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Question 16: Show that Corollary 3 of Theorem 16.2 is false for any commutative ring that has a zero divisor. | Question 16: Show that Corollary 3 of Theorem 16.2 is false for any commutative ring that has a zero divisor. | ||
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--[[User:Robertsr|Robertsr]] 19:44, 5 November 2008 (UTC) | --[[User:Robertsr|Robertsr]] 19:44, 5 November 2008 (UTC) | ||
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| + | I don't understand what you did for the last two lines. You may be attempting to deal with the case if a=b which then lets your original polynomial satisfy the corrolary. (not good) An easy way to construct a polynomial that shows the cor is false in a ring with zero divisors is to look at the polynomial ax=0. This has roots 0, and b. There are two roots to this regardless if a=b because a does not equal zero by the definition of a zero divisor. | ||
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| + | -Allen | ||
Latest revision as of 07:29, 9 November 2008
Question 16: Show that Corollary 3 of Theorem 16.2 is false for any commutative ring that has a zero divisor.
Corollary 3 states that a polynomial of degree n over a field has a most n zeros (roots) counting multiplicity. In other words, if have zero divisors Corollary 3 fails.
PROOF:
If the field R has zero divisors then there is a polynomial with more than deg(7) roots. a, b are contained in R a * b = 0 which make a and b zero divisors since a,b do not equal 0
(x-a) (x-b) has roots a,b
f: x^2 -ax-bx+ab
x(x-a-b) = 0 this has 4 roots (a,b,0, a+b) and degree of polynomial f is 2
x = a + b
commutative used -ax = -xa
--Robertsr 19:44, 5 November 2008 (UTC)
I don't understand what you did for the last two lines. You may be attempting to deal with the case if a=b which then lets your original polynomial satisfy the corrolary. (not good) An easy way to construct a polynomial that shows the cor is false in a ring with zero divisors is to look at the polynomial ax=0. This has roots 0, and b. There are two roots to this regardless if a=b because a does not equal zero by the definition of a zero divisor.
-Allen
