Let $R$ be a ring with $1$, which contains $\mathbb{Q}$, and generated over $\mathbb{Q}$ by two elements $x,y$ such that $yx-xy=1$. Show that $R$ is simple.
What i did? Certainly $x, y \in R$ as $1\in R$. Consider $0\neq I$ to be a two-sided ideal of $R$. If one among $x,y\in I$, say $x\in I$ then $yx\in I$ and $xy\in I$ so $1=yx-xy\in I$ thus $I=R$.
If $x,y\notin I$ then for any $0\neq r\in I$ and $\frac{yx}{r}, \frac{xy}{r} \in R$ we have $1 = \frac{yx}{r} r - \frac{xy}{r} r\in I$ then $I=R$ as $1\in I$.
Can i use the rationals in this way? Please suggest me. Thank you.
By re-denoting $y$ as $\frac{d}{dx}$ you can see that your ring is nothing else but polynomial differential operators on the affine line $\mathbb{A}^1_{\mathbb{Q}}$, where $x$ act by multiplication by $x$, and $\frac{d}{dx}$ acts by differentiation. It is easy to see that any differential operator can be put into the form $p_n(x) \frac{d^n}{dx^n}+(\text{lower order terms})$ using the commutator relation you have. Then you can prove by induction the following. First, you show that if there is a polynomial $p(x)$ in your ideal $I$, then $I$ must contain $1$ (just commute this polynomial with $\frac{d}{dx}$ to obtain $p'(x)$ inside $I$). By the same easy argument you can show that if $\frac{d^n}{dx^n}\in I$ then $1\in I$. From these two steps you can conclude the statement for any operator $D\neq 0\in I$.