# Irreducible

#### Poirot

##### Banned
Consider the polynomial f=x^2+1. Regarded as a polynomial over which of the following coefficient rings is this polynomial irreducible?

(i) Z, (ii) R, (iii) C, (iv) Z mod 2. Give Reasons for your answer.

To be irreducible means (i) not a unit and (ii) if f=ab then a|f and b|f. As to the first, I don't think f is a unit in any of these.

Thanks

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Consider the polynomial f=x^2+1. Regarded as a polynomial over which of the following coefficient rings is this polynomial irreducible?

(i) Z, (ii) R, (iii) C, (iv) Z mod 2. Give Reasons for your answer.

To be irreducible means (i) not a unit and (ii) if f=ab then a|f and b|f. As to the first, I don't think f is a unit in any of these.

Thanks
I don't think the way you have defined an irreducible polynomial is correct.
Condition (ii) is wrong.

edit: In fact, condition (ii) contains no information in it.

#### hmmm16

##### Member
Should your definition of irreducible not be as follows:

For an integral domain $\mathcal{R}$,if $a\in\mathcal{R}$ and a is neither zero nor a unit;
we say that a is irreducible in $\mathcal{R}$ if whenever $a = cd$ for $c,d\in\mathcal{R}$ it follows that either c or d is a unit.

#### chisigma

##### Well-known member
A polynomial $p(x)$ is said to be irreducible if don't exist two polynomials $a(x)$ and $b(x)$, both different from $p(x)$, so that $p(x)=a(x)\ b(x)$. In our case is $p(x)=1+x^{2}$ and we have...

(i) if the coefficients of $p(x)$ are in $\mathbb{Z}$, then $p(x)$ is irreducible...

(ii) if the coefficients of $p(x)$ are in $\mathbb{R}$, then $p(x)$ is irreducible...

(iii) if the coefficients of $p(x)$ are in $\mathbb{C}$, then $p(x)$ is not irreducible because is $1+x^{2}= (1-i\ x)\ (1+i\ x)$...

(iv) if the coefficients of $p(x)$ are in $\mathbb{Z}\ \text{mod}\ 2$, then $p(x)$ is not irreducible because is $1+x^{2}= (1+x)\ (1+x)$...

Kind regards

$\chi$ $\sigma$

#### Poirot

##### Banned
I just got it myeslf but thanks anyway.