Mixing
Proving $R-S$ contains a prime ideal when $S$ is a multiplicative set
$begingroup$
I'm mainly trying to prove that
If $0not in Ssubseteq R$ is a multiplicative subset of a
commutative ring $R$ with identity. Then $R-S$ contains a prime ideal.
Now, by using Zorn's lemma, one can show that $R-S$ contains a maximal ideal I. In fact, I is a prime ideal as well, so we need to show this.To do that, I assumed the opposite, i.e $exists a,b in R$ s.t $abin I$, but $a,b not in I$. Then looked each case. However, I stuck when $a,b in R-[Scup I]$. The following is what I've done for this case;
Proof:
In this case, our assumptions are $a,bin R-[Scup I]$ and $I + (a) = J$, where $J$ is an ideal in $R$ and $J cap S not = emptyset$.
These imply that; $S^{-1}(I + (a)) = S^{-1} J$, but since $J cap S not = emptyset$, by a theorem that we proved in the class, $S^{-1} J = S^{-1}R$.Therefore, it must be true that $exists (iin I, rin R, s_1,s in S)$ s.t
$$frac{i+ra}{s_1} =frac{1}{s}$$
but this implies $exists s_2 in S$ s.t
$$ s_2(is + sra -1 s_1) = 0 quad Rightarrow quad (s_1 s )i + (s_1 s r) a = (s_1 s )1,$$
,so
$$(b s_1 s )i + (s_1 s r) ab = (s_1 s )b.$$
Observe that by our assumption $abin I$, so $(s_1 s)b in I$.
Now, at the point, I don't know how to proceed, so I would appreciate any help or hint because I stuck at this point for over a week now.
Edit
Thanks for the great answers; however, but I'm specifically interested in how to continue from the point where I stuck.
abstract-algebra ring-theory ideals maximal-and-prime-ideals localization
asked Jan 17 at 9:21
onurcanbektasonurcanbektas
3,42111036
$endgroup$
add a comment |
$begingroup$
I'm mainly trying to prove that
If $0not in Ssubseteq R$ is a multiplicative subset of a
commutative ring $R$ with identity. Then $R-S$ contains a prime ideal.
Now, by using Zorn's lemma, one can show that $R-S$ contains a maximal ideal I. In fact, I is a prime ideal as well, so we need to show this.To do that, I assumed the opposite, i.e $exists a,b in R$ s.t $abin I$, but $a,b not in I$. Then looked each case. However, I stuck when $a,b in R-[Scup I]$. The following is what I've done for this case;
Proof:
In this case, our assumptions are $a,bin R-[Scup I]$ and $I + (a) = J$, where $J$ is an ideal in $R$ and $J cap S not = emptyset$.
These imply that; $S^{-1}(I + (a)) = S^{-1} J$, but since $J cap S not = emptyset$, by a theorem that we proved in the class, $S^{-1} J = S^{-1}R$.Therefore, it must be true that $exists (iin I, rin R, s_1,s in S)$ s.t
$$frac{i+ra}{s_1} =frac{1}{s}$$
but this implies $exists s_2 in S$ s.t
$$ s_2(is + sra -1 s_1) = 0 quad Rightarrow quad (s_1 s )i + (s_1 s r) a = (s_1 s )1,$$
,so
$$(b s_1 s )i + (s_1 s r) ab = (s_1 s )b.$$
Observe that by our assumption $abin I$, so $(s_1 s)b in I$.
Now, at the point, I don't know how to proceed, so I would appreciate any help or hint because I stuck at this point for over a week now.
Edit
Thanks for the great answers; however, but I'm specifically interested in how to continue from the point where I stuck.
abstract-algebra ring-theory ideals maximal-and-prime-ideals localization
asked Jan 17 at 9:21
onurcanbektasonurcanbektas
3,42111036
$endgroup$
$begingroup$
Every maximal ideal is prime.
$endgroup$
– Math_QED
Jan 17 at 10:05
1
$begingroup$
@Math_QED You're right for maximal ideals in rings but the intention in this question is that $;I;$ is maximal with respect to be an ideal not contained in $;S;$ ...
$endgroup$
– DonAntonio
Jan 17 at 10:06
add a comment |
$begingroup$
I'm mainly trying to prove that
If $0not in Ssubseteq R$ is a multiplicative subset of a
commutative ring $R$ with identity. Then $R-S$ contains a prime ideal.
Now, by using Zorn's lemma, one can show that $R-S$ contains a maximal ideal I. In fact, I is a prime ideal as well, so we need to show this.To do that, I assumed the opposite, i.e $exists a,b in R$ s.t $abin I$, but $a,b not in I$. Then looked each case. However, I stuck when $a,b in R-[Scup I]$. The following is what I've done for this case;
Proof:
In this case, our assumptions are $a,bin R-[Scup I]$ and $I + (a) = J$, where $J$ is an ideal in $R$ and $J cap S not = emptyset$.
These imply that; $S^{-1}(I + (a)) = S^{-1} J$, but since $J cap S not = emptyset$, by a theorem that we proved in the class, $S^{-1} J = S^{-1}R$.Therefore, it must be true that $exists (iin I, rin R, s_1,s in S)$ s.t
$$frac{i+ra}{s_1} =frac{1}{s}$$
but this implies $exists s_2 in S$ s.t
$$ s_2(is + sra -1 s_1) = 0 quad Rightarrow quad (s_1 s )i + (s_1 s r) a = (s_1 s )1,$$
,so
$$(b s_1 s )i + (s_1 s r) ab = (s_1 s )b.$$
Observe that by our assumption $abin I$, so $(s_1 s)b in I$.
Now, at the point, I don't know how to proceed, so I would appreciate any help or hint because I stuck at this point for over a week now.
Edit
Thanks for the great answers; however, but I'm specifically interested in how to continue from the point where I stuck.
abstract-algebra ring-theory ideals maximal-and-prime-ideals localization
asked Jan 17 at 9:21
onurcanbektasonurcanbektas
3,42111036
$endgroup$
I'm mainly trying to prove that
If $0not in Ssubseteq R$ is a multiplicative subset of a
commutative ring $R$ with identity. Then $R-S$ contains a prime ideal.
Now, by using Zorn's lemma, one can show that $R-S$ contains a maximal ideal I. In fact, I is a prime ideal as well, so we need to show this.To do that, I assumed the opposite, i.e $exists a,b in R$ s.t $abin I$, but $a,b not in I$. Then looked each case. However, I stuck when $a,b in R-[Scup I]$. The following is what I've done for this case;
Proof:
In this case, our assumptions are $a,bin R-[Scup I]$ and $I + (a) = J$, where $J$ is an ideal in $R$ and $J cap S not = emptyset$.
These imply that; $S^{-1}(I + (a)) = S^{-1} J$, but since $J cap S not = emptyset$, by a theorem that we proved in the class, $S^{-1} J = S^{-1}R$.Therefore, it must be true that $exists (iin I, rin R, s_1,s in S)$ s.t
$$frac{i+ra}{s_1} =frac{1}{s}$$
but this implies $exists s_2 in S$ s.t
$$ s_2(is + sra -1 s_1) = 0 quad Rightarrow quad (s_1 s )i + (s_1 s r) a = (s_1 s )1,$$
,so
$$(b s_1 s )i + (s_1 s r) ab = (s_1 s )b.$$
Observe that by our assumption $abin I$, so $(s_1 s)b in I$.
Now, at the point, I don't know how to proceed, so I would appreciate any help or hint because I stuck at this point for over a week now.
Edit
Thanks for the great answers; however, but I'm specifically interested in how to continue from the point where I stuck.
abstract-algebra ring-theory ideals maximal-and-prime-ideals localization
abstract-algebra ring-theory ideals maximal-and-prime-ideals localization
asked Jan 17 at 9:21
onurcanbektasonurcanbektas
3,42111036
asked Jan 17 at 9:21
onurcanbektasonurcanbektas
3,42111036
edited Jan 17 at 10:54
onurcanbektas
asked Jan 17 at 9:21
onurcanbektasonurcanbektas
3,42111036
asked Jan 17 at 9:21
onurcanbektasonurcanbektas
3,42111036
asked Jan 17 at 9:21
onurcanbektasonurcanbektas
3,42111036
3,42111036
$begingroup$
Every maximal ideal is prime.
$endgroup$
– Math_QED
Jan 17 at 10:05
1
$begingroup$
@Math_QED You're right for maximal ideals in rings but the intention in this question is that $;I;$ is maximal with respect to be an ideal not contained in $;S;$ ...
$endgroup$
– DonAntonio
Jan 17 at 10:06
add a comment |
$begingroup$
Every maximal ideal is prime.
$endgroup$
– Math_QED
Jan 17 at 10:05
1
$begingroup$
@Math_QED You're right for maximal ideals in rings but the intention in this question is that $;I;$ is maximal with respect to be an ideal not contained in $;S;$ ...
$endgroup$
– DonAntonio
Jan 17 at 10:06
$begingroup$
Every maximal ideal is prime.
$endgroup$
– Math_QED
Jan 17 at 10:05
$begingroup$
Every maximal ideal is prime.
$endgroup$
– Math_QED
Jan 17 at 10:05
1
1
$begingroup$
@Math_QED You're right for maximal ideals in rings but the intention in this question is that $;I;$ is maximal with respect to be an ideal not contained in $;S;$ ...
$endgroup$
– DonAntonio
Jan 17 at 10:06
$begingroup$
@Math_QED You're right for maximal ideals in rings but the intention in this question is that $;I;$ is maximal with respect to be an ideal not contained in $;S;$ ...
$endgroup$
– DonAntonio
Jan 17 at 10:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $;I;$ isn't a prime idea of $;R;$ , then there exist $;a,bin R;$ s.t. $;abin I;$ but $;a,bnotin I;$ and from here we get
$$Ilneq begin{cases}I+aR,\I+bRend{cases}implies exists,i,jin I,,,,x,y,in R;;s.t.;;i+ax,,,j+bynotin Rsetminus Simplies$$
$$begin{cases}i+ax=s_1in S\{}\j+by=s_2in Send{cases}implies Sni s_1s_2=ij+iby+jax+abxyin Iimplies Icap Sneq emptyset$$
and this is a contradiction.
answered Jan 17 at 9:30
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Here's an answer that uses techniques that you might have not seen, but can be a motivation for learning them.
Since $S$ is a multiplicative subset we can form the localization $S^{-1}R$, a nonzero ring together with a natural map
$$f: R rightarrow S^{-1}R$$
such that $f(s)$ is a unit in $S^{-1}R$ for all $s in S$.
Then $S^{-1}R$ contains a prime ideal $mathfrak{p}$, and pulling back gives a prime ideal $mathfrak{q}= f^{-1}(mathfrak{p})$. If $sin S$ is contained in $mathfrak{q}$ then $f(s)$ would be a unit contained in $mathfrak{p}$, which is a contradiction and shows that $mathfrak{q}$ is the prime you want.
answered Jan 17 at 9:42
Jef LJef L
2,866617
$endgroup$
$begingroup$
As you say, most probably this is stuff the OP hasn't yet learnt, yet you could perhaps make it a little easier by saying $;mathcal q;$ is the inverse image of $;mathcal p;$ under the natural map (homomorphism). This he must surely have already studied: inverse homomorphic images of prime ideas are prime ideals.
$endgroup$
– DonAntonio
Jan 17 at 9:46
$begingroup$
No, I have seen them, but how can you argue that $S^{-1} R $ containes a prime ideal ?
$endgroup$
– onurcanbektas
Jan 17 at 10:16
2
$begingroup$
Every nonzero ring contains a maximal ideal by Zorn's lemma, hence a prime ideal
$endgroup$
– Jef L
Jan 17 at 13:55
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $;I;$ isn't a prime idea of $;R;$ , then there exist $;a,bin R;$ s.t. $;abin I;$ but $;a,bnotin I;$ and from here we get
$$Ilneq begin{cases}I+aR,\I+bRend{cases}implies exists,i,jin I,,,,x,y,in R;;s.t.;;i+ax,,,j+bynotin Rsetminus Simplies$$
$$begin{cases}i+ax=s_1in S\{}\j+by=s_2in Send{cases}implies Sni s_1s_2=ij+iby+jax+abxyin Iimplies Icap Sneq emptyset$$
and this is a contradiction.
answered Jan 17 at 9:30
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
If $;I;$ isn't a prime idea of $;R;$ , then there exist $;a,bin R;$ s.t. $;abin I;$ but $;a,bnotin I;$ and from here we get
$$Ilneq begin{cases}I+aR,\I+bRend{cases}implies exists,i,jin I,,,,x,y,in R;;s.t.;;i+ax,,,j+bynotin Rsetminus Simplies$$
$$begin{cases}i+ax=s_1in S\{}\j+by=s_2in Send{cases}implies Sni s_1s_2=ij+iby+jax+abxyin Iimplies Icap Sneq emptyset$$
and this is a contradiction.
answered Jan 17 at 9:30
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
If $;I;$ isn't a prime idea of $;R;$ , then there exist $;a,bin R;$ s.t. $;abin I;$ but $;a,bnotin I;$ and from here we get
$$Ilneq begin{cases}I+aR,\I+bRend{cases}implies exists,i,jin I,,,,x,y,in R;;s.t.;;i+ax,,,j+bynotin Rsetminus Simplies$$
$$begin{cases}i+ax=s_1in S\{}\j+by=s_2in Send{cases}implies Sni s_1s_2=ij+iby+jax+abxyin Iimplies Icap Sneq emptyset$$
and this is a contradiction.
answered Jan 17 at 9:30
DonAntonioDonAntonio
179k1494230
$endgroup$
If $;I;$ isn't a prime idea of $;R;$ , then there exist $;a,bin R;$ s.t. $;abin I;$ but $;a,bnotin I;$ and from here we get
$$Ilneq begin{cases}I+aR,\I+bRend{cases}implies exists,i,jin I,,,,x,y,in R;;s.t.;;i+ax,,,j+bynotin Rsetminus Simplies$$
$$begin{cases}i+ax=s_1in S\{}\j+by=s_2in Send{cases}implies Sni s_1s_2=ij+iby+jax+abxyin Iimplies Icap Sneq emptyset$$
and this is a contradiction.
answered Jan 17 at 9:30
DonAntonioDonAntonio
179k1494230
edited Jan 17 at 9:43
answered Jan 17 at 9:30
DonAntonioDonAntonio
179k1494230
answered Jan 17 at 9:30
DonAntonioDonAntonio
179k1494230
answered Jan 17 at 9:30
DonAntonioDonAntonio
179k1494230
179k1494230
add a comment |
add a comment |
$begingroup$
Here's an answer that uses techniques that you might have not seen, but can be a motivation for learning them.
Since $S$ is a multiplicative subset we can form the localization $S^{-1}R$, a nonzero ring together with a natural map
$$f: R rightarrow S^{-1}R$$
such that $f(s)$ is a unit in $S^{-1}R$ for all $s in S$.
Then $S^{-1}R$ contains a prime ideal $mathfrak{p}$, and pulling back gives a prime ideal $mathfrak{q}= f^{-1}(mathfrak{p})$. If $sin S$ is contained in $mathfrak{q}$ then $f(s)$ would be a unit contained in $mathfrak{p}$, which is a contradiction and shows that $mathfrak{q}$ is the prime you want.
answered Jan 17 at 9:42
Jef LJef L
2,866617
$endgroup$
$begingroup$
As you say, most probably this is stuff the OP hasn't yet learnt, yet you could perhaps make it a little easier by saying $;mathcal q;$ is the inverse image of $;mathcal p;$ under the natural map (homomorphism). This he must surely have already studied: inverse homomorphic images of prime ideas are prime ideals.
$endgroup$
– DonAntonio
Jan 17 at 9:46
$begingroup$
No, I have seen them, but how can you argue that $S^{-1} R $ containes a prime ideal ?
$endgroup$
– onurcanbektas
Jan 17 at 10:16
2
$begingroup$
Every nonzero ring contains a maximal ideal by Zorn's lemma, hence a prime ideal
$endgroup$
– Jef L
Jan 17 at 13:55
add a comment |
$begingroup$
Here's an answer that uses techniques that you might have not seen, but can be a motivation for learning them.
Since $S$ is a multiplicative subset we can form the localization $S^{-1}R$, a nonzero ring together with a natural map
$$f: R rightarrow S^{-1}R$$
such that $f(s)$ is a unit in $S^{-1}R$ for all $s in S$.
Then $S^{-1}R$ contains a prime ideal $mathfrak{p}$, and pulling back gives a prime ideal $mathfrak{q}= f^{-1}(mathfrak{p})$. If $sin S$ is contained in $mathfrak{q}$ then $f(s)$ would be a unit contained in $mathfrak{p}$, which is a contradiction and shows that $mathfrak{q}$ is the prime you want.
answered Jan 17 at 9:42
Jef LJef L
2,866617
$endgroup$
$begingroup$
As you say, most probably this is stuff the OP hasn't yet learnt, yet you could perhaps make it a little easier by saying $;mathcal q;$ is the inverse image of $;mathcal p;$ under the natural map (homomorphism). This he must surely have already studied: inverse homomorphic images of prime ideas are prime ideals.
$endgroup$
– DonAntonio
Jan 17 at 9:46
$begingroup$
No, I have seen them, but how can you argue that $S^{-1} R $ containes a prime ideal ?
$endgroup$
– onurcanbektas
Jan 17 at 10:16
2
$begingroup$
Every nonzero ring contains a maximal ideal by Zorn's lemma, hence a prime ideal
$endgroup$
– Jef L
Jan 17 at 13:55
add a comment |
$begingroup$
Here's an answer that uses techniques that you might have not seen, but can be a motivation for learning them.
Since $S$ is a multiplicative subset we can form the localization $S^{-1}R$, a nonzero ring together with a natural map
$$f: R rightarrow S^{-1}R$$
such that $f(s)$ is a unit in $S^{-1}R$ for all $s in S$.
Then $S^{-1}R$ contains a prime ideal $mathfrak{p}$, and pulling back gives a prime ideal $mathfrak{q}= f^{-1}(mathfrak{p})$. If $sin S$ is contained in $mathfrak{q}$ then $f(s)$ would be a unit contained in $mathfrak{p}$, which is a contradiction and shows that $mathfrak{q}$ is the prime you want.
answered Jan 17 at 9:42
Jef LJef L
2,866617
$endgroup$
Here's an answer that uses techniques that you might have not seen, but can be a motivation for learning them.
Since $S$ is a multiplicative subset we can form the localization $S^{-1}R$, a nonzero ring together with a natural map
$$f: R rightarrow S^{-1}R$$
such that $f(s)$ is a unit in $S^{-1}R$ for all $s in S$.
Then $S^{-1}R$ contains a prime ideal $mathfrak{p}$, and pulling back gives a prime ideal $mathfrak{q}= f^{-1}(mathfrak{p})$. If $sin S$ is contained in $mathfrak{q}$ then $f(s)$ would be a unit contained in $mathfrak{p}$, which is a contradiction and shows that $mathfrak{q}$ is the prime you want.
answered Jan 17 at 9:42
Jef LJef L
2,866617
answered Jan 17 at 9:42
Jef LJef L
2,866617
answered Jan 17 at 9:42
Jef LJef L
2,866617
answered Jan 17 at 9:42
Jef LJef L
2,866617
2,866617
$begingroup$
As you say, most probably this is stuff the OP hasn't yet learnt, yet you could perhaps make it a little easier by saying $;mathcal q;$ is the inverse image of $;mathcal p;$ under the natural map (homomorphism). This he must surely have already studied: inverse homomorphic images of prime ideas are prime ideals.
$endgroup$
– DonAntonio
Jan 17 at 9:46
$begingroup$
No, I have seen them, but how can you argue that $S^{-1} R $ containes a prime ideal ?
$endgroup$
– onurcanbektas
Jan 17 at 10:16
2
$begingroup$
Every nonzero ring contains a maximal ideal by Zorn's lemma, hence a prime ideal
$endgroup$
– Jef L
Jan 17 at 13:55
add a comment |
$begingroup$
As you say, most probably this is stuff the OP hasn't yet learnt, yet you could perhaps make it a little easier by saying $;mathcal q;$ is the inverse image of $;mathcal p;$ under the natural map (homomorphism). This he must surely have already studied: inverse homomorphic images of prime ideas are prime ideals.
$endgroup$
– DonAntonio
Jan 17 at 9:46
$begingroup$
No, I have seen them, but how can you argue that $S^{-1} R $ containes a prime ideal ?
$endgroup$
– onurcanbektas
Jan 17 at 10:16
2
$begingroup$
Every nonzero ring contains a maximal ideal by Zorn's lemma, hence a prime ideal
$endgroup$
– Jef L
Jan 17 at 13:55
$begingroup$
As you say, most probably this is stuff the OP hasn't yet learnt, yet you could perhaps make it a little easier by saying $;mathcal q;$ is the inverse image of $;mathcal p;$ under the natural map (homomorphism). This he must surely have already studied: inverse homomorphic images of prime ideas are prime ideals.
$endgroup$
– DonAntonio
Jan 17 at 9:46
$begingroup$
As you say, most probably this is stuff the OP hasn't yet learnt, yet you could perhaps make it a little easier by saying $;mathcal q;$ is the inverse image of $;mathcal p;$ under the natural map (homomorphism). This he must surely have already studied: inverse homomorphic images of prime ideas are prime ideals.
$endgroup$
– DonAntonio
Jan 17 at 9:46
$begingroup$
No, I have seen them, but how can you argue that $S^{-1} R $ containes a prime ideal ?
$endgroup$
– onurcanbektas
Jan 17 at 10:16
$begingroup$
No, I have seen them, but how can you argue that $S^{-1} R $ containes a prime ideal ?
$endgroup$
– onurcanbektas
Jan 17 at 10:16
2
2
$begingroup$
Every nonzero ring contains a maximal ideal by Zorn's lemma, hence a prime ideal
$endgroup$
– Jef L
Jan 17 at 13:55
$begingroup$
Every nonzero ring contains a maximal ideal by Zorn's lemma, hence a prime ideal
$endgroup$
– Jef L
Jan 17 at 13:55
add a comment |
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$begingroup$
Every maximal ideal is prime.
$endgroup$
– Math_QED
Jan 17 at 10:05
1
$begingroup$
@Math_QED You're right for maximal ideals in rings but the intention in this question is that $;I;$ is maximal with respect to be an ideal not contained in $;S;$ ...
$endgroup$
– DonAntonio
Jan 17 at 10:06