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Jan
31
asked The kernel of homomorphism of a local ring into a field is its maximal ideal?
Jan
31
asked If a neibourhood of the origin shrinks to the origin then its closure also shurinks to the origin?
Jan
21
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
We can prove that $n=m$ by utilizing the isomorphism of $F(\alpha_1,\cdots,\alpha_k,\alpha_{k+1})[t]$ and $\sigma_{k+1}(F(\alpha_1,\cdots,\alpha_k,\alpha_{k+1}))[t]$, since the latter is a uniquely factrizable domain.
Jan
21
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
After $\alpha_{k+1}$ a root of $f(t)$ is adjoined to $F(\alpha_1,\cdots,\alpha_k)$,and the extension $\sigma_{k+1}$ of $\sigma:F \rightarrow L$ to $F(\alpha_1,\cdots,\alpha_k,\alpha_{k+1})$ is obtained.$f(t)$ splits off a new factor $(x-\alpha_{k+1})^m$ in $F(\alpha_1,\cdots,\alpha_k,\alpha_{k+1})$. At the same time $\sigma_{k+1}f$ splits off a new factor $(x-\beta_{k+1})^n$ in $\sigma_{k+1}(F(\alpha_1,\cdots,\alpha_k,\alpha_{k+1}))$, where $\beta_{k+1}=\sigma_{k+1}(\alpha_{k+1})$. We can prove that $n=m$ (continues to next comment)
Jan
21
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
It seems that your argument is based on the existence of the extension proved by using Zorn's lemma. It is right. But, the proposition I asked about is logically prior to the existence of an embedding of algebraic extension of $k$ in L as user121926 pointed out. That's why I was interested in your second comment, because it seemed you suggested a constructive way to prove the second part of the proposition. I have no objection to your suggestion. I only wanted the explanation that process can be completed successfully. I've come up with the explanation. I post it the next comment.
Jan
21
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
In your second comment after your answer, you mentioned that $\alpha_i$'s are roots of $p(t)$. If they are not, how do you construct the extension of $\sigma$ to the splitting field for $p(t)$ over $k$ ?
Jan
20
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
I'm sorry again. I failed to post the previous comment. If $p(t) \in k[t]$ is the irreducible polynomial of $\alpha_1$ and $\tau_1:k(\alpha_1)\rightarrow L$ is the extension of $\sigma$, and $\tau_2:k(\alpha_1,\alpha_2)\rightarrow L$ is the extension of $\tau_1$, then the destination of $\alpha_2$ must be a root of $p^\sigma$. For $0=\tau_2(p(\alpha_2))=p^\sigma(\tau_2(\alpha_2))$.
Jan
20
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
I'm sorry. I assumed that $\beta_1$ is the destination of $\alpha_1$. If $\tau_2:k(\alpha_1,\alpha_2)\rightarrow L$ is the extension of $\tau_1:k(\alpha_1)\rightarrow L$.
Jan
20
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
When you extend $k(\alpha_1)\rightarrow L$ to $k(\alpha_1,\alpha_2)\rightarrow L$, you have to choose the destination of $\alpha_2$. If $\alpha_1 \neq \alpha_2$, that must be different from $\beta_1$.
Jan
20
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
It seems to me that to construct the embedding $\mu$ you need the same number of distinct roots of $p^\sigma(t)$ in $L$ as $p(t)$ in $k^a$. But that is what we want to prove in the first place.
Jan
19
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
Sorry for my repeated questions. When you extend the embedding $k(\alpha_1)\rightarrow L$ to $k(\alpha_1,\alpha_2)\rightarrow L$, you have to find another root of $p^\sigma$ in $L$. Is this always possible ?
Jan
18
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
Your argument is clear. I understand that the second part of the proposition is proved in that way. Do you think that it is impossible to prove it without using the extension of $\sigma$ to an embedding of $k^a$ in $L$ ?
Jan
18
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
Could you elaborate on the 4th part ?
Jan
18
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
This is exactly what is shown in Theorem 2.8. There is the following sentence imediately after the Proposition 2.7.: This is an important fact, which we shall analyse more closely later. Does this mean that the second part of the proposition will become clear later ? Or, is there other way to prove it ?
Jan
17
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
In Theorem 2.8. the existence of an extension of $\sigma$ to an embedding of arbitrary algebraic extension of $k$ is proved by using Proposition 2.7.
Jan
17
comment The number of possible extensions of an embedding of a field into a algebraically closed field.
The edition is revised third.
Jan
17
asked The number of possible extensions of an embedding of a field into a algebraically closed field.
Jan
13
comment Triangle inequality of a metric on a quotient space of a topological vector space
inf A + inf B = inf (A+B). See for example Exercise 1.3.9. of understanding analysis by Stephen Abbott.
Dec
7
comment Question about extensions of homomorphisms
Thank you for answering my question. Am I right in thinking that an extension of $\psi$ is defined as a unique homomorphism of $A/\mathfrak{m}$-algebras $\tilde{\psi}:B/\mathfrak{P} \rightarrow L$ such that $\tilde{\psi}(x^{-1}+\mathfrak{P})=\lambda$ for any $\lambda\in L$.
Dec
6
asked Question about extensions of homomorphisms