Yes/ No : The sum of two ideals of a ring R is an ideal of R Is the  following statement is true/false ?
The sum of two  ideals of a ring  $R $  is an  ideal  of $ R$
My attempt : I thinks this statement is false .
Take  $S=\{ \begin{pmatrix} a& 0  \\ b & 0 \end{pmatrix} : a, b \in  \mathbb{Z} \}$ , 
$T=\{\begin{pmatrix} 0& c  \\ 0 & 0 \end{pmatrix} ,c \in \mathbb{Z}\}$
Here $S +T =\{\begin{pmatrix} a& c  \\ b & 0 \end{pmatrix} , a, b, c  \in \mathbb{Z}\}$
Now  take $P =\begin{pmatrix} 1& 1  \\ 1 & 0 \end{pmatrix}$ ,$Q=\begin{pmatrix} 2& 2  \\ 2 & 0 \end{pmatrix} \in  S+  T $ 
But $PQ  \notin S + T$ ,that is its  contradicts  
so this statement is false
 A: Ideal $I\subseteq R$ must be a subgroup that also has the property that $\forall r\in R,\;\forall i\in I,\;r\cdot i,\:i\cdot r\in I$. Clearly, if $I,J\subseteq R$ are ideals, then $I+J$ is a subgroup of $R$. 
Let $r\in R$, and let $i+j\in I+J$. 
$r\cdot(i+j)=r\cdot i+r\cdot j$, where $r\cdot i\in I,\;r\cdot j\in J$.
Likewise, $(i+j)\cdot r = i\cdot r+j\cdot r$, where $i\cdot r\in I,\;j\cdot r\in J$.
Hence, $I+J$ is an ideal of $R$.
A: Your counterexample does not work, since $S$ is not an ideal. To see this, consider $$\begin{pmatrix}a&0\\b&0\end{pmatrix}\begin{pmatrix}x&y\\z&w\end{pmatrix}=\begin{pmatrix}ax&ay\\bx&by\end{pmatrix}\not\in S$$

If $I,J\subseteq R$ are two ideals in the ring $R$, then $I+J$ is defined as $$I+J=\{i+j\mid i\in I, j\in J\}$$
To show that $I+J\subseteq R$ is an ideal you need to show:


*

*If $a,b\in I+J$, then $a+b\in I+J$.


Hint:

 How can we represent $a$, if we know that $a\in I+J$? Use the fact that both $I,J$ are ideals.



*

*If $c\in I+J$, then $rc\in I+J$ for any $r\in R$
A: In your example, $S$ is indeed a left ideal, but $T$ is not:
$$\begin{pmatrix}
0&1\\1&0
\end{pmatrix}\begin{pmatrix}
0&c\\0&0
\end{pmatrix}=\begin{pmatrix}
0&0\\0&c
\end{pmatrix}$$
so your ‘counterexample’ is not valid.
Actually, it is true that the sum of two left ideals is a left ideal: $D+T$ is indeed an (additive) subgroup of $R$, ans, for left multiplication
$$a(s+t)=\underbrace{as}_{\in S}+\underbrace{at}_{\in T}$$
is indeed in $S+T$.
