# 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

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

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 $$r\in R$$

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 $$R$$.