Is $\mathbb{R}[x^2,x^3]=\{f=\sum_{i=0}^n a_ix^i\in\mathbb{R}[x]:a_1=0\}$ is Euclidean Domain?

Question

Is $\mathbb{R}[x^2,x^3]=\{f=\sum_{i=0}^n a_ix^i\in\mathbb{R}[x]:a_1=0\}$ an Euclidean Domain ?

My answer : I know that it is integral domain ,by theorem R is integral domain then $\Bbb R[x]$ is integral domain.

Now I am confused that here how can I claimed that it is Euclidean domain or not ?

Any hints...

Please help me.

It is not even a UFD : $x^6 = x^3 \cdot x^3 = x^2 \cdot x^2 \cdot x^2$, and the factors are irreducible elements. — Watson, May 25 '18 at 19:52
@LordSharktheUnknownit is $ 0+ a_1x + a_2 x^2+ a_3x^3 + a_4x^4 +a_5x^5 +a_6x^6 $ ...by this i can say No, — , May 25 '18 at 19:54

Watson · Accepted Answer · 2018-05-25T20:02:20.223

1

As you may know, any euclidean domain is a PID, and any PID is a UFD. But your ring $R$ is not even a UFD : $x^6 = x^3 \cdot x^3 = x^2 \cdot x^2 \cdot x^2$, and the factors can be checked to be irreducible elements.

Your ring is isomorphic to $\Bbb R[X,Y] / (Y^2-X^3)$ (see here), so you may find this related question or this one. As mentioned there, $R$ is not even integrally closed, for $a := x^3 / x^2 \in \mathrm{Frac}(R)$ is integral over $R$ (since $a^2 - x^2 = 0$), but $a \not\in R$.

edited May 25 '18 at 20:02

answered May 25 '18 at 19:56

Watson

23,793

thanks.. @watson – May 25 '18 at 19:58

Is $\mathbb{R}[x^2,x^3]=\{f=\sum_{i=0}^n a_ix^i\in\mathbb{R}[x]:a_1=0\}$ is Euclidean Domain?

1 Answers1