Question

Prove that the only units in are 1 and -1. Theorem 4.2

If , show that its only associates are and.

Short answer

It is proved that the only units in${\mathbb{Q}}_{\mathbb{Z}}\left[x\right]$ are 1 and -1.

It is shown that the only associates of $\mathrm{f}\left(\mathrm{x}\right)\in {\mathrm{\mathbb{Q}}}_{\mathrm{\mathbb{Z}}}\left[\mathrm{x}\right]$are $f\left(x\right)$and $-f\left(x\right)$.

Step-by-step solution

Referring to Theorem 4.2

Theorem 4.2

If R is an integral domain, and $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, $\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$are non-zero polynomials in$\mathit{R}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$ , then$\mathit{d}\mathit{e}\mathit{g}\mathbf{\left[}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]}\mathbf{=}\mathit{d}\mathit{e}\mathit{g}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{d}\mathit{e}\mathit{g}\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

Proving that the only units in  ℚℤ[x] are 1 and -1

To prove that the only unit in ${\mathbb{Q}}_{\mathbb{Z}}\left[x\right]$are 1 and -1 firstly, we need to prove that 1 and -1 are units of ${\mathbb{Q}}_{\mathbb{Z}}\left[x\right]$.

Since $1\cdot 1=1$and$(-1).(-1)=1$ , 1 and -1 both are invertible.

Hence, 1 and -1 are units of${\mathbb{Q}}_{\mathbb{Z}}\left[x\right]$ .

Now, we must prove that they are the only unit.

Suppose$f\left(x\right)\in {\mathbb{Q}}_{\mathbb{Z}}\left[x\right]$ is a unit, and$g\left(x\right)$ is its inverse.

Then,$f\left(x\right)g\left(x\right)=1$ .

So, according to Theorem 4.2, find $\mathrm{deg}\left[f\left(x\right)g\left(x\right)\right]$as:

$\begin{array}{c}\mathrm{deg}\left[f\left(x\right)g\left(x\right)\right]=\mathrm{deg}f\left(x\right)+\mathrm{deg}g\left(x\right)\\ =\mathrm{deg}\left(1\right)\\ =0\end{array}$

Therefore, degree of both$f\left(x\right)$ and $g\left(x\right)$must be zero.

Hence, both polynomials must be constant.

As we know that only constants in ${\mathbb{Q}}_{\mathbb{Z}}\left[x\right]$are 1 and -1 hence, it is proved that the only unit in ${\mathbb{Q}}_{\mathbb{Z}}\left[x\right]$are 1 and -1.

Showing that the only associates of  f(x) ∈ ℚℤ[x] are role="math" localid="1659538977726" f(x)  and -f(x)

The associates of a ring element are obtained by multiplying it with a unit.

As we have already proved in part (a) that the only units of${\mathbb{Q}}_{\mathbb{Z}}\left[x\right]$ are 1 and -1.

Therefore, the required associates can be written as:

$\begin{array}{c}\mathrm{f}\left(\mathrm{x}\right)\cdot 1=\mathrm{f}\left(\mathrm{x}\right)\\ \mathrm{f}\left(\mathrm{x}\right)\cdot (-1)=-\mathrm{f}\left(\mathrm{x}\right)\end{array}$

Hence, it is shown that the only associates of$f\left(x\right)\in {\mathbb{Q}}_{\mathbb{Z}}\left[x\right]$ are$f\left(x\right)$ and $-f\left(x\right)$.