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

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $S+T$.


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 $a+b\in I+J$.


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 $r\in R$

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 $R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.