Archive for the ‘Quotient Rings’ Category

Here you can see part (1). In the following, $Z \langle x,y \rangle$ is the algebra generated by non-commuting variables $x,y$ and with coefficients in $Z.$ So an element of  $Z \langle x,y \rangle$ is a finite sum of terms in the form $\alpha x_1x_2 \ldots x_n,$ where $n$ is any positive integer, $\alpha \in Z$ and each $x_i$ is either $1$ or $x$ or $y.$ For example $f = \alpha_0 + \alpha_1 xyx + \alpha_2 yx^3y^2$ where $\alpha_i \in Z.$

Lemma. Let $Z$ be the center of a domain $R.$ If $R$ is not left (resp. right) Ore, then $R$ contains a copy of $Z \langle x,y \rangle.$

Proof. Choose $0 \neq r_1,r_2 \in R$ such that $Rr_1 \cap Rr_2 = (0).$ Then $r_1,r_2$ are left linearly independent over $R$ and we’re done by Jategaonkar’s lemma. $\Box$

Corollary. Let $A$ be an algebra over a field. If $A$ is PI or has a finite GK-dimension, then $A$ is both left and right Ore.

Proof. Otherwise, by the lemma, $A$ would contain a free algebra and we know that such algebras are neither PI nor have a finite GK dimension. $\Box$

A left Ore domain which is not right Ore. Let $D$ be a division ring and $\sigma : D \longrightarrow D$ be a homomorphism which is not onto.  Since $\sigma (1_D)=1_D \neq 0,$ we have $\ker \sigma \neq D$ and thus $\ker \sigma = (0),$ because $D$ has no non-zero proper ideal. Therefore $\sigma$ is injective and hence the left skew polynomial ring $R=D[x,\sigma]$ is a domain.

Claim 1. $R$ is not right Ore.

Proof. Suppose $R$ is right Ore. Since we’ve assumed that $\sigma$ is not onto, there exists $d \in D - \sigma (D).$ Now we must have $xR \cap dxR \neq (0).$ So, there exist non-zero elements $g, h \in R$ such that $xg=dxh.$ Clearly $g, h$ must have the same degree, say $n.$ Let $d_1x^n, \ d_2x^n$ be the leading terms of $g, h$ respectively. So the leading terms of $xg , dxh,$ which have to be equal, are  $\sigma (d_1)x^{n+1}, \ d \sigma (d_2) x^{n+1}$ respectively. Equating these leading terms gives us $d=\sigma (d_1 d_2^{-1}) \in \sigma (D),$ which is a contradiction. $\Box$

Claim 2. Euclidean algorithm holds in $R.$

Proof. That means we need to prove for every $f, g \in R,$ there exists $r, s \in R$ such that $f=sg + r,$ where either $r = 0$ or $\deg r < \deg g.$ The proof is by induction over $\deg f$: if $\deg f < \deg g,$ then we may choose $s = 0$ and $r = f.$ Let $d_1 x^n, \ d_2 x^m, \ n \geq m,$ be the leading terms of $f, g$ respectively. Then $f$ and $d_1x^{n-m}d_2^{-1}g$ have the same leading terms and thus $h = f - d_1x^{n-m}d_2^{-1}g$ has a degree $< n.$ Thus, by induction, there exist $r_1, s_1 \in R$ such that $h = s_1g + r_1,$ where either $r_1 = 0$ or $\deg r_1 < \deg g.$ Finally we have $h=s_1g + r_1 = f - d_1 x^{n-m}d_2^{-1}g$ and so $f = (s_1 + d_1x^{n-m}d_2^{-1})g + r_1. \ \Box$

Claim 3. $R$ is left Ore.

Proof. Let $I$ be a non-zero left ideal of $R$ and choose a non-zero $g \in I$ with the smallest degree. For every $f \in I,$ by Claim 2, there exist $r, s \in R$ such that $f = sg + r,$ where either $r = 0$ or $\deg r < \deg g.$ But $r = f - sg \in I$ and thus we cannot have $\deg r < \deg g.$ Therefore $r=0$ and $f \in Rg,$ i.e. $I=Rg.$ So every left ideal of $R$ is principal and, as a result, $R$ is left Noetherian and therefore left Ore. $\Box$

Ore domains (1)

Posted: January 3, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
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Definition. Let $R$ be a domain and $S=R \setminus \{0\}.$ If $S$ is left (resp. right) Ore, then $R$ is called a left (resp. right) Ore domain.

Remark 1. Note that a domain $R$ is a left Ore domain if and only if $Rr_1 \cap Rr_2 \neq (0),$ for all non-zero elements $r_1,r_2 \in R.$ Similarly $R$ is a right Ore domain if and only if $r_1R \cap r_2R \neq (0),$ for all non-zero elements $r_1,r_2 \in R.$

Remark 2. The left (resp. right) quotient ring of a left (resp. right) Ore domain is a division ring. The reason is that every element of $Q(R)$ is in the form $x = s^{-1}r,$ where $0 \neq s \in R, \ r \in R.$ Now if $x \neq 0,$ then $r \neq 0$ and then $x^{-1}=r^{-1}s.$

Theorem. Every left (resp. right) Noetherian domain is left (resp. right) Ore.

Proof. We will prove the theorem for left Noetherian domains only. The “right” version, as usual, has the same argument. Let $r_1, r_2 \in R$ be non-zero. We need to show that $Rr_1 \cap Rr_2 \neq (0).$ So suppose, on the contrary, that $Rr_1 \cap Rr_2 = (0).$ We will show that the sum $\sum_{n=0}^{\infty} R r_1r_2^n$ is direct and thus $R$ cannot be left Noetherian because then we would have the non-stopping increasing chain of left ideals

$Rr_1 \subset Rr_1 \oplus Rr_1r_2 \subset Rr_1 \oplus Rr_1r_2 \oplus Rr_1r_2^2 \subset \cdots.$

To prove that the sum is direct, suppose that the sum is not direct and choose $n$ to be the smallest postive integer for which there exist $a_j \in R, \ j=0, \cdots , n,$ not all zero, such that $\sum_{j=0}^n a_jr_1r_2^j = 0.$ Clearly $n \geq 1.$ Thus

$-a_0r_1=\left (\sum_{j=0}^{n-1} a_{j+1}r_1r_2^{j} \right)r_2 \in Rr_1 \cap Rr_2 = (0).$

Therefore, since $r_1, r_2$ are non-zero and $R$ is a domain, we must have $a_0=0$ and $\sum_{j=0}^{n-1} a_{j+1}r_1r_2^j = 0,$ contradicting the minimality of $n. \ \Box$

Quotient rings; some facts (2)

Posted: January 2, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
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For the first part see here.

6) If $R$ is simple, then $Z(R)=Z(Q).$

Proof. Let $x=s^{-1}a \in Z(Q).$ Then from $xs=sx$ we get $sa=as$ and thus $s^{-1}a=as^{-1}.$ Hence for every $b \in R$ we’ll have $s^{-1}ab=bs^{-1}a=bas^{-1},$ which gives us $abs=sba.$ Also, since $R$ is simple, $RsR=R,$ which means $\sum_{i=1}^n b_isc_i = 1,$ for some $b_i, \ c_i \in R.$ Thus $\sum_{i=1}^n sb_iac_i = \sum_{i=1}^n ab_isc_i = a=sx$ and so $x=\sum_{i=1}^n b_iac_i \in R.$ Therefore $x \in Z(R)$ which proves $Z(Q) \subseteq Z(R).$ Conversely, let $b \in Z(R)$ and $x=s^{-1}a \in Q$. Since $bs=sb,$ we have $s^{-1}b=bs^{-1}$ and thus $bx=bs^{-1}a=s^{-1}ba=s^{-1}ab=xb$ and so $b \in Z(Q). \Box$

7) The left uniform dimension of $R$ and $Q$ are equal.

Proof. We saw in the previous section that the left ideals of $Q$ are exactly in the form $QI,$ where $I$ is a left ideal of $R.$ Clearly $\sum QI_i$ is direct iff $\sum I_i$ is direct.

8) Let $N$ be a nilpotent ideal of $R$ and let $I$ be the right annihilator of $N$ in $R.$ Then $I$ is an essential left ideal of $R$ and hence $QI$ is an essential left ideal of $Q.$

Proof. Let $I$ be the right annihilator of $N$ in $R.$ For an essential left ideal $J$ of $R$ the left ideal $QJ$ of $Q$ is essential in $Q$ because for every non-zero left ideal $K$ of $R : (0) \neq \ Q(J \cap K) \subseteq QJ \cap QK.$ So we only need to prove the first part of the claim. Let $J$ be any non-zero left ideal of $R$ and put $n=\min \{k \geq 0 : \ N^k J \neq (0) \}.$ Then $(0) \neq N^n J \subseteq I \cap J.$

9) If $Q$ is semisimple, then $R$ is semiprime.

Proof. So we need to prove that $R$ has no non-zero nilpotent ideal. Suppose that $N$ is a nilpotent ideal of $R$ and let $I$ be the right annihilator of $N$ in $R.$ Since $Q$ is semisimple, $QI \oplus A = Q,$ for some left ideal $A$ of $Q.$ But, from the previous fact, we know that $QI$ is essential in $Q$ and thus $A=(0),$ i.e. $QI=Q.$ Thus $s^{-1}a=1,$ for some $a \in I=\text{r.ann}_R N.$ So $s=a$ and $Ns=Na=(0).$ Thus $N=Nss^{-1}=(0).$

We proved, in the previous section, that if $R$ is prime, then $Q$ is prime too.

10) If $Q$ is simple, then $R$ is prime.

Proof. Let $I,J$ be two non-zero ideals of $R.$ We need to show that $IJ \neq (0).$ We have $QIQ=Q,$ because $I \neq (0)$ and $Q$ is simple. Therefore $1=\sum_{i=1}^n x_ia_iy_i,$ for some $x_i,y_i \in Q$ and $a_i \in I.$ We can write $x_i = s^{-1}b_i,$ for some $b_i \in R.$ Then $s=\sum_{i=1}^n b_ia_iy_i \in IQ.$ So $IQ$ is a right ideal of $Q$ which contains a unit. Thus $IQ=Q.$ Similarly $JQ=Q$ and hence $IJQ=Q.$ As a result, $IJ \neq (0). \Box$

Quotient rings; some facts (1)

Posted: January 1, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
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Throughout $R$ is a ring with unity, $S \subset R$ is a regular submonoid which is left Ore. So $Q,$ the left quotient ring of $R$ with respect to $S,$ exists and $R$ is a subring of $Q.$ The following facts also holds for the “right” version:

1) If $I \lhd_{\ell} Q$, then $I \cap R \lhd_{\ell} R$ and if $J \lhd_{\ell} Q,$ then $I \cap R = J \cap R \Longleftrightarrow I=J.$ Also $I=Q(I \cap R).$

Proof. The first part is trivial. Now suppose that $s^{-1}a=x \in I.$ Then $a = sx \in I \cap R = J \cap R$ and so $x=s^{-1}a \in J.$ Hence $I \subseteq J.$ Similarly $J \subseteq I.$ For the last part, if $x = s^{-1}a \in I,$ then $a=sx \in I \cap R$ and so $x=s^{-1}a \in Q(I \cap R).$

2) If $I \lhd_{\ell} R,$ then $QI \lhd_{\ell} Q$ and $QI=\{s^{-1}a : \ s \in S, \ a \in I \}.$ It is also true that $QI \cap R = \{a \in R : \ sa \in I \ \text{for some} \ s \in S \}.$

Proof.  Again, the first part is trivial. To prove the non-trivial side of the second part, let $x = \sum_{i=1}^n x_i a_i \in QI,$ where $x_i \in Q$ and $a_i \in I.$ As we’ve showed before, there exist some $s \in S$ and $b_i \in R$ such that $x_i = s^{-1}b_i$ and thus $x = s^{-1}a,$ where $a=\sum_{i=1}^n b_ia_i \in I.$ For the last part, suppose that $s^{-1}b \in QI \cap R,$ where $s \in S, \ b \in I,$ by the second part. So we have $s^{-1}b=a \in R$ and thus $sa = b \in I.$

3) For any $I, J \lhd_{\ell} R$ we have $I \cap J = (0) \Longrightarrow QI \cap QJ = (0).$

Proof. Let $x=s^{-1}a = t^{-1}b \in QI \cap QJ,$ where $a \in I, \ b \in J.$ We have $ts^{-1} \in Q$ and thus $ts^{-1}=u^{-1}c,$ for some $u \in S, \ c \in R.$ Then, since $ts^{-1}a=b,$ we get $ca=ub \in I \cap J = (0),$ which gives us $b=0$ because $u$ is a unit in $Q$ (or because $u$ is regular in $R$). Thus $x = 0.$

4) If $R$ is left Noetherian, then so is $Q.$ If $I \lhd R$ and $Q$ is left Noetherian, then $QI \lhd Q.$

Proof. For the first part, first note that, by 1) and 2), left ideals of $Q$ are exactly in the form $QI,$ where $I$ is a left ideal of $R.$ Let $I \lhd_{\ell} R$ and $I=\sum_{i=1}^n Ra_i.$ Then $QI=\sum_{i=1}^n Qa_i.$ For the second part, let $I$ be a two-sided ideal of $R$ and $s \in S.$ Then $I \subseteq Is^{-1} \subseteq Is^{-2} \subseteq \cdots$ and thus $QI \subseteq QIs^{-1} \subseteq QIs^{-2} \subseteq \cdots.$ So, since $Q$ is left Noetherian, there exists some $n \geq 0$ such that $QIs^{-n} = QIs^{-n-1},$ and thus $QI=QIs^{-1}.$ Therefore $QIs^{-1}a = QIa \subseteq QI,$ for all $s \in S, \ a \in R.$ Hence $QIQ \subseteq QI$ and so $QI$ is a two-sided ideal of $Q.$ (In fact $QIQ=QI.$)

5) If $R$ is prime, then so is $Q.$

Proof. Suppose that $x_1Qx_2=(0),$ for some $x_1=s_1^{-1}a_1, x_2=s_2^{-1}a_2 \in Q.$ Then $s_1^{-1}a_1(as_2)s_2^{-1}a_2 = 0,$ for all $a \in R.$ Thus $s_1^{-1}a_1aa_2 = 0,$ which gives us $a_1aa_2=0,$ i.e. $a_1Ra_2=(0).$ Therefore either $a_1=0$ or $a_2=0,$ which means either $x_1=0$ or $x_2=0.$ This also proves that if $R$ is semiprime, then so is $Q.$

Throughout $S$ is a regular submonoid of the ring $R,$ i.e.$S$ is multiplicatively closed, $1_R \in S$ and every element of $S$ is regular.

Remarks:

1) $S^{-1}R$ (resp. $RS^{-1}$) is defined if and only if $S$ is left (resp. right) Ore. In this case, since every element of $S$ is regular, the corresponding map $\varphi : R \longrightarrow S^{-1}R$ (resp. $\varphi : R \longrightarrow RS^{-1}$) defined by  $\varphi (r) = 1_R^{-1}r$ (resp. $\varphi (r) = r1_R^{-1}$) is one-to-one and hence $S^{-1}R$ (resp. $RS^{-1}$) contains a copy of $R.$

2) If every element of $S$ is a unit, then $Rs = R,$ for all $s \in S$ and thus $Rs \cap Sr = Sr \neq \emptyset,$ for all $s \in S, \ r \in R.$ So, $S$ is left (and right) Ore and therefore $S^{-1}R$ (and $RS^{-1}$) both exist and are equal to $R.$ The reason that they are equal to $R$ is that $s^{-1} \in R,$ for all $s \in S$ and thus $s^{-1}r \in R,$ for all $s \in S, \ r \in R.$

3) An example of 2): let $R$ be a left Artinian ring and $s$ a regular element of $R.$ Then the chain $Rs \supseteq Rs^2 \supseteq \cdots$ must terminate, i.e. there exists $n \geq 1$ such that $Rs^n = Rs^{n+1}.$ Thus $s^n = rs^{n+1},$ for some $r \in R.$ Therefore $(1-rs)s^n = 0,$ which gives us $rs = 1.$ Obviously the same result holds for right Artinian rings.

4) If $S$ contains all the regular elements of $R$ and it is left (resp. right) Ore, then the left (resp. right) quotient ring of $R$ is denoted by $Q(R).$

5) If $S$ is central, i.e. it is contained in the center of $R$, then it’s clearly both left and right Ore and thus $S^{-1}R \cong RS^{-1}.$

Lemma. Suppose that $S$ is central and $A=\{ s^{-1}r_1, \cdots , s^{-1}r_n \} \subset S^{-1}R.$ Let $B=\{r_1, \cdots , r_n \}.$ Then $A^m = (s^{-1})^m B^m,$ for all $m \in \mathbb{N}.$

Proof. Since $S$ is central, we have $s^{-1}x = xs^{-1}$ for all $s \in S, \ x \in R.$ Therefore $s^m(s^{-1}x_1 s^{-1}x_2 \cdots s^{-1}x_m)=x_1x_2 \cdots x_m,$ for all $x_j \in R. \ \Box$

We will use the above lemma to prove that the GK-dimension of an algebra is invariant under localization with respect to a central submonoid:

Theorem. If $R$ is an algebra over a field $F$ and $S$ is central, then $\text{GK}(S^{-1}R) = \text{GK}(R).$

Proof. Let $W=\sum_{j=1}^n Fq_j,$ where $q_j \in S^{-1}R.$ As we’ve seen before, there exit $s \in S$ and $r_1, \cdots , r_n \in R$ such that $q_j = s^{-1}r_j,$ for all $j.$ Let $V=\sum_{j=1}^n F r_j.$ Then, by the lemma, $s^m W^m = V^m,$ for all integers $m \geq 0$ (note that for $m = 0$ both sides are equal to $F$). Therefore $W^m = (s^{-1})^m V^m$ and hence $\dim_F W^m = \dim_F (s^{-1})^m V^m \leq \dim_F V^m.$ Thus $\text{GK}(S^{-1}A) \leq \text{GK}(A).$ The other direction of the inequality is trivial because $A \subseteq S^{-1}A. \ \Box$

Existence of quotient rings

Posted: December 27, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
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We will assume throughout that $R$ is a ring with unity and $S \neq \emptyset$ a multiplicatively closed subset of $R$ with $0 \notin S$ and $1_R \in S.$

Definition 1. $S$ is called a left (resp. right) denominator set if:

1) $S$ is left (resp. right) Ore;

2) for $r \in R$ and $s \in S$ with $rs = 0,$ there exists $s' \in S$ such that $s'r=0$  (resp. for $r \in R$ and $s \in S$ with $sr = 0,$ there exists $s' \in S$ such that $rs'=0$).

Theorem. The left (resp. right) quotient ring of $R$ with respect to $S$ exists if and only if $S$ is a left (resp. right) denominator set.

Proof. (Sketch) We’ve already proved that if the left (resp. right) quotient ring of $R$ with respect to $S$ exists, then $S$ is a left (resp. right) denominator set. For the converse, assuming that $S$ is a left denominator (the “right” version is the same), we’ll construct the left quotient ring of $R$ by first defining a relation on $S \times R$: we say $(s_1,r_1) \sim (s_2,r_2)$ if and only if there exist $r_1',r_2' \in R$ such that $r_1'r_1=r_2'r_2$ and $r_1's_1=r_2' s_2 \in S.$

Claim. The relation $\sim$ is an equivalence relation.

Proof. The reflexive property is proved by choosing $r_1'=r_2'=1_R.$ Obviously $\sim$ is symmetric. In order to prove that $\sim$ is transitive suppose that $(s_1,r_1) \sim (s_2,r_2) \sim (s_3,r_3).$ So there exist $r_1',r_2'$  such that $r_1'r_1=r_2'r_2$ and $r_1's_1 = r_2's_2 \in S.$ There also exist $r_2'',r_3'' \in R$ such that $r_2''r_2=r_3''r_3$ and $r_2''s_2=r_3''s_3 \in S.$ On the other hand, since $S$ is left Ore, we have $Rr_2's_2 \cap S r_2''s_2 \neq\emptyset$ and thus there exist $r \in R$ and $s \in S$ such that $rr_2's_2 = sr_2''s_2.$ So $(rr_2' - sr_2'')s_2=0$ and hence there exists $t \in S$ such that $t(rr_2' - sr_2'')=0,$ i.e. $trr_2' = tsr_2''.$ Finally if we put $x=trr_1', \ y=tsr_3'',$ then $xr_1=yr_3, \ xs_1=ys_3 \in S,$ i.e. $(r_1,s_1) \sim (r_3,s_3). \ \Box$

Next, we will show the equivalence class of the element $(s,r) \in S \times R$ by $s^{-1}r$ and we let $Q$ be the set of all $s^{-1}r.$ We are going to put a ring structure on $Q.$ Let $\alpha = s_1^{-1}r_1, \ \beta = s_2^{-1}r_2$ be two elements of $Q.$ By left Ore condition, $Rs_1 \cap Ss_2 \neq \emptyset$ and thus there exist $r \in R$ and $s \in S$ such that $rs_1=ss_2=t \in S.$ Now define $\alpha + \beta = t^{-1}(rr_1 + sr_2).$ Also, since $Rs_2 \cap Sr_1 \neq \emptyset,$ there exist $r' \in R$ and $s' \in S$ such that $r's_2=s'r_1.$ Let $s's_1=t' \in S$ and define $\alpha \beta = t'^{-1}r'r_2.$ It is proved that the addition and multiplication that we’ve defined are well-defined and satisfy all the conditions needed to make $Q$ a ring. Let $1=1_R.$ It’s easy to see that $1^{-1}0 = 0_Q$ and $1^{-1}1 = 1_Q.$ Finally define $\varphi : R \longrightarrow Q$ by $\varphi (r)=1^{-1}r.$ See that $\varphi$ is a ring homomorphism. Now $r \in \ker \varphi$ if and only if $(1,r) \sim (1,0)$ if and only if there exist $r_1,r_2 \in R$ such that $r_1r=0$ and $r_1=r_2 = s \in S.$ Thus $\ker \varphi = \{r \in R : \ sr = 0, \text{for some} \ s \in S \}.$ Therefore $Q=S^{-1}R$ is the left quotient ring of $R$ with respect to $S. \ \Box$

Remark. Every finite subset of $S^{-1}R$ can be written as $\{s^{-1}x_1, \cdots , s^{-1}x_n \}.$ This is easy to see: let $\{s_1^{-1}y_1, \cdots s_n^{-1}y_n \} \subset S^{-1}R.$ By Remark 3 in this post, there exist $r_1, \cdots , r_n \in R$ such that $r_1s_1 = \cdots = r_ns_n = s \in S.$ Let $x_j = r_jy_j,$ for every $1 \leq j \leq n$, and see that $(s_j , y_j) \sim (s , x_j).$  So $s_j^{-1}y_j = s^{-1}x_j,$ for all $j$.

Ore condition

Posted: December 26, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
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In this section, $R$ is a unitary ring, $S \subset R$ is multiplicatively closed with $0 \notin S$ and $1_R \in S.$

Definition. $S$ is called a left Ore set if $Rs \cap Sr \neq \emptyset,$ for all $r \in R, \ s \in S.$ Similarly $S$ is called a right Ore set if $sR \cap rS \neq \emptyset,$ for all $r \in R, \ s \in S.$

Remark 1. If $Q$ is the left (resp., right) quotient ring of $R$ with respect to $S,$ then  $S$ is a left (resp., right) Ore set. The reason is that if $\varphi : R \longrightarrow Q$ is the corresponding map, then for any $r \in R, \ s \in S$ we must have $\varphi (r) (\varphi (s))^{-1}=(\varphi (s'))^{-1} \varphi (r'),$ for some $r' \in R, \ s' \in S.$ Thus $\varphi (s'r)=\varphi (r's)$ and hence $s'r - rs \in \ker \varphi.$ So $s''(s'r - r's)=0,$ for some $s'' \in S.$ Hence $s''s'r=s''r's \in Rs \cap Sr.$

Remark 2. If $S$ is a left Ore set, then $I=\{r \in R : \ sr = 0, \ \text{for some} \ s \in S \}$ is an ideal of $R.$ (Recall that if $R$ has the left quotient ring $Q$, then $I$ is nothing but the kernel of the corresponding map from $R$ to $Q$). To see this, we note that $I$ is clearly closed under right multiplication by elements of $R.$ also:

i) for every $r,r' \in I: \ r+r' \in I.$ To prove this, we have $sr=s'r'=0,$ for some $s,s' \in S.$ By Remark 1, $Rs \cap Ss' \neq \emptyset.$ Therefore $r_1s=s''s',$ for some $r_1 \in R, \ s'' \in S.$ So $s''s'r=r_1sr=0, \ s''s'r'=0$ and hence $s''s'(r+r')=0,$ i.e. $r+r' \in I.$

ii) for every $r \in I, \ r' \in R: \ r'r \in I.$ To prove this one, we have $sr=0,$ for some $s \in S.$ Now, by Remark 1: $Rs \cap Sr' \neq \emptyset.$ So $r''s=s'r',$ for some $r'' \in R, \ s' \in S.$ Thus $0=r''sr=s'r'r,$ i.e. $r'r \in I.$

iii) if $S$ is a right Ore set, then $J=\{r \in R : \ rs = 0, \ \text{for some} \ s \in S \} \lhd R.$

Remark 3. Suppose $S$ is left Ore and $s_1, \cdots , s_n \in S.$ There exist $r_1, \cdots , r_n \in R$ such that $r_1s_1 = \cdots = r_ns_n \in S.$ The proof is by induction over $n$: if $n = 1,$ then we may choose $r_1=1.$ Suppose $n > 1$ and the claim is true for $n-1.$ Choose $r_1', \cdots , r_{n-1}'$ so that $r_1's_1 = \cdots = r_{n-1}' s_{n-1}=s \in S.$ Also there exist $r_n \in R$ and $t \in S$ such that $r_ns_n = ts,$ since $Rs_n \cap Ss \neq \emptyset.$ Let $r_j = tr_j'$ for $j=1, \cdots , n-1.$ Then for all $1 \leq j \leq n-1$ we have $r_j s_j = t r_j' s_j = ts = r_ns_n$ and $ts \in S.$ Similarly, if $S$ is right Ore and $s_1, \cdots , s_n \in S,$ then there exist $r_1, \cdots , r_n \in R$ such that $s_1r_1 = \cdots = s_n r_n \in S.$