Let $P$ be prime and contain $IJ$, the product ideal. Then $I subset P$ or $J subset P$
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So I already witness the solution. It is this: Assume $I notsubset P$, then there is $i notin P$. Then the product $ij in IJ subset P$, but since $P$ is prime, $i in P$ or $j in P$, so $J subset P$.
here is what I don't get. Doesn't this only show a finite truncation belongs to $P$? Don't we have to show that for any finite sum $i_1j_1 + dots + i_nj_n in P$, $i_n in P$ for every $n$ (assuming $|J| = |I| = n$ for ease of argument)
abstract-algebra ring-theory proof-explanation
asked Dec 22 '18 at 9:08
HawkHawk
5,5851140110
$endgroup$
add a comment |
$begingroup$
So I already witness the solution. It is this: Assume $I notsubset P$, then there is $i notin P$. Then the product $ij in IJ subset P$, but since $P$ is prime, $i in P$ or $j in P$, so $J subset P$.
here is what I don't get. Doesn't this only show a finite truncation belongs to $P$? Don't we have to show that for any finite sum $i_1j_1 + dots + i_nj_n in P$, $i_n in P$ for every $n$ (assuming $|J| = |I| = n$ for ease of argument)
abstract-algebra ring-theory proof-explanation
asked Dec 22 '18 at 9:08
HawkHawk
5,5851140110
$endgroup$
$begingroup$
What does $|I|$ mean? At any rate, while it is true that such finite sums belong to $IJ$, you don't need to use that fact - it surely is true that $ijin IJ$ and this is enough.
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– Wojowu
Dec 22 '18 at 9:14
1
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What is your definition of prime ideal? This definition is the one that works in non-commutative settings.
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– user370967
Dec 22 '18 at 9:14
1
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@Math_QED $ab in P$, then $a in P$ or $b in P.$
$endgroup$
– Hawk
Dec 22 '18 at 9:15
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@Hawk If you are working with commutative rings, always be sure to use the [commutative-algebra] tag and/or mention commutativity in the post.
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– rschwieb
Dec 22 '18 at 16:56
add a comment |
$begingroup$
So I already witness the solution. It is this: Assume $I notsubset P$, then there is $i notin P$. Then the product $ij in IJ subset P$, but since $P$ is prime, $i in P$ or $j in P$, so $J subset P$.
here is what I don't get. Doesn't this only show a finite truncation belongs to $P$? Don't we have to show that for any finite sum $i_1j_1 + dots + i_nj_n in P$, $i_n in P$ for every $n$ (assuming $|J| = |I| = n$ for ease of argument)
abstract-algebra ring-theory proof-explanation
asked Dec 22 '18 at 9:08
HawkHawk
5,5851140110
$endgroup$
So I already witness the solution. It is this: Assume $I notsubset P$, then there is $i notin P$. Then the product $ij in IJ subset P$, but since $P$ is prime, $i in P$ or $j in P$, so $J subset P$.
here is what I don't get. Doesn't this only show a finite truncation belongs to $P$? Don't we have to show that for any finite sum $i_1j_1 + dots + i_nj_n in P$, $i_n in P$ for every $n$ (assuming $|J| = |I| = n$ for ease of argument)
abstract-algebra ring-theory proof-explanation
abstract-algebra ring-theory proof-explanation
asked Dec 22 '18 at 9:08
HawkHawk
5,5851140110
asked Dec 22 '18 at 9:08
HawkHawk
5,5851140110
asked Dec 22 '18 at 9:08
HawkHawk
5,5851140110
asked Dec 22 '18 at 9:08
HawkHawk
5,5851140110
asked Dec 22 '18 at 9:08
HawkHawk
5,5851140110
5,5851140110
$begingroup$
What does $|I|$ mean? At any rate, while it is true that such finite sums belong to $IJ$, you don't need to use that fact - it surely is true that $ijin IJ$ and this is enough.
$endgroup$
– Wojowu
Dec 22 '18 at 9:14
1
$begingroup$
What is your definition of prime ideal? This definition is the one that works in non-commutative settings.
$endgroup$
– user370967
Dec 22 '18 at 9:14
1
$begingroup$
@Math_QED $ab in P$, then $a in P$ or $b in P.$
$endgroup$
– Hawk
Dec 22 '18 at 9:15
$begingroup$
@Hawk If you are working with commutative rings, always be sure to use the [commutative-algebra] tag and/or mention commutativity in the post.
$endgroup$
– rschwieb
Dec 22 '18 at 16:56
add a comment |
$begingroup$
What does $|I|$ mean? At any rate, while it is true that such finite sums belong to $IJ$, you don't need to use that fact - it surely is true that $ijin IJ$ and this is enough.
$endgroup$
– Wojowu
Dec 22 '18 at 9:14
1
$begingroup$
What is your definition of prime ideal? This definition is the one that works in non-commutative settings.
$endgroup$
– user370967
Dec 22 '18 at 9:14
1
$begingroup$
@Math_QED $ab in P$, then $a in P$ or $b in P.$
$endgroup$
– Hawk
Dec 22 '18 at 9:15
$begingroup$
@Hawk If you are working with commutative rings, always be sure to use the [commutative-algebra] tag and/or mention commutativity in the post.
$endgroup$
– rschwieb
Dec 22 '18 at 16:56
$begingroup$
What does $|I|$ mean? At any rate, while it is true that such finite sums belong to $IJ$, you don't need to use that fact - it surely is true that $ijin IJ$ and this is enough.
$endgroup$
– Wojowu
Dec 22 '18 at 9:14
$begingroup$
What does $|I|$ mean? At any rate, while it is true that such finite sums belong to $IJ$, you don't need to use that fact - it surely is true that $ijin IJ$ and this is enough.
$endgroup$
– Wojowu
Dec 22 '18 at 9:14
1
1
$begingroup$
What is your definition of prime ideal? This definition is the one that works in non-commutative settings.
$endgroup$
– user370967
Dec 22 '18 at 9:14
$begingroup$
What is your definition of prime ideal? This definition is the one that works in non-commutative settings.
$endgroup$
– user370967
Dec 22 '18 at 9:14
1
1
$begingroup$
@Math_QED $ab in P$, then $a in P$ or $b in P.$
$endgroup$
– Hawk
Dec 22 '18 at 9:15
$begingroup$
@Math_QED $ab in P$, then $a in P$ or $b in P.$
$endgroup$
– Hawk
Dec 22 '18 at 9:15
$begingroup$
@Hawk If you are working with commutative rings, always be sure to use the [commutative-algebra] tag and/or mention commutativity in the post.
$endgroup$
– rschwieb
Dec 22 '18 at 16:56
$begingroup$
@Hawk If you are working with commutative rings, always be sure to use the [commutative-algebra] tag and/or mention commutativity in the post.
$endgroup$
– rschwieb
Dec 22 '18 at 16:56
add a comment |
2 Answers
2
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Well, suppose $Inotsubset P$. Then there is $iin I$ such that $inotin P$. For EACH $jin J$, $ijin IJsubseteq P$ by hypothesis. But $P$ is prime and so $jin P$. Hence $Jsubseteq P$.
answered Dec 22 '18 at 9:15
WuestenfuxWuestenfux
5,5581513
$endgroup$
add a comment |
$begingroup$
The proof as written is phrased a little awkwardly and missing important details. Here is a correct proof (with everything spelled out):
Suppose $IJ subset P$ and $I subsetneq P$. We wish to show that for all $j in J$, we have $j in P$. Fix $j in J$ $i in I setminus P$ and note that $ij in IJ$. Since $IJ subset P$, we have that $ij in P$ but $P$ is prime, so we must have that either $i in P$ or $j in P$. Since $i notin P$ (remember, $i in I setminus P$), we conclude that $j in J$. This shows that $J subset P$.
As far as the finite sums go, they belong to $IJ$ and thus to $P$ by assumption.
answered Dec 22 '18 at 9:16
Aniruddh AgarwalAniruddh Agarwal
1218
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add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
Well, suppose $Inotsubset P$. Then there is $iin I$ such that $inotin P$. For EACH $jin J$, $ijin IJsubseteq P$ by hypothesis. But $P$ is prime and so $jin P$. Hence $Jsubseteq P$.
answered Dec 22 '18 at 9:15
WuestenfuxWuestenfux
5,5581513
$endgroup$
add a comment |
$begingroup$
Well, suppose $Inotsubset P$. Then there is $iin I$ such that $inotin P$. For EACH $jin J$, $ijin IJsubseteq P$ by hypothesis. But $P$ is prime and so $jin P$. Hence $Jsubseteq P$.
answered Dec 22 '18 at 9:15
WuestenfuxWuestenfux
5,5581513
$endgroup$
add a comment |
$begingroup$
Well, suppose $Inotsubset P$. Then there is $iin I$ such that $inotin P$. For EACH $jin J$, $ijin IJsubseteq P$ by hypothesis. But $P$ is prime and so $jin P$. Hence $Jsubseteq P$.
answered Dec 22 '18 at 9:15
WuestenfuxWuestenfux
5,5581513
$endgroup$
Well, suppose $Inotsubset P$. Then there is $iin I$ such that $inotin P$. For EACH $jin J$, $ijin IJsubseteq P$ by hypothesis. But $P$ is prime and so $jin P$. Hence $Jsubseteq P$.
answered Dec 22 '18 at 9:15
WuestenfuxWuestenfux
5,5581513
answered Dec 22 '18 at 9:15
WuestenfuxWuestenfux
5,5581513
answered Dec 22 '18 at 9:15
WuestenfuxWuestenfux
5,5581513
answered Dec 22 '18 at 9:15
WuestenfuxWuestenfux
5,5581513
5,5581513
add a comment |
add a comment |
$begingroup$
The proof as written is phrased a little awkwardly and missing important details. Here is a correct proof (with everything spelled out):
Suppose $IJ subset P$ and $I subsetneq P$. We wish to show that for all $j in J$, we have $j in P$. Fix $j in J$ $i in I setminus P$ and note that $ij in IJ$. Since $IJ subset P$, we have that $ij in P$ but $P$ is prime, so we must have that either $i in P$ or $j in P$. Since $i notin P$ (remember, $i in I setminus P$), we conclude that $j in J$. This shows that $J subset P$.
As far as the finite sums go, they belong to $IJ$ and thus to $P$ by assumption.
answered Dec 22 '18 at 9:16
Aniruddh AgarwalAniruddh Agarwal
1218
$endgroup$
add a comment |
$begingroup$
The proof as written is phrased a little awkwardly and missing important details. Here is a correct proof (with everything spelled out):
Suppose $IJ subset P$ and $I subsetneq P$. We wish to show that for all $j in J$, we have $j in P$. Fix $j in J$ $i in I setminus P$ and note that $ij in IJ$. Since $IJ subset P$, we have that $ij in P$ but $P$ is prime, so we must have that either $i in P$ or $j in P$. Since $i notin P$ (remember, $i in I setminus P$), we conclude that $j in J$. This shows that $J subset P$.
As far as the finite sums go, they belong to $IJ$ and thus to $P$ by assumption.
answered Dec 22 '18 at 9:16
Aniruddh AgarwalAniruddh Agarwal
1218
$endgroup$
add a comment |
$begingroup$
The proof as written is phrased a little awkwardly and missing important details. Here is a correct proof (with everything spelled out):
Suppose $IJ subset P$ and $I subsetneq P$. We wish to show that for all $j in J$, we have $j in P$. Fix $j in J$ $i in I setminus P$ and note that $ij in IJ$. Since $IJ subset P$, we have that $ij in P$ but $P$ is prime, so we must have that either $i in P$ or $j in P$. Since $i notin P$ (remember, $i in I setminus P$), we conclude that $j in J$. This shows that $J subset P$.
As far as the finite sums go, they belong to $IJ$ and thus to $P$ by assumption.
answered Dec 22 '18 at 9:16
Aniruddh AgarwalAniruddh Agarwal
1218
$endgroup$
The proof as written is phrased a little awkwardly and missing important details. Here is a correct proof (with everything spelled out):
Suppose $IJ subset P$ and $I subsetneq P$. We wish to show that for all $j in J$, we have $j in P$. Fix $j in J$ $i in I setminus P$ and note that $ij in IJ$. Since $IJ subset P$, we have that $ij in P$ but $P$ is prime, so we must have that either $i in P$ or $j in P$. Since $i notin P$ (remember, $i in I setminus P$), we conclude that $j in J$. This shows that $J subset P$.
As far as the finite sums go, they belong to $IJ$ and thus to $P$ by assumption.
answered Dec 22 '18 at 9:16
Aniruddh AgarwalAniruddh Agarwal
1218
answered Dec 22 '18 at 9:16
Aniruddh AgarwalAniruddh Agarwal
1218
answered Dec 22 '18 at 9:16
Aniruddh AgarwalAniruddh Agarwal
1218
answered Dec 22 '18 at 9:16
Aniruddh AgarwalAniruddh Agarwal
1218
1218
add a comment |
add a comment |
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$begingroup$
What does $|I|$ mean? At any rate, while it is true that such finite sums belong to $IJ$, you don't need to use that fact - it surely is true that $ijin IJ$ and this is enough.
$endgroup$
– Wojowu
Dec 22 '18 at 9:14
1
$begingroup$
What is your definition of prime ideal? This definition is the one that works in non-commutative settings.
$endgroup$
– user370967
Dec 22 '18 at 9:14
1
$begingroup$
@Math_QED $ab in P$, then $a in P$ or $b in P.$
$endgroup$
– Hawk
Dec 22 '18 at 9:15
$begingroup$
@Hawk If you are working with commutative rings, always be sure to use the [commutative-algebra] tag and/or mention commutativity in the post.
$endgroup$
– rschwieb
Dec 22 '18 at 16:56