# checking answer on units of integer

• Oct 26th 2009, 04:05 PM
knguyen2005
checking answer on units of integer
U(Z_4) ={ 1, 3}. Z_4 is integers mod n
In general, U(Z_n) = { a in Z_n: hcf(a,n) =1}
And
U(Z_4[X]) = { 0, 1, 2 ,3}. Is this correct?
where Z_n[X] is polynomial ring over Z generated by one variable
How do you generate the formula of U(Z_n[X])?

Thank you
• Oct 26th 2009, 04:16 PM
tonio
Quote:

Originally Posted by knguyen2005
U(Z_4) ={ 1, 3}. Z_4 is integers mod n
In general, U(Z_n) = { a in Z_n: hcf(a,n) =1}
And
U(Z_4[X]) = { 0, 1, 2 ,3}. Is this correct?

Nop. In general, $U(R[x])=U(R)$ , when R is a ring, so in this case the units of the polynomial ring are the same as the ring's.

Tonio

where Z_n[X] is polynomial ring over Z generated by one variable
How do you generate the formula of U(Z_n[X])?

Thank you

.
• Oct 27th 2009, 01:24 AM
knguyen2005
Thank you Tonio, but I think you misunderstood my question
I know in general http://www.mathhelpforum.com/math-he...0333e387-1.gif but what I meant was how do you write down the elements of the set
Example: U(Z_4)={1, 3} since these elements of the set are coprime to 4
Ok, how about the polynomial of the ring over Z_4. Can we find the elements in U(Z_4[x]) ?
is it U(Z_4[X]) = { 0, 1, 2 ,3} or U(Z_4[X]) = { 1, 2 ,3} ?
• Oct 27th 2009, 02:14 AM
NonCommAlg
if $R$ is a commutative ring with identity element, then $U(R[x])=U(R)$ holds only for reduced rings, i.e. rings with no non-zero nilpotent elements.

in general $f(x)=a_0 + a_1x + \cdots + a_n x^n \in U(R[x])$ if and only if $a_0 \in U(R)$ and $a_j$ is nilpotent for all $j \geq 1.$ see my post in here: http://www.mathhelpforum.com/math-he...ial-rings.html

for example if $R=\mathbb{Z}/4\mathbb{Z},$ then $f(x)=a_0+a_1x + \cdots + a_nx^n \in U(R[x])$ if and only if $a_0 \in \{\bar{1}, \bar{3} \}$ and $a_j \in \{\bar{0}, \bar{2} \}, \ \forall j \geq 1.$