(New page: Corollary 3 of THM 16.2 : "A polynomial of degree ''n'' over a field has at most ''n'' zeros, counting multiplicity" Fields and an finite integral domains are one and the same. (THM 13.2)...) |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 9: | Line 9: | ||
''Not sure if this is sound. Comments?'' | ''Not sure if this is sound. Comments?'' | ||
--[[User:Bcaulkin|Bcaulkin]] 21:27, 1 April 2009 (UTC) | --[[User:Bcaulkin|Bcaulkin]] 21:27, 1 April 2009 (UTC) | ||
| + | |||
| + | |||
| + | So, I'm not entirely sure what angle the question is going at, but I think taking x^3 in Z mod 8Z will work as an example showing that the corollary does not hold. --[[User:Jcromer|Jcromer]] 22:19, 1 April 2009 (UTC) | ||
| + | |||
| + | ---- | ||
| + | Just showing a counterexample will not suffice, as the question asks to prove Corollary 3 false for ''any'' commutative ring that has a zero divisor. | ||
| + | |||
| + | Consider <math>\scriptstyle f(x)</math> and <math>\scriptstyle g(x)</math> such that: | ||
| + | <math>\scriptstyle f(x)\ =\ a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0</math>, | ||
| + | <math>\scriptstyle g(x)\ =\ b_mx^m+b_{m-1}x^{m-1}+\ldots+b_1x+b_0</math>, | ||
| + | and both <math>\scriptstyle f(x)</math> and <math>\scriptstyle g(x)</math> are over a commutative ring <math>\scriptstyle R</math> with zero-divisors <math>\scriptstyle a_0</math> and <math>\scriptstyle b_0</math>. Say the number of zeroes of <math>\scriptstyle f(x)</math> is <math>\scriptstyle n</math> and the number of zeroes of <math>\scriptstyle g(x)</math> is <math>\scriptstyle m</math>. Note that 0 is not a root of either <math>\scriptstyle f(x)</math> or <math>\scriptstyle g(x)</math> because <math>\scriptstyle a_0</math> and <math>\scriptstyle b_0</math> are nonzero (by the definition of a zero-divisor). Now consider | ||
| + | |||
| + | <math>\scriptstyle f(x)\cdot g(x)\ =\ c_{m+n}x^{m+n}+c{m+n-1}x^{m+n-1}+\ldots+c_1x+c_0</math>, where | ||
| + | <math>\scriptstyle c_k\ =\ a_kb_0+a_{k-1}b_1+\ldots+a_1b_{k-1}+a_0b_k</math>. | ||
| + | |||
| + | Say the number of zeroes of <math>\scriptstyle f(x)\cdot g(x)</math> is <math>\scriptstyle t</math>. Clearly <math>\scriptstyle t\ \geq\ m+n</math>, because any root of either <math>\scriptstyle f(x)</math> or <math>\scriptstyle g(x)</math> is a root of <math>\scriptstyle f(x)\cdot g(x)</math>. But since <math>\scriptstyle c_0\ =\ a_0b_0\ =\ 0</math>, | ||
| + | |||
| + | <math>\scriptstyle f(x)\cdot g(x)\ =\ c_{m+n}x^{m+n}+c_{m+n-1}x^{m+n-1}+\ldots+c_1x\ =\ x(c_{m+n}x^{m+n-1}+c_{m+n-1}x^{m+n-2}+\ldots+c_1)</math>. | ||
| + | |||
| + | Now, 0 is a zero of <math>\scriptstyle f(x)\cdot g(x)</math>, and <math>\scriptstyle t</math> must be at least <math>\scriptstyle m+n+1</math>. However, the degree of <math>\scriptstyle f(x)\cdot g(x)</math> is only m+n! Corollary 3 therefore cannot hold. <math>\scriptstyle\Box</math> | ||
| + | :--[[User:Narupley|Nick Rupley]] 03:35, 2 April 2009 (UTC) | ||
Latest revision as of 22:35, 1 April 2009
Corollary 3 of THM 16.2 : "A polynomial of degree n over a field has at most n zeros, counting multiplicity"
Fields and an finite integral domains are one and the same. (THM 13.2)
Finite integral domains are commutative rings with unity and no zero-divisors (Definition of integral domain)
So, if the commutative ring has zero divisors, it cannot be a field, thus no polynomials may over it, thus Corollary 3 is false for any ring with zero-divisors.
Not sure if this is sound. Comments? --Bcaulkin 21:27, 1 April 2009 (UTC)
So, I'm not entirely sure what angle the question is going at, but I think taking x^3 in Z mod 8Z will work as an example showing that the corollary does not hold. --Jcromer 22:19, 1 April 2009 (UTC)
Just showing a counterexample will not suffice, as the question asks to prove Corollary 3 false for any commutative ring that has a zero divisor.
Consider $ \scriptstyle f(x) $ and $ \scriptstyle g(x) $ such that: $ \scriptstyle f(x)\ =\ a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0 $, $ \scriptstyle g(x)\ =\ b_mx^m+b_{m-1}x^{m-1}+\ldots+b_1x+b_0 $, and both $ \scriptstyle f(x) $ and $ \scriptstyle g(x) $ are over a commutative ring $ \scriptstyle R $ with zero-divisors $ \scriptstyle a_0 $ and $ \scriptstyle b_0 $. Say the number of zeroes of $ \scriptstyle f(x) $ is $ \scriptstyle n $ and the number of zeroes of $ \scriptstyle g(x) $ is $ \scriptstyle m $. Note that 0 is not a root of either $ \scriptstyle f(x) $ or $ \scriptstyle g(x) $ because $ \scriptstyle a_0 $ and $ \scriptstyle b_0 $ are nonzero (by the definition of a zero-divisor). Now consider
$ \scriptstyle f(x)\cdot g(x)\ =\ c_{m+n}x^{m+n}+c{m+n-1}x^{m+n-1}+\ldots+c_1x+c_0 $, where $ \scriptstyle c_k\ =\ a_kb_0+a_{k-1}b_1+\ldots+a_1b_{k-1}+a_0b_k $.
Say the number of zeroes of $ \scriptstyle f(x)\cdot g(x) $ is $ \scriptstyle t $. Clearly $ \scriptstyle t\ \geq\ m+n $, because any root of either $ \scriptstyle f(x) $ or $ \scriptstyle g(x) $ is a root of $ \scriptstyle f(x)\cdot g(x) $. But since $ \scriptstyle c_0\ =\ a_0b_0\ =\ 0 $,
$ \scriptstyle f(x)\cdot g(x)\ =\ c_{m+n}x^{m+n}+c_{m+n-1}x^{m+n-1}+\ldots+c_1x\ =\ x(c_{m+n}x^{m+n-1}+c_{m+n-1}x^{m+n-2}+\ldots+c_1) $.
Now, 0 is a zero of $ \scriptstyle f(x)\cdot g(x) $, and $ \scriptstyle t $ must be at least $ \scriptstyle m+n+1 $. However, the degree of $ \scriptstyle f(x)\cdot g(x) $ is only m+n! Corollary 3 therefore cannot hold. $ \scriptstyle\Box $
- --Nick Rupley 03:35, 2 April 2009 (UTC)
