Ideal $I=(x^3,x^5)$ in $\mathbb Q[x]$ . It is true that $I=(x^3)$ ?What about $I=(x^5)$?

41 Views Asked by At

Ideal $~I=(x^3,x^5)~$ in $~\mathbb Q[x]~$ . It is true that $~I=(x^3)~$ ?What about $~I=(x^5)~$ ?

My work : I was thinking to solve this with something like:

$~(x^3)(1+x^2)=x^3+x^5~$ and $~(1+x^2)\in \mathbb Q[x]~$

and for $~I=(x^5)\implies (x^5)*(1+1/x^2)=x^3+x^5~$ but $~1/x^2~$ is not always in $~\mathbb Q[x]~$, I think.

How close to the answer am I ?

Thanks for help.

1

There are 1 best solutions below

2
On BEST ANSWER

The polynomial ring $K[x]$, where $K$ is a field, is a principal ideal domain. Each nonzero ideal is generated by the small monic polynomial containing it (smallest in the sense of degree). The reason is that in $K[x]$ we have division with remainder.

In your case, $I=\langle x^3,x^5\rangle = \langle x^3\rangle$ with $x^5 = x^2\cdot x^3$.

Moreover, the ideal $\langle x^5 \rangle$ is contained in $\langle x^3\rangle$.