Prove that a module is projective or not 
  
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*Let $R=\left(\begin{array}{cc}\mathbb{Q}&\mathbb{Q}\\0&\mathbb{Q}\end{array}\right), J=\left(\begin{array}{cc}0&\mathbb{Q}\\0&0\end{array}\right)$. Prove that $R/J$ is not a projective $R$-module.
  

I'm really misunderstand about $R/J$, about it's equivalent class. I can understand $\mathbb{Z}/ 2\mathbb{Z}=\mathbb{Z}_2$, but in general I can't imagine $A/B$.

  
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*Let $R=\left(\begin{array}{cc}\mathbb{R}&\mathbb{R}\\\mathbb{R}&\mathbb{R}\end{array}\right), M=\left(\begin{array}{cc}\mathbb{R}&\mathbb{R}\\0&0\end{array}\right)$. Prove that $M$ is a right projective module $R$-module but not free.
  

Prove that with universal property really hard with me because of maps (homomorphisms) constructive.
Thanks for regarding!
 A: Hints:
1) One characterization of projective modules you could use is that in the exact sequence
$$
0\to A\to B\to P\to 0
$$
with $P$ projective, the sequence splits. That is, $B\cong A\oplus P$ as $R$ modules.
In your case: $R\cong J\oplus (R/J)$. But for $J$ to be a direct summand of $R$, it would have to contain a nonzero idempotent... does it?
2) Here, $R$ is a four dimensional $\Bbb R$-algebra. $M$ is only a two dimensional $\Bbb R$ vector space. Could $M$ be the direct sum of copies of $R$? On the other hand, it is relatively easy to see another module $N$ such that $M\oplus N=R$, showing that $M$ is projective. In case you don't have this fact yet, it would also be a good thing to prove: a module is projective iff it is a direct summand of a free module. This follows immediately from the lemma I gave in the hint for 1).

In general, any quotient $A/B$ can be thought of as "like $A$, except anything that I see from $B$ is considered zero." For example, $\begin{bmatrix} a&b\\0&c\end{bmatrix}=\begin{bmatrix} a&0\\0&c\end{bmatrix}+\begin{bmatrix} 0&b\\0&0\end{bmatrix}=\begin{bmatrix} a&0\\0&c\end{bmatrix}$ mod $J$.
So in your case, $R/J\cong \begin{bmatrix}\Bbb Q&0\\0&\Bbb Q\end{bmatrix}\cong \Bbb Q\oplus \Bbb Q$.

Actually, it is true in general for rings with identity that the Jacobson radical $J$ (which is what $J$ is in the case above) is never a right or left summand of $R$ (unless it's zero.) The reason is that the Jacobson radical is a superfluous submodule of $R_R$ and $_RR$. By definition, that is that $B+J=R$ implies that $B=R$.
