0}$, so we only need to prove that $f_0(t) = \prod_{i=1}^{m_0} (t- \tau_i(\alpha))^{n_0/m_0}$.

From now, suppose we are in the case $K = K(\alpha)$. Let $K_s$ be the separable closure of $F$ in $K$, so by Proposition 50, $K/K_s$ is purely inseparable and $K_s/F$ is separable. By corollary 47, there is an integer $a\ge 0$ such that $K_s = K(\alpha^{p^a})$. Let $g(t)$ be the minimal polynomial of $\alpha^{p^a}$ over $F$. Then $f(t) = g(t^{p^a})$, because $g(\alpha^{p^a}) = 0$ and they have the same degree $n = p^a m$. Meanwhile, $(t - \sigma_i(\alpha))^{p^a} = (t^{p^a} - \sigma_i(\alpha^{p^a}))$ so we only need to prove that $g(t) = \prod_{i=1}^m (t - \sigma_i(\alpha^{p^a}))$.

So we only need to prove the theorem when $\alpha$ is separable. But then this is clear : the $\sigma_i(\alpha)$ are roots of $f(t)$ because $\sigma_i$ are embeddings of $F$-algebras, and every factor in the product is distinct, so the product has to divide $f(t)$, and both polynomials have the same degree, so they must be equal.

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Thank you, Mercio. Your proof is basically the same as I had in my mind. By the way, if $\alpha \in F$, since $\sigma(\alpha) = \alpha$, $N_{K/F}(\alpha) = (\prod_{\sigma \in S}\sigma(\alpha))^{[K : F]_i}$ still holds. –  Makoto Kato May 25 '12 at 21:20

I don't think the downvotes are helping the situation. In case anyone is downvoting the OP's question out of some sense of solidarity with / loyalty to me: thanks, but please don't.

On the other hand, exactly what the OP wants to ask is not clear, even to me. "I don't understand the proof. Would any one please enlighten me?" isn't specific enough to be helpful. In fact, more than a couple of days ago now, this started with a comment the OP left to me elsewhere on this site:

"@Pete I think the proof of Theorem 60 is incorrect (I'm not talking about some obvious typos there). Please forgive me if this is my mistake."

Let me start by saying that I most certainly do want errors in my mathematical writings to be brought to my attention. At this point though I have about 2000 pages of such things available on the internet, so it is a nontrivial matter as to how to best deal with adding / editing / correcting these things. Generally at any point in time I am more mentally present and interested in making these changes on some documents than others (if anyone cares, at the moment it is honors calculus / undergraduate analysis, geometry of numbers and quadratic forms which have most of my attention).

What I'm trying to say is: at the moment I don't really have a well thought out system to deal with modifications to the product I've already made available. People write me to suggest changes and corrections fairly frequently (a few times a month, on average), and I'm sorry to say that I address some but not all of these suggestions in a timely fashion. For now, if you want to maximize the chances that a correction gets made sooner rather than later, then it helps to give me as much information as possible and give me as concrete a task as possible. For instance:

1) Please include specific information about which notes you are looking at: a link to a particular file on my webpage is ideal.

2) If you think that I've made a mistake, please say exactly what you think the mistake is. If you think you know the correction, of course please tell me.

3) If you just didn't understand something in my notes, or have a natural followup question, or are seeking references on something I alluded to too casually (N.B.: I have gotten much better in recent years at including precise, formal references even in lecture notes for my undergraduate classes), then you might want to try asking someone else first, or asking on this site. When it comes to teaching, I will naturally put my own students, and students at my own university, first. Anyway, when it comes to questions about my expository notes, I promise you that I am not uniquely or best qualified to answer them. On the contrary, for almost anything that I have written about, there are people here who are more qualified to address your questions than I am.

If you do ask a question here that comes from my own writings, please ping me. I will be interested to read it, even if I am not one of the ones to answer it.

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Pete, see the 20 comments at math.stackexchange.com/questions/136456/… along with those at mathoverflow.net/questions/95225/… and perhaps my email to Franz if your curiosity extends that far. I did not downvote either, but there is a pattern. –  Will Jagy May 18 '12 at 19:35
@Will Somehow you seem to have developed ill feelings toward me. Would you mind to tell me what made you so if that is the case? And if you want to say something to me, say it straight to me. –  Makoto Kato May 19 '12 at 11:24
@Will "but there is a pattern" So what is the pattern? And what's wrong with that? –  Makoto Kato May 19 '12 at 15:21
@MakotoKato, please email me. Click on my profile. If you do not see an email address for me, search at ams.org/cml and use the gmail address if possible. I really do not know enough about you at this stage. –  Will Jagy May 19 '12 at 18:00
@Will I want you to answer me here. If you think this is personal, don't write personal matters here in the first place. –  Makoto Kato May 19 '12 at 19:49
What exactly you don't understand in that proof? It is heavily based on former results in the same notes, and if you know already that the conjugates of certain element are the images of this element under the different embeddings $\,K\to \overline{K}\,$, each being repeated precisely $\,[K:F]_i\,$ , then...well, then we're done!
Perhaps Mr. Clarke's defintion of $N_{K/F}(\alpha)$ and yours are different. Let $\alpha_{L}$ be the $F$-linear map K → K definded by x → $\alpha.x$. Then $N_{K/F}(\alpha)$ is the determinant of $\alpha_{L}$. –  Makoto Kato May 18 '12 at 3:22