The krull dimension of $BbbZ$ and artinian rings The Next CEO of Stack OverflowWhy are Artinian rings of Krull dimension 0?Krull DimensionWhy do Artinian rings have dimension 0 and not 1?Artinian rings have finite lengthKrull Dimension of Direct Limits of Zero-Dimensional RingsPreservation of Krull dimension under inverse limitlocal-global rings and its examplesWhy is this a corollary of [Noetherian of Krull dimension 0 iff Artinian]?Why intermediate rings of Krull have dimension $leq 1$?On “homogeneous” height and “homogeneous” Krull-dimension?
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The krull dimension of $BbbZ$ and artinian rings
The Next CEO of Stack OverflowWhy are Artinian rings of Krull dimension 0?Krull DimensionWhy do Artinian rings have dimension 0 and not 1?Artinian rings have finite lengthKrull Dimension of Direct Limits of Zero-Dimensional RingsPreservation of Krull dimension under inverse limitlocal-global rings and its examplesWhy is this a corollary of [Noetherian of Krull dimension 0 iff Artinian]?Why intermediate rings of Krull have dimension $leq 1$?On “homogeneous” height and “homogeneous” Krull-dimension?
$begingroup$
On page thirty of Matsumura, it says that $BbbZ$ has krull dimension 1 because every prime ideal is maximal. I understand this because for any prime p you have $0 subset p$.
However, for artinian rings, it says that the dimesion is zero because every prime ideal is maximal. This is where I'm confused. As above, we have $0 subset p$ for any nonzero prime ideal...but then why isn't he dimension 1?
abstract-algebra ring-theory commutative-algebra
asked Oct 15 '14 at 15:07
ArtusArtus
452612
$endgroup$
add a comment |
$begingroup$
On page thirty of Matsumura, it says that $BbbZ$ has krull dimension 1 because every prime ideal is maximal. I understand this because for any prime p you have $0 subset p$.
However, for artinian rings, it says that the dimesion is zero because every prime ideal is maximal. This is where I'm confused. As above, we have $0 subset p$ for any nonzero prime ideal...but then why isn't he dimension 1?
abstract-algebra ring-theory commutative-algebra
asked Oct 15 '14 at 15:07
ArtusArtus
452612
$endgroup$
1
$begingroup$
"...because every nonzero prime ideal is maximal..."
$endgroup$
– Orest Bucicovschi
Oct 15 '14 at 15:24
1
$begingroup$
Note that $mathbbZ$ is not artinian. See Thomas Andrews' answer below.
$endgroup$
– anomaly
Oct 15 '14 at 15:37
add a comment |
$begingroup$
On page thirty of Matsumura, it says that $BbbZ$ has krull dimension 1 because every prime ideal is maximal. I understand this because for any prime p you have $0 subset p$.
However, for artinian rings, it says that the dimesion is zero because every prime ideal is maximal. This is where I'm confused. As above, we have $0 subset p$ for any nonzero prime ideal...but then why isn't he dimension 1?
abstract-algebra ring-theory commutative-algebra
asked Oct 15 '14 at 15:07
ArtusArtus
452612
$endgroup$
On page thirty of Matsumura, it says that $BbbZ$ has krull dimension 1 because every prime ideal is maximal. I understand this because for any prime p you have $0 subset p$.
However, for artinian rings, it says that the dimesion is zero because every prime ideal is maximal. This is where I'm confused. As above, we have $0 subset p$ for any nonzero prime ideal...but then why isn't he dimension 1?
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
asked Oct 15 '14 at 15:07
ArtusArtus
452612
asked Oct 15 '14 at 15:07
ArtusArtus
452612
asked Oct 15 '14 at 15:07
ArtusArtus
452612
asked Oct 15 '14 at 15:07
ArtusArtus
452612
asked Oct 15 '14 at 15:07
ArtusArtus
452612
452612
1
$begingroup$
"...because every nonzero prime ideal is maximal..."
$endgroup$
– Orest Bucicovschi
Oct 15 '14 at 15:24
1
$begingroup$
Note that $mathbbZ$ is not artinian. See Thomas Andrews' answer below.
$endgroup$
– anomaly
Oct 15 '14 at 15:37
add a comment |
1
$begingroup$
"...because every nonzero prime ideal is maximal..."
$endgroup$
– Orest Bucicovschi
Oct 15 '14 at 15:24
1
$begingroup$
Note that $mathbbZ$ is not artinian. See Thomas Andrews' answer below.
$endgroup$
– anomaly
Oct 15 '14 at 15:37
1
1
$begingroup$
"...because every nonzero prime ideal is maximal..."
$endgroup$
– Orest Bucicovschi
Oct 15 '14 at 15:24
$begingroup$
"...because every nonzero prime ideal is maximal..."
$endgroup$
– Orest Bucicovschi
Oct 15 '14 at 15:24
1
1
$begingroup$
Note that $mathbbZ$ is not artinian. See Thomas Andrews' answer below.
$endgroup$
– anomaly
Oct 15 '14 at 15:37
$begingroup$
Note that $mathbbZ$ is not artinian. See Thomas Andrews' answer below.
$endgroup$
– anomaly
Oct 15 '14 at 15:37
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Is $0$ a prime ideal in all rings? If $0$ is a prime ideal, and it is maximal, is there any other prime ideal?
answered Oct 15 '14 at 15:13
Thomas AndrewsThomas Andrews
130k12147298
$endgroup$
add a comment |
$begingroup$
Every prime ideal $mathfrak p$ in an Artinian ring $R$ is maximal. To see this, let $x in R / mathfrak p$, $x ne 0$. Consider the descending chain $(x) supset (x^2) supset cdots$ to show that $x$ is invertible.
On the other hand, $mathbb Z$ is a principal ideal domain. Every nonzero prime ideal in a PID is maximal. The word "nonzero" here is the crucial difference.
It follows that $mathbb Z$ has dimension $1$ whereas Artinian rings have dimension $0$.
answered Oct 15 '14 at 17:13
Ayman HouriehAyman Hourieh
31.3k466119
$endgroup$
add a comment |
$begingroup$
Fields are obviously artinian, as they have only trivial ideals.
Suppose $R$ is an artinian ring which is not artinian and let $xin R$ an element which is neither $0$ nor invertible. Consider the descending chain of ideals
$$
(x)supset(x^2)supsetcdotssupset(x^n)supset(x^n+1)supset cdots.
$$
The chain stabilizes by assumption.
If $(x^n)=(0)$ eventually, $x$ is a nilpotent element.
If $(x^n)=(x^n+1)neq(0)$ eventually, we can write $x^n=yx^n+1$ for some $yin R$ from which we get $x^n(xy-1)=0$, i.e. $x$ is a $0$-divisor.
In either case $R$ is not a domain, hence the ideal $(0)$ isn't prime.
answered Mar 19 at 9:10
Andrea MoriAndrea Mori
20.1k13466
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Is $0$ a prime ideal in all rings? If $0$ is a prime ideal, and it is maximal, is there any other prime ideal?
answered Oct 15 '14 at 15:13
Thomas AndrewsThomas Andrews
130k12147298
$endgroup$
add a comment |
$begingroup$
Hint: Is $0$ a prime ideal in all rings? If $0$ is a prime ideal, and it is maximal, is there any other prime ideal?
answered Oct 15 '14 at 15:13
Thomas AndrewsThomas Andrews
130k12147298
$endgroup$
add a comment |
$begingroup$
Hint: Is $0$ a prime ideal in all rings? If $0$ is a prime ideal, and it is maximal, is there any other prime ideal?
answered Oct 15 '14 at 15:13
Thomas AndrewsThomas Andrews
130k12147298
$endgroup$
Hint: Is $0$ a prime ideal in all rings? If $0$ is a prime ideal, and it is maximal, is there any other prime ideal?
answered Oct 15 '14 at 15:13
Thomas AndrewsThomas Andrews
130k12147298
answered Oct 15 '14 at 15:13
Thomas AndrewsThomas Andrews
130k12147298
answered Oct 15 '14 at 15:13
Thomas AndrewsThomas Andrews
130k12147298
answered Oct 15 '14 at 15:13
Thomas AndrewsThomas Andrews
130k12147298
130k12147298
add a comment |
add a comment |
$begingroup$
Every prime ideal $mathfrak p$ in an Artinian ring $R$ is maximal. To see this, let $x in R / mathfrak p$, $x ne 0$. Consider the descending chain $(x) supset (x^2) supset cdots$ to show that $x$ is invertible.
On the other hand, $mathbb Z$ is a principal ideal domain. Every nonzero prime ideal in a PID is maximal. The word "nonzero" here is the crucial difference.
It follows that $mathbb Z$ has dimension $1$ whereas Artinian rings have dimension $0$.
answered Oct 15 '14 at 17:13
Ayman HouriehAyman Hourieh
31.3k466119
$endgroup$
add a comment |
$begingroup$
Every prime ideal $mathfrak p$ in an Artinian ring $R$ is maximal. To see this, let $x in R / mathfrak p$, $x ne 0$. Consider the descending chain $(x) supset (x^2) supset cdots$ to show that $x$ is invertible.
On the other hand, $mathbb Z$ is a principal ideal domain. Every nonzero prime ideal in a PID is maximal. The word "nonzero" here is the crucial difference.
It follows that $mathbb Z$ has dimension $1$ whereas Artinian rings have dimension $0$.
answered Oct 15 '14 at 17:13
Ayman HouriehAyman Hourieh
31.3k466119
$endgroup$
add a comment |
$begingroup$
Every prime ideal $mathfrak p$ in an Artinian ring $R$ is maximal. To see this, let $x in R / mathfrak p$, $x ne 0$. Consider the descending chain $(x) supset (x^2) supset cdots$ to show that $x$ is invertible.
On the other hand, $mathbb Z$ is a principal ideal domain. Every nonzero prime ideal in a PID is maximal. The word "nonzero" here is the crucial difference.
It follows that $mathbb Z$ has dimension $1$ whereas Artinian rings have dimension $0$.
answered Oct 15 '14 at 17:13
Ayman HouriehAyman Hourieh
31.3k466119
$endgroup$
Every prime ideal $mathfrak p$ in an Artinian ring $R$ is maximal. To see this, let $x in R / mathfrak p$, $x ne 0$. Consider the descending chain $(x) supset (x^2) supset cdots$ to show that $x$ is invertible.
On the other hand, $mathbb Z$ is a principal ideal domain. Every nonzero prime ideal in a PID is maximal. The word "nonzero" here is the crucial difference.
It follows that $mathbb Z$ has dimension $1$ whereas Artinian rings have dimension $0$.
answered Oct 15 '14 at 17:13
Ayman HouriehAyman Hourieh
31.3k466119
answered Oct 15 '14 at 17:13
Ayman HouriehAyman Hourieh
31.3k466119
answered Oct 15 '14 at 17:13
Ayman HouriehAyman Hourieh
31.3k466119
answered Oct 15 '14 at 17:13
Ayman HouriehAyman Hourieh
31.3k466119
31.3k466119
add a comment |
add a comment |
$begingroup$
Fields are obviously artinian, as they have only trivial ideals.
Suppose $R$ is an artinian ring which is not artinian and let $xin R$ an element which is neither $0$ nor invertible. Consider the descending chain of ideals
$$
(x)supset(x^2)supsetcdotssupset(x^n)supset(x^n+1)supset cdots.
$$
The chain stabilizes by assumption.
If $(x^n)=(0)$ eventually, $x$ is a nilpotent element.
If $(x^n)=(x^n+1)neq(0)$ eventually, we can write $x^n=yx^n+1$ for some $yin R$ from which we get $x^n(xy-1)=0$, i.e. $x$ is a $0$-divisor.
In either case $R$ is not a domain, hence the ideal $(0)$ isn't prime.
answered Mar 19 at 9:10
Andrea MoriAndrea Mori
20.1k13466
$endgroup$
add a comment |
$begingroup$
Fields are obviously artinian, as they have only trivial ideals.
Suppose $R$ is an artinian ring which is not artinian and let $xin R$ an element which is neither $0$ nor invertible. Consider the descending chain of ideals
$$
(x)supset(x^2)supsetcdotssupset(x^n)supset(x^n+1)supset cdots.
$$
The chain stabilizes by assumption.
If $(x^n)=(0)$ eventually, $x$ is a nilpotent element.
If $(x^n)=(x^n+1)neq(0)$ eventually, we can write $x^n=yx^n+1$ for some $yin R$ from which we get $x^n(xy-1)=0$, i.e. $x$ is a $0$-divisor.
In either case $R$ is not a domain, hence the ideal $(0)$ isn't prime.
answered Mar 19 at 9:10
Andrea MoriAndrea Mori
20.1k13466
$endgroup$
add a comment |
$begingroup$
Fields are obviously artinian, as they have only trivial ideals.
Suppose $R$ is an artinian ring which is not artinian and let $xin R$ an element which is neither $0$ nor invertible. Consider the descending chain of ideals
$$
(x)supset(x^2)supsetcdotssupset(x^n)supset(x^n+1)supset cdots.
$$
The chain stabilizes by assumption.
If $(x^n)=(0)$ eventually, $x$ is a nilpotent element.
If $(x^n)=(x^n+1)neq(0)$ eventually, we can write $x^n=yx^n+1$ for some $yin R$ from which we get $x^n(xy-1)=0$, i.e. $x$ is a $0$-divisor.
In either case $R$ is not a domain, hence the ideal $(0)$ isn't prime.
answered Mar 19 at 9:10
Andrea MoriAndrea Mori
20.1k13466
$endgroup$
Fields are obviously artinian, as they have only trivial ideals.
Suppose $R$ is an artinian ring which is not artinian and let $xin R$ an element which is neither $0$ nor invertible. Consider the descending chain of ideals
$$
(x)supset(x^2)supsetcdotssupset(x^n)supset(x^n+1)supset cdots.
$$
The chain stabilizes by assumption.
If $(x^n)=(0)$ eventually, $x$ is a nilpotent element.
If $(x^n)=(x^n+1)neq(0)$ eventually, we can write $x^n=yx^n+1$ for some $yin R$ from which we get $x^n(xy-1)=0$, i.e. $x$ is a $0$-divisor.
In either case $R$ is not a domain, hence the ideal $(0)$ isn't prime.
answered Mar 19 at 9:10
Andrea MoriAndrea Mori
20.1k13466
answered Mar 19 at 9:10
Andrea MoriAndrea Mori
20.1k13466
answered Mar 19 at 9:10
Andrea MoriAndrea Mori
20.1k13466
answered Mar 19 at 9:10
Andrea MoriAndrea Mori
20.1k13466
20.1k13466
add a comment |
add a comment |
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$begingroup$
"...because every nonzero prime ideal is maximal..."
$endgroup$
– Orest Bucicovschi
Oct 15 '14 at 15:24
1
$begingroup$
Note that $mathbbZ$ is not artinian. See Thomas Andrews' answer below.
$endgroup$
– anomaly
Oct 15 '14 at 15:37