Show that ##f(x)=2',1',2'## in the irreducible Polynomial

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The discussion centers on proving that the polynomial f(x) = x^3 + x^2 + 2' is irreducible over the field Z_3. Participants demonstrate that f(0'), f(1'), and f(2') yield non-zero results, indicating the polynomial has no roots in Z_3. This absence of roots suggests that if f(x) were reducible, it would imply the existence of linear factors, which is excluded by the calculations. The conversation also touches on the characteristics of integral domains, emphasizing that the absence of zero divisors is crucial for irreducibility. Overall, the reasoning supports the conclusion that f(x) is indeed irreducible in Z_3.
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Homework Statement
See attached
Relevant Equations
Ring Theory
1679216081033.png


My interest is on the highlighted; my understanding is that,

let

##f(x)=x^3+x^2+2^{'}##

then

##f(1^{'})=1{'}+1{'}+2^{'}=4^{'} ##

we know that in ##\mathbb{z_3} ## that ##\dfrac{4}{3}=1^{'}##

##f(2^{'})=8^{'}+4^{'}+2^{'}=14{'} ##

we know that in ##\dfrac{14^{'}}{3}=1^{'}##...

I hope that is the correct reasoning for the highlighted part indicated in red.
 
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chwala said:
Homework Statement:: See attached
Relevant Equations:: Ring Theory

View attachment 323813

My interest is on the highlighted; my understanding is that,

let

##f(x)=x^3+x^2+2^{'}##

then
##f(0')=2'.##
chwala said:
##f(1^{'})=1{'}+1{'}+2^{'}=4^{'} ##

we know that in ##\mathbb{z_3} ## that ##\dfrac{4}{3}=1^{'}##
No fractions, please. We have ##4'=1' ## in ##\mathbb{Z}_3##.
chwala said:
##f(2^{'})=8^{'}+4^{'}+2^{'}=14{'} ##

we know that in ##\dfrac{14^{'}}{3}=1^{'}##...

Same here, only that ##14'=2'##

chwala said:
I hope that is the correct reasoning for the highlighted part indicated in red.
These three equations show that the polynomial has no zero. If it was reducible, say ##f(x)=g(x)\cdot h(x)##, then either ##g(x)## or ##h(x)## had to be linear, say ##g(x)=x-c'.## But this means ##g(c')=0## and ##f(c')=g(c')\cdot h(c')=0## which we excluded.
 
fresh_42 said:
##f(0')=2'.##

No fractions, please. We have ##4'=1' ## in ##\mathbb{Z}_3##.Same here, only that ##14'=2'##These three equations show that the polynomial has no zero. If it was reducible, say ##f(x)=g(x)\cdot h(x)##, then either ##g(x)## or ##h(x)## had to be linear, say ##g(x)=x-c'.## But this means ##g(c')=0## and ##f(c')=g(c')\cdot h(c')=0## which we excluded.
Noted @fresh_42 ...on the fraction bit. Cheers...
 
fresh_42 said:
##f(0')=2'.##

No fractions, please. We have ##4'=1' ## in ##\mathbb{Z}_3##.Same here, only that ##14'=2'##These three equations show that the polynomial has no zero. If it was reducible, say ##f(x)=g(x)\cdot h(x)##, then either ##g(x)## or ##h(x)## had to be linear, say ##g(x)=x-c'.## But this means ##g(c')=0## and ##f(c')=g(c')\cdot h(c')=0## which we excluded.
If a given polynomial has ##f(x)=0##, then it would imply existence of zero divisors- hence no integral domain...correct?
 
chwala said:
If a given polynomial, say with two variables, ##x## and ##y## has ##f(x)=0##,...
What do you mean? You say two variables but write only one.
chwala said:
then it would imply existence of zero divisors- hence no integral domain...correct?

like in this case, we are having ##x^{s}## and ##2## as our ##y##...
The definition of an integral domain is simple. It means that no non-zero elements can be multiplied to zero.
In formulas: ##(a\cdot b= 0 \Longrightarrow a=0 \text{ or }b=0) \Longleftrightarrow (a\neq 0 \text{ and }b\neq 0 \Longrightarrow a\cdot b\neq 0).##

E.g., the 12 hour marks on a clock's face ##\{0,1,2,3,\ldots,10,11\}## has zero divisors: ##3\cdot 4 = 12 = 0## or ##2\cdot 6 = 0.## The light switch ##\{0,1\}## has no zero-divisors. Although we have ##1+1=0## we do not have ##1\cdot 1=0,## and ##2## does not exist (or equals ##0##, depending on how we define it).

##\mathbb{Z}_n## is an integral domain if and only if ##n## is prime. In this case, it is even a field.
 
Note that in \mathbb{Z}_3, 2' = -1' and (-1')^n is 1' if n is even and -1' if n is odd. Therefore (2')^3 + (2')^2 = 0'.
 
Well, strictly speaking, ##\frac{a}{b}= ab^{-1}##. Though ##b^{-1}## may not exist in ##\mathbb Z_n## if ##n## is not a prime.
 
chwala said:
Homework Statement: See attached
Relevant Equations: Ring Theory

View attachment 323813

My interest is on the highlighted; my understanding is that,

let

##f(x)=x^3+x^2+2^{'}##

then

##f(1^{'})=1{'}+1{'}+2^{'}=4^{'} ##

we know that in ##\mathbb{z_3} ## that ##\dfrac{4}{3}=1^{'}##

##f(2^{'})=8^{'}+4^{'}+2^{'}=14{'} ##

we know that in ##\dfrac{14^{'}}{3}=1^{'}##...

I hope that is the correct reasoning for the highlighted part indicated in red.
Just to point out something in Fresh_44 comment. The method used by the author is a common technique used for ℤn , when n is prime.

ie., exhaust all the possible cases {1,2,..., n-1}. If f does not qual zero for any of these elements, then f is irreducible over ℤn. If f does equal zero for one of these elements, say an element a, then f is reducible, since this value is a zero (root) which is equivalent to saying x-a is a factor.

These problems become more interesting, when we are not working with finite integral domains.
 
Thanks @MidgetDwarf ...noted, I will look at this/get back on forum in a few weeks...trying to check on the health of a family member at moment. Cheers man!
 
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