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 $\mathbb{Z}_3$, $2' = -1'$ and $(-1')^n$ is $1'$ if $n$ is even and $-1'$ if $n$ is odd. Therefore $(2')^3 + (2')^2 = 0'$.

 

[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!