Yes/ No : The sum of two ideals of a ring R is an ideal of Recu
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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1$\\begingroup$ How is $S$ an ideal in your example? $\\endgroup$ – Arnaud Mortier 9 hours ago
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$\\begingroup$ @ArnaudMortier it is a subgroup of R $\\endgroup$ – jasmine 9 hours ago
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1$\\begingroup$ Sure, but an ideal is more than just a subgroup, right? $\\endgroup$ – Arnaud Mortier 9 hours ago
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$\\begingroup$ ya @ArnaudMortier we can take as subgroup $\\endgroup$ – jasmine 9 hours ago
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$\\begingroup$ each ideal S of a ring R is a subgroup $\\endgroup$ – jasmine 9 hours ago
3 Answers
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 $S+T$.
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 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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$\\begingroup$ Your second bullet point is a special case of the third, so it does not need to be checked separately. $\\endgroup$ – Misha Lavrov 56 mins ago
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$\\begingroup$ That’s true, thank you @MishaLavrov $\\endgroup$ – cansomeonehelpmeout 55 mins ago