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

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In summary: If a given polynomial has ##f(x)=0##, then it would imply existence of zero divisors- hence no integral domain...correct?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.
  • #1
chwala
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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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  • #2
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.
 
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  • #3
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...
 
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  • #4
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?
 
  • #5
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.
 
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  • #6
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].
 
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  • #7
Well, strictly speaking, ##\frac{a}{b}= ab^{-1}##. Though ##b^{-1}## may not exist in ##\mathbb Z_n## if ##n## is not a prime.
 
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  • #8
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.
 
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  • #9
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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1. What does it mean for a polynomial to be irreducible?

A polynomial is considered irreducible if it cannot be factored into polynomials of lower degree with coefficients in the same field. In other words, there are no non-constant polynomials that can divide evenly into the given polynomial.

2. How do you show that a polynomial is irreducible?

To show that a polynomial is irreducible, one method is to use the Rational Root Theorem to check for any possible rational roots. If there are no rational roots, then the polynomial is irreducible. Another method is to use the Eisenstein's Criterion, which states that if a polynomial has a prime number that divides all but the leading coefficient and the constant term, then the polynomial is irreducible.

3. What does it mean for a polynomial to have a degree of 2',1',2'?

A polynomial's degree is the highest power of the variable in the polynomial. In this case, the polynomial has a degree of 2, meaning the highest power of the variable is 2. The 1' and 2' represent the coefficients of the polynomial.

4. Can an irreducible polynomial have multiple degrees?

No, an irreducible polynomial can only have one degree. This is because the definition of an irreducible polynomial is that it cannot be factored into polynomials of lower degree. If it had multiple degrees, it could be factored into polynomials of those lower degrees.

5. How is the irreducibility of a polynomial important in mathematics?

The concept of irreducible polynomials is important in many areas of mathematics, such as algebra, number theory, and cryptography. It allows for the simplification of complex expressions and equations, and also has applications in determining the solvability of certain equations. In cryptography, irreducible polynomials are used to generate finite fields, which are crucial in many encryption algorithms.

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