For this question, How do I show that the only 2 sided ideals are 0 and R? I understand the theory behind ideals but I have a very theoretical understanding and I was hoping someone could help solve this question?
From my understanding, $ P(x) = C0 + C1X + ... + CkX^k $ and I said $ Q[x] $ could be polynomails in the variable X with coefficinets in Q.
For $ b) $ I know I need to find a P(x) where $\begin{bmatrix} P(0) & 0 \\ P'(0)& P(0) \\ \end{bmatrix}$
is equal to $\begin{bmatrix} a & 0 \\ b & a \\ \end{bmatrix}$
For $c)$ I need a P(x)=x. p(0)=0 and P'(x)=1. How would I go about doing this, Ive been working on it for a few days and thats all I have so far so any help would be apprecaited!
Hint to 2: If $F$ is a surjective ring homomorphism, the kernel $I$ of $F$ is an ideal of ${\Bbb Q}[x]$ such that ${\Bbb Q}[x]/I$ is isomorphic to $S$. (Homomorphism theorem)
Hint to 3: From the quotient structure of $S$ it is clear that $S$ is not an integral domain. Moreover, the ideals of $S$ correspond exactly to the ideals of ${\Bbb Q}[x]$ containing $I=\langle x^2\rangle$. (First isomorphism theorem)