Question

Show that there are infinitely many distinct congruence classes modulo x²- 2 in $\mathbf{\mathbb{Q}}\left[x\right]$. Describe them.

Short answer

It is proved that there are many distinct congruence classesx² - 2 in $\mathrm{\mathbb{Q}}\left[x\right]$.

Step-by-step solution

Corollary statement

Assume that p (x) is a non-zero polynomial of degree n in and also assume that congruence modulo p (x) .

1. When f(x) is divided by p (x) , we have $\left[f\left[x\right]\right]=\left[r\left(x\right)\right]$, where r(x) is the remainder and $f\left(x\right)\in F\left[x\right]$.
2. Let S be the set of all polynomials whose degrees are less then the degree of p (x), then each congruence class modulo p (x) is also the class of some polynomial in S, where each class is distinct.

Proof part

The given expression is x² - 2 in $\mathrm{\mathbb{Q}}\left[x\right]$.

As per the corollary stated in step 1, each polynomial $f\left(x\right)\in F\left[x\right]$ is congruent modulo to the remainder r(x) after the division by x² - 2 and the polynomials with different remainders which belong to different equivalent classes.

For $a_{1}x+a_0\in \mathrm{\mathbb{Q}}\left[x\right]$, there are infinitely many polynomials of degree at most 1.

As the degree of the polynomial $a_{1}x+a_0\in \mathrm{\mathbb{Q}}\left[x\right]$is less than the degree of x² - 2 , where the distinct polynomials of degree 1 are in distinct conjugacy classes.

Let, $p\left(x\right)=x^2-\mathbf{2}$, then by division algorithm, we have, $f\left(x\right)=q\left(x\right)\left(x^2-\mathbf{2}\right)+r\left(x\right)$having the remainder as 0, or we can say that,$\mathrm{deg}r\left(x\right)<\text{deg}\left(x^2-\mathbf{2}\right)=2$$\mathrm{deg}r\left(x\right)<\text{deg}\left(x^2-\mathbf{2}\right)=2$

Hence, every polynomial is equivalent modulo x² - 2 to one of degree at most 1.