Two-Factor Authentication
by djmathman, Feb 19, 2017, 3:49 AM
So recently CMU asked us (or at least all faculty and student employees) to switch over to a 2fa login system, which basically means that we need to have our phones with us to log in for security purposes
unfortunately my parents did not realize this, and so now I am temporarily blocked from my CMU account due to excessive login attempts
RIP
unfortunately my parents did not realize this, and so now I am temporarily blocked from my CMU account due to excessive login attempts
RIP
21-238 Math Studies Algebra II Homework 2 Problem 5 wrote:
(a) Suppose that 
is a commutative ring with 
and that 
is a nonzero -module such that for all 
and 
, we have that 
if and only if 
or 
. Show that is an integral domain.
(b) Suppose that and are as in part (a), and that 
is an -module homomorphism. Suppose also that we have a set 
and an injection 
such that for all 
, 
. Show that 
is -linearly independent.
is a commutative ring with 
and that 
is a nonzero -module such that for all 
and 
, we have that 
if and only if 
or 
. Show that is an integral domain.(b) Suppose that
and are as in part (a), and that 
is an -module homomorphism. Suppose also that we have a set 
and an injection 
such that for all 
, 
. Show that 
is -linearly independent.
for 
.
.![\[0 = (ab) \cdot x = a\cdot (b\cdot x),\]](http://latex.artofproblemsolving.com/f/f/7/ff762bf0cf62cc076d4dcc7f15eb9916aabc9e04.png)
so either 
or 
.
implies either 
,
practically by definition. It suffices to show that 
,
,
,![\[\sum_{j=0}^{k-1}r_j\cdot x_j = 0\quad\implies\quad r_j = 0\text{ for all }j.\]](http://latex.artofproblemsolving.com/1/a/7/1a79a5eef7a47512a740d71387bbb2b53736690e.png)
To do this, we will use induction on 
. For the base case of 
,
;
,
.
and 
such that 
.
.![\[0 = f(x_0)\cdot 0 = f(x_0)\cdot\sum_{j=0}^k(r_j\cdot x_j) = \sum_{j=0}^kf(x_0)\cdot(r_j\cdot x_j) = \sum_{j=0}^k(f(x_0)r_j)\cdot x_j.\]](http://latex.artofproblemsolving.com/c/f/c/cfc78021e2027ad496b505641d3104c6c5c1a333.png)
Additionally, since 
is a 
,
Combining both of these observations yields that ![\[\sum_{j=0}^k(f(x_0)r_j)\cdot x_j = \sum_{j=0}^k(r_jf(x_j))\cdot x_j\quad\implies\quad \sum_{j=0}^k\left[r_j(f(x_0) - f(x_j))\right]\cdot x_j = 0.\]](http://latex.artofproblemsolving.com/f/6/1/f6119f6502700f1ecc6c5aa96a4dc862efc2a060.png)
Note that in fact the 
term goes away, since 
.
for all 
.
or 
.
is an injection, the latter case cannot occur. Hence This post has been edited 2 times. Last edited by djmathman, Feb 19, 2017, 4:09 AM
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