Is the ideal $I = \langle x^2,x^3 \rangle $ in $\mathbb{R}[X]$ equal with $ \langle x^3 \rangle $? Are the two ideals in the title equal ? 

What I think:

I'm not sure if it's true but I think the two ideals are equal, considering the elements in the ideal $I = \langle x^3 \rangle \{ \sum_{i=1} ^{n} f(x) \cdot x^3 \cdot h(x) | f(x),h(x) \in \mathbb{R}[X] \}$ if we take $f(x) = \frac{1}{x^3}$ and $h(x) = x^2$ we can get $x^2 = f(x) \cdot x^3 \cdot h(x)$ so that implies that $x^2 \in \langle x^3 \rangle$ and $ \langle x^2,x^3 \rangle =  \langle x^3 \rangle  $. I don't know if I can chose $f(x) = \frac 1 {x^3}$ since x can be 0.
Indeed, you cannot take $1/x^2$ as this is not a polynomial. Yet that $x$ can be $0$ is not the problem. You could not take, say, $1/(1+x^2)$ either. The elements of $\mathbb{R}[X]$ are the polynomials, that is expressions of the form $a_0 + a_1x + a_2 x^2 + \dots a_nx^n$.  

If you restrict to polynomials correctly you can show that $x^2$ cannot be in the ideal generated by $x^3$. (For example consider the rank.) Thus the two ideals are not equal. 

Would you allow expressions such as $1/x^2$ as elements of your ring, that is you consider rational functions, then indeed the ideals would be equal. Yet in the ring of rational functions $\mathbb{R}(X)$ &ndash; note the parenthesis instead of brackets &ndash; almost all ideals are equal as it is a field and the only ideals are $0$ and the ring itself.
Indeed, you cannot take $1/x^2$ as this is not a polynomial. Yet that $x$ can be $0$ is not the problem. You could not take, say, $1/(1+x^2)$ either. The elements of $\mathbb{R}[X]$ are the polynomials, that is expressions of the form $a_0 + a_1x + a_2 x^2 + \dots a_nx^n$.  

If you restrict to polynomials correctly you can show that $x^2$ cannot be in the ideal generated by $x^3$. (For example consider the rank.) Thus the two ideals are not equal. 

Would you allow expressions such as $1/x^2$ as elements of your ring, that is you consider rational functions, then indeed the ideals would be equal. Yet in the ring of rational functions $\mathbb{R}(X)$ &ndash; note the parenthesis instead of brackets &ndash; almost all ideals are equal as it is a field and the only ideals are $0$ and the ring itself.