Modules vs Vector Spaces

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Please go through this article once, it will clear your viewpoint about these two structures.

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Compactness in Metric Spaces

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  The idea of compactness comes from the fact that these are the nice set next to the finite sets. i.e. It shares many properties of finite sets in a topological space. The first thing common between finite and compact sets is, these are closed and bounded. Every sequence has a limit point. A continuous image of a compact set is compact and many more. Compactness has a very close relation to completeness . Let X be a topological space then the definition of compactness is defined in the following way-: X is said to be compact if every open covering of X has a finite sub-covering. Heine-Borel Theorem -: Heine-Borel make our work easy for Euclidean spaces by generalizing that, ‘ K is a compact subset of R if and only if it is closed and bounded. ’ But above theorem doesn’t work well for all metric spaces , for example -: Take N with discrete metric. What is missing here? Completeness? No! Boundedness? No! Closedness? No! Then what? The thing that is missing is totally bounded
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Hilbert's Irreduicibility Theorem

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Hilbert's Irreduicibility Theorem :  Let $\Q$ be the field of rational numbers and $f_1(x_1, x_2, ..., x_n, y_1, y_2, ..., y_m), f_1(x_1, x_2, ..., x_n, y_1, y_2, ..., y_m),...,f_r(x_1, x_2, ..., x_n, y_1, y_2, ..., y_m)$ be a set of irreducible polynomials in $\Q(x_1, x_2, ..., x_n)[y_1, y_2, ..., y_m]$ . Then Hilbert's Irreduicibility Theorem tells that there exists an n-tuple $(a_1, a_2, ..., a_n)$ in $\Q^n$ such that $f_1(a_1, a_2, ..., a_n, y_1, y_2, ..., y_m), f_2(a_1, a_2, ..., a_n, y_1, y_2, ..., y_m),..., f_r (a_1, a_2, ..., a_n, y_1, y_2, ..., y_m)$ are irreducible in $\Q[ y_1, y_2, ..., y_m]$. A cute application of above theorem is that, If a polynomial $p(x)$ in $\Q[x]$ takes square value at every points in $\Q$ then $p(x)$ is itself a square in $\Q[x]$. (Hint : Apply Hilbert's irreducibilty to f(x, y) = y^2-g(x).)
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