Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Mathematics
Linear and Abstract Algebra
Primitive Elements and Free Modules .... ....
Reply to thread
Message
[QUOTE="andrewkirk, post: 6044486, member: 265790"] That ##\{x_1,...,x_n\}## is a basis for ##F## means the same as that $$F=x_1R\oplus...\oplus x_nR$$ and since ##M\triangleq x_1R\oplus...\oplus x_{n-1}R## we have ##F=M\oplus x_nR##. Since the induction hypothesis gives us that ##\{x, x'_2,...,x'_{n-1}\}## is a basis of ##M## we have $$M = xR\oplus x'_2R\oplus x'_3R\oplus...\oplus x'_{n-1}$$ So we have $$F=M\oplus x_nR = xR\oplus x'_2R\oplus x'_3R\oplus...\oplus x'_{n-1}\oplus x_nR$$ which is equivalent to saying that ##\{x,x'_2,...,x'_{n-1},x_n\}## is a basis for ##F##. [/QUOTE]
Insert quotes…
Post reply
Forums
Mathematics
Linear and Abstract Algebra
Primitive Elements and Free Modules .... ....
Back
Top