Andreas Caranti
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 Apr15 comment every finitely generated vector space has a basis. Question about the proof Yes, proceeding by induction means exactly that, keep discarding elements until you get an independent set of generators. Apr15 revised every finitely generated vector space has a basis. Question about the proof Fixed brackets Apr15 answered every finitely generated vector space has a basis. Question about the proof Apr15 comment Why $K = (X_1, X_2, …)$, the ideal generated by $X_1, X_2, …$ not finitely generated as a R-module? @user136266,$X_t$ is an indeterminate. Apr14 revised Definition of a Finite Normal Extension added 331 characters in body Apr14 answered Definition of a Finite Normal Extension Apr14 revised find the minimal polynomial over $\mathbb{Q}$ of the algebraic number $\mathbb{(1+\sqrt{5})/2}$ added 6 characters in body; edited title Apr14 comment Multiple Group Representations using Cayley's Thm I believe what you are saying is correct. Apr14 answered Multiple Group Representations using Cayley's Thm Apr14 revised Why $K = (X_1, X_2, …)$, the ideal generated by $X_1, X_2, …$ not finitely generated as a R-module? added 4 characters in body Apr14 answered Why $K = (X_1, X_2, …)$, the ideal generated by $X_1, X_2, …$ not finitely generated as a R-module? Apr14 revised Why $K = (X_1, X_2, …)$, the ideal generated by $X_1, X_2, …$ not finitely generated as a R-module? added 13 characters in body Apr14 comment Show that $A∩B∩C= ∅$ is only true when $A∩B = ∅, A∩C = ∅$ or $B∩C = ∅$ or show a counterexample. So, as advised also by @egreg, try to find three sets $A, B, C$ which satisfy $A \cap B \cap C = \emptyset$ but not (any of) the $A \cap B = A \cap C = B \cap C = \emptyset$. You don't have to look far, you can choose $A, B, C \subset \{1, 2, 3\}$. Apr14 comment Show that $A∩B∩C= ∅$ is only true when $A∩B = ∅, A∩C = ∅$ or $B∩C = ∅$ or show a counterexample. I believe you are confused about the question itself, which I believe to be the following: Suppose $A, B, C$ are three sets, such that $A \cap B \cap C = \emptyset$. Is it true that $A \cap B = A \cap C = B \cap C = \emptyset$? Apr14 comment Showing a linear combination of matrices is nilpotent for any constants Thanks a lot @abel Apr14 revised Showing a linear combination of matrices is nilpotent for any constants deleted 382 characters in body Apr14 revised Showing a linear combination of matrices is nilpotent for any constants deleted 382 characters in body Apr14 revised Showing a linear combination of matrices is nilpotent for any constants added 6 characters in body Apr13 comment Showing a companion matrix is normal $\overline{a_{0} \bar{a_{1}}} = \bar{a_{0}} a_{1}$, and you're done. This is because $\bar{\bar{a}} = a$, and $\overline{a b} = \bar{a} \bar{b}$. Apr13 revised Exercise in group action blocks deleted 1 character in body