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

[chwala]

Homework Statement

See attached

Relevant Equations

Ring Theory

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.

 

[fresh_42]

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'.##

##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##.

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

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

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

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.

 

[chwala]

##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...

 

[chwala]

##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?

 

[fresh_42]

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.

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.

 

[pasmith]

Note that in [itex]\mathbb{Z}_3[/itex], [itex]2' = -1'[/itex] and [itex](-1')^n[/itex] is [itex]1'[/itex] if [itex]n[/itex] is even and [itex]-1'[/itex] if [itex]n[/itex] is odd. Therefore [itex](2')^3 + (2')^2 = 0'[/itex].

 

[WWGD]

Well, strictly speaking, ##\frac{a}{b}= ab^{-1}##. Though ##b^{-1}## may not exist in ##\mathbb Z_n## if ##n## is not a prime.

 

[MidgetDwarf]

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.

 

[chwala]

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!