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As an example, I sometimes see the statement that we fix any embedding $\bar{K} \hookrightarrow \bar{K_v}$ where $K$ is algebraic number field, $K_{v}$ is the completion by the place $v$ and $\bar{k}$ is the separable closure of $K$. I can understand the intuitive meaning of "fix any embedding", but I can't comprehend what do formally. What does "fix any embedding" mean? Thanks in advance!

user947740
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    Even though your question makes sense on a more general level, can you explain your notation? So what are $\bar{K}$ and $\bar{K}_v$? This will help in providing a better answer. – Mark Kamsma Jul 06 '21 at 13:59
  • In the book I saw, $K$ is algebraic number field, $K_{v}$ is the completion by the place $v$ and $\bar{K}$ is the separable closure of $K$. Indeed this is an example so the word "fix" may have used more generically. – user947740 Jul 06 '21 at 14:10
  • It is probably best to edit that into your question as well, so that the question itself contains all the information. – Mark Kamsma Jul 06 '21 at 14:24
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    Probably it is only a way of saying: let E an embedding whatever; then blah blah. If the proof does not assume any specific fact about E, then the result will hold for every embedding. – Mauro ALLEGRANZA Jul 06 '21 at 14:26
  • @MarkKamsma Thank you for your comment! I edited my question and added the information of the example. – user947740 Jul 06 '21 at 14:31
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    It's probably the same as when saying "Fix n an integer" or "Fix t a real number". That is so that you don't think that whatever you "fix" can vary later during the proof – Stefan Octavian Jul 06 '21 at 14:34
  • @MauroALLEGRANZA After all, do that means stating "for any embedding ..." as the formula? – user947740 Jul 06 '21 at 14:43
  • @StefanOctavian I appreciate your good example. Indeed, I can't also comprehend "fix" in these examples. – user947740 Jul 06 '21 at 14:54
  • For example, the following question in Mathematics StackExchange use "Fix an integer n": Fix an integer n>2 . If |G|=n<∞ , prove that G has no subgroup of order n−1. I think this can be paraphrased as follows: "For all $n>2$, $|G|=n<\infty \rightarrow G$ has no subgroup of order $n−1$." Similarly, how about the answer that "Fix any embedding" is "For any embedding"? – user947740 Jul 06 '21 at 15:27
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    What do you mean by "formally"? Is there some particular formal proof system you want to express this in? In any case, "fix ..." (like "let ...") just means you are introducing a free variable which satisfies some formula, though the exact rule of formal logic you are using to do so will vary based on context. – Eric Wofsey Jul 06 '21 at 19:37

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Formally, "Fix $A$. Then ..." means the same thing as "For all $A$, we have ...". However, one chooses that formulation in order to emphasize that, in the specific situation, one does not really care that it holds for all $A$. We just choose some $A$ (so actually, we need to assume that some $A$ exists) and work with it, and ignore that there are other $A$s well. For example (this is a little bit more elementary than your example), if $K$ is a field and $L$ is an algebraic extension of $K$, we can fix an embedding $L \to \overline{K}$ to develop basic field theory. Of course, there are lots of embeddings, but when we are only interested to prove something about $L$, sometimes the existence of one embedding is enough.

Martin Brandenburg
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