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 Jul 30 comment Question on Forster's proof of the residue theorem Thank you very much for your detailed explanation. I forgot to write that the $U_k$'s are chosen so that $U_j \cap U_k = \emptyset$ for $j\neq k$. From this I think that $\epsilon \rightarrow 0$ is not necessary in the last equality because $a_k$ is the only singular point in $U_k$. Am I wrong ? Jun 5 comment Why is it Artinian? I didn't know the alternative definition. Thank you for that. Jun 5 comment Why is it Artinian? Thank you for your help. I understand that for each product $x_1 x_2 \cdots x_n$ where $x_i \in m_i$ there exists $i$ such that $x_i \in p$. But how do we know that $p \supset m_i$ for some $i$. May 6 comment Proof of a corollary of the Noether normalisation lemma @Cantlog : There is an addendum between the lemma and the corollary saying "under the condition of the lemma, if furthermore $k$ is algebraiccaly closed, and $A$ is an integral domain with field of fractions $K$ then ...". So, they may be the additional conditions. 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.