338 reputation
15
bio website
location
age
visits member for 2 years, 6 months
seen 18 hours ago

Apr
11
comment If $X \times \{0\} \cup A \times I$ is closed in $X \times I$. Then, is $A$ closed in $X$?
Am I right in thinking that $B\times (0,1]$ is the complement of $X \times \{0\} \cup A \times I$ in $X \times I$. If it is true, then it follows that $B\times (0,1]$ is open.
Apr
11
comment Is the length of the composition series of a free module identical to the number of its bases?
It is the page 118 of 1969 version.
Apr
11
comment Is the length of the composition series of a free module identical to the number of its bases?
It works. Then, the description of the example in this famous book is not correct. Anyway, thank you for your help.
Apr
11
comment Is the length of the composition series of a free module identical to the number of its bases?
In the example, $A_n$ is the set of the homogeneous polynomials of degree $n$.
Apr
11
comment Is the length of the composition series of a free module identical to the number of its bases?
$A$ is a Noetherian graded ring $A=\oplus_{n=0}^\infty A_n$. $M$ is a finitely-generated graded $A$-module $M=\oplus_{n=0}^\infty M_n$.
Mar
25
comment If a chain of distinct irreducible closed subsets of a quasi-affine variety $Y$ is maximal, a chain of their closures is maximal in $\overline{Y}$?
Thank you again. I understand completely this time.
Mar
24
comment If a chain of distinct irreducible closed subsets of a quasi-affine variety $Y$ is maximal, a chain of their closures is maximal in $\overline{Y}$?
Thank you for answering my question. I didn't notice that $W'$ is an open subset of W. One thing is not clear to me. You wrote $Z_i$ is not only irreducible in $Y$ but also in $X$. Does the irreducibility of a subset depend on the space containing it ?
Jan
31
comment If a neibourhood of the origin shrinks to the origin then its closure also shurinks to the origin?
The definition in the book is as follows: $X$ is an F-space if its topology $\tau$ is induced by a complete invariant metric $d$.
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 ?