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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

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    $\begingroup$ How is $S$ an ideal in your example? $\endgroup$ – Arnaud Mortier Aug 17 at 13:55
  • $\begingroup$ @ArnaudMortier it is a subgroup of R $\endgroup$ – jasmine Aug 17 at 13:57
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    $\begingroup$ Sure, but an ideal is more than just a subgroup, right? $\endgroup$ – Arnaud Mortier Aug 17 at 13:57
  • $\begingroup$ ya @ArnaudMortier we can take as subgroup $\endgroup$ – jasmine Aug 17 at 13:58
  • $\begingroup$ each ideal S of a ring R is a subgroup $\endgroup$ – jasmine Aug 17 at 14:00
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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 $S+T$.

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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 $r\in R$
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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 $R$.

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