Show that $k[x,y,z]/(xz-y^2)$ is not a UFD.
$begingroup$
I am trying to show that $k[x,y,z]/(xz-y^2)notcong k[x,y]$.
The latter is a UFD, so I am trying to show the former is not. Clearly $x$ is not prime, since $xmid xz$ which implies $xmid y^2$, but $xmid y$.
So if I can show that $x$ is irreducible then $k[x,y,z]/(xz-y^2)$ is a not a UFD, because irreducible implies prime in UFDs.
Attempt: Assume $x$ is reduced into $ab$ in $k[x,y,z]/(xz-y^2)$, then we know there exists $cin k[x,y,z]$ such that $x=ab+c(xz-y^2)$ in $k[x,y,z]$, or more usefully as $x-ab=(xz-y^2)c$. Someone suggested that I write $a,b$ such they are degree at most 1 in $y$, that is, replace every $y^n$ using $xz$. This means the LHS is degree at most 2 in $y$ and the RHS is degree at least two in $y$ in $x-ab=(xz-y^2)c$. I am not sure where to go from here just by knowing that $c$ has no $y$ terms though. Ultimately I want to show that this forces $x=ab$ in $k[x,y,z]$ which, since $x$ is irreducible there, shows that either $a,b$ is a unit, which means $x$ is irreducible in $k[x,y,z]/(xz-y^2)$.
ring-theory unique-factorization-domains
edited Jul 29 '16 at 16:07
user26857
39.5k124283
asked Apr 3 '13 at 20:50
Steven-OwenSteven-Owen
2,42052047
$endgroup$
add a comment |
$begingroup$
I am trying to show that $k[x,y,z]/(xz-y^2)notcong k[x,y]$.
The latter is a UFD, so I am trying to show the former is not. Clearly $x$ is not prime, since $xmid xz$ which implies $xmid y^2$, but $xmid y$.
So if I can show that $x$ is irreducible then $k[x,y,z]/(xz-y^2)$ is a not a UFD, because irreducible implies prime in UFDs.
Attempt: Assume $x$ is reduced into $ab$ in $k[x,y,z]/(xz-y^2)$, then we know there exists $cin k[x,y,z]$ such that $x=ab+c(xz-y^2)$ in $k[x,y,z]$, or more usefully as $x-ab=(xz-y^2)c$. Someone suggested that I write $a,b$ such they are degree at most 1 in $y$, that is, replace every $y^n$ using $xz$. This means the LHS is degree at most 2 in $y$ and the RHS is degree at least two in $y$ in $x-ab=(xz-y^2)c$. I am not sure where to go from here just by knowing that $c$ has no $y$ terms though. Ultimately I want to show that this forces $x=ab$ in $k[x,y,z]$ which, since $x$ is irreducible there, shows that either $a,b$ is a unit, which means $x$ is irreducible in $k[x,y,z]/(xz-y^2)$.
ring-theory unique-factorization-domains
edited Jul 29 '16 at 16:07
user26857
39.5k124283
asked Apr 3 '13 at 20:50
Steven-OwenSteven-Owen
2,42052047
$endgroup$
add a comment |
$begingroup$
I am trying to show that $k[x,y,z]/(xz-y^2)notcong k[x,y]$.
The latter is a UFD, so I am trying to show the former is not. Clearly $x$ is not prime, since $xmid xz$ which implies $xmid y^2$, but $xmid y$.
So if I can show that $x$ is irreducible then $k[x,y,z]/(xz-y^2)$ is a not a UFD, because irreducible implies prime in UFDs.
Attempt: Assume $x$ is reduced into $ab$ in $k[x,y,z]/(xz-y^2)$, then we know there exists $cin k[x,y,z]$ such that $x=ab+c(xz-y^2)$ in $k[x,y,z]$, or more usefully as $x-ab=(xz-y^2)c$. Someone suggested that I write $a,b$ such they are degree at most 1 in $y$, that is, replace every $y^n$ using $xz$. This means the LHS is degree at most 2 in $y$ and the RHS is degree at least two in $y$ in $x-ab=(xz-y^2)c$. I am not sure where to go from here just by knowing that $c$ has no $y$ terms though. Ultimately I want to show that this forces $x=ab$ in $k[x,y,z]$ which, since $x$ is irreducible there, shows that either $a,b$ is a unit, which means $x$ is irreducible in $k[x,y,z]/(xz-y^2)$.
ring-theory unique-factorization-domains
edited Jul 29 '16 at 16:07
user26857
39.5k124283
asked Apr 3 '13 at 20:50
Steven-OwenSteven-Owen
2,42052047
$endgroup$
I am trying to show that $k[x,y,z]/(xz-y^2)notcong k[x,y]$.
The latter is a UFD, so I am trying to show the former is not. Clearly $x$ is not prime, since $xmid xz$ which implies $xmid y^2$, but $xmid y$.
So if I can show that $x$ is irreducible then $k[x,y,z]/(xz-y^2)$ is a not a UFD, because irreducible implies prime in UFDs.
Attempt: Assume $x$ is reduced into $ab$ in $k[x,y,z]/(xz-y^2)$, then we know there exists $cin k[x,y,z]$ such that $x=ab+c(xz-y^2)$ in $k[x,y,z]$, or more usefully as $x-ab=(xz-y^2)c$. Someone suggested that I write $a,b$ such they are degree at most 1 in $y$, that is, replace every $y^n$ using $xz$. This means the LHS is degree at most 2 in $y$ and the RHS is degree at least two in $y$ in $x-ab=(xz-y^2)c$. I am not sure where to go from here just by knowing that $c$ has no $y$ terms though. Ultimately I want to show that this forces $x=ab$ in $k[x,y,z]$ which, since $x$ is irreducible there, shows that either $a,b$ is a unit, which means $x$ is irreducible in $k[x,y,z]/(xz-y^2)$.
ring-theory unique-factorization-domains
ring-theory unique-factorization-domains
edited Jul 29 '16 at 16:07
user26857
39.5k124283
asked Apr 3 '13 at 20:50
Steven-OwenSteven-Owen
2,42052047
edited Jul 29 '16 at 16:07
user26857
39.5k124283
asked Apr 3 '13 at 20:50
Steven-OwenSteven-Owen
2,42052047
edited Jul 29 '16 at 16:07
user26857
39.5k124283
edited Jul 29 '16 at 16:07
user26857
39.5k124283
edited Jul 29 '16 at 16:07
user26857
39.5k124283
39.5k124283
asked Apr 3 '13 at 20:50
Steven-OwenSteven-Owen
2,42052047
asked Apr 3 '13 at 20:50
Steven-OwenSteven-Owen
2,42052047
asked Apr 3 '13 at 20:50
Steven-OwenSteven-Owen
2,42052047
2,42052047
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's easier just to observe that $xz=y^2$ in your ring gives a decomposition into irreducibles that is not unique ($x$, $y$, and $z$ are irreducible because they are of degree one; your quotient is homogeneous). Of course, this also shows $x$ is not prime since it divides $y^2=y y$ but divides neither $y$ nor $y$.
answered Apr 3 '13 at 20:57
StephenStephen
10.7k12439
$endgroup$
add a comment |
$begingroup$
The algebra $k[x,y,z]/(xz-y^2)$ is a graded domain, with (the classes of) $x$, $y$ and $z$ all in degree $1$.
In a graded domain, the product of two non-homogeneous elements is non-homogeneous. So if $x$ is a product of two elements, these must be homogeneous. Since $x$ has degree $1$, it can only be a product of an element of degree $1$ and an element of degree $0$. Since the only elements of degree zero are units, we are done.
answered Apr 3 '13 at 20:58
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
$endgroup$
$begingroup$
For OP: If you're wondering how Mariano deduced that $xz - y^2$ is an irreducible polynomial in $k[x,y,z]$, you can apply Eisenstein's criterion with the prime element $x$ to $xz - y^2$ considered as a polynomial in $y$ with coefficients in $k[x,z]$.
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– user38268
Apr 3 '13 at 22:31
add a comment |
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2 Answers
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2 Answers
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active
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active
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votes
$begingroup$
It's easier just to observe that $xz=y^2$ in your ring gives a decomposition into irreducibles that is not unique ($x$, $y$, and $z$ are irreducible because they are of degree one; your quotient is homogeneous). Of course, this also shows $x$ is not prime since it divides $y^2=y y$ but divides neither $y$ nor $y$.
answered Apr 3 '13 at 20:57
StephenStephen
10.7k12439
$endgroup$
add a comment |
$begingroup$
It's easier just to observe that $xz=y^2$ in your ring gives a decomposition into irreducibles that is not unique ($x$, $y$, and $z$ are irreducible because they are of degree one; your quotient is homogeneous). Of course, this also shows $x$ is not prime since it divides $y^2=y y$ but divides neither $y$ nor $y$.
answered Apr 3 '13 at 20:57
StephenStephen
10.7k12439
$endgroup$
add a comment |
$begingroup$
It's easier just to observe that $xz=y^2$ in your ring gives a decomposition into irreducibles that is not unique ($x$, $y$, and $z$ are irreducible because they are of degree one; your quotient is homogeneous). Of course, this also shows $x$ is not prime since it divides $y^2=y y$ but divides neither $y$ nor $y$.
answered Apr 3 '13 at 20:57
StephenStephen
10.7k12439
$endgroup$
It's easier just to observe that $xz=y^2$ in your ring gives a decomposition into irreducibles that is not unique ($x$, $y$, and $z$ are irreducible because they are of degree one; your quotient is homogeneous). Of course, this also shows $x$ is not prime since it divides $y^2=y y$ but divides neither $y$ nor $y$.
answered Apr 3 '13 at 20:57
StephenStephen
10.7k12439
answered Apr 3 '13 at 20:57
StephenStephen
10.7k12439
answered Apr 3 '13 at 20:57
StephenStephen
10.7k12439
answered Apr 3 '13 at 20:57
StephenStephen
10.7k12439
10.7k12439
add a comment |
add a comment |
$begingroup$
The algebra $k[x,y,z]/(xz-y^2)$ is a graded domain, with (the classes of) $x$, $y$ and $z$ all in degree $1$.
In a graded domain, the product of two non-homogeneous elements is non-homogeneous. So if $x$ is a product of two elements, these must be homogeneous. Since $x$ has degree $1$, it can only be a product of an element of degree $1$ and an element of degree $0$. Since the only elements of degree zero are units, we are done.
answered Apr 3 '13 at 20:58
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
$endgroup$
$begingroup$
For OP: If you're wondering how Mariano deduced that $xz - y^2$ is an irreducible polynomial in $k[x,y,z]$, you can apply Eisenstein's criterion with the prime element $x$ to $xz - y^2$ considered as a polynomial in $y$ with coefficients in $k[x,z]$.
$endgroup$
– user38268
Apr 3 '13 at 22:31
add a comment |
$begingroup$
The algebra $k[x,y,z]/(xz-y^2)$ is a graded domain, with (the classes of) $x$, $y$ and $z$ all in degree $1$.
In a graded domain, the product of two non-homogeneous elements is non-homogeneous. So if $x$ is a product of two elements, these must be homogeneous. Since $x$ has degree $1$, it can only be a product of an element of degree $1$ and an element of degree $0$. Since the only elements of degree zero are units, we are done.
answered Apr 3 '13 at 20:58
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
$endgroup$
$begingroup$
For OP: If you're wondering how Mariano deduced that $xz - y^2$ is an irreducible polynomial in $k[x,y,z]$, you can apply Eisenstein's criterion with the prime element $x$ to $xz - y^2$ considered as a polynomial in $y$ with coefficients in $k[x,z]$.
$endgroup$
– user38268
Apr 3 '13 at 22:31
add a comment |
$begingroup$
The algebra $k[x,y,z]/(xz-y^2)$ is a graded domain, with (the classes of) $x$, $y$ and $z$ all in degree $1$.
In a graded domain, the product of two non-homogeneous elements is non-homogeneous. So if $x$ is a product of two elements, these must be homogeneous. Since $x$ has degree $1$, it can only be a product of an element of degree $1$ and an element of degree $0$. Since the only elements of degree zero are units, we are done.
answered Apr 3 '13 at 20:58
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
$endgroup$
The algebra $k[x,y,z]/(xz-y^2)$ is a graded domain, with (the classes of) $x$, $y$ and $z$ all in degree $1$.
In a graded domain, the product of two non-homogeneous elements is non-homogeneous. So if $x$ is a product of two elements, these must be homogeneous. Since $x$ has degree $1$, it can only be a product of an element of degree $1$ and an element of degree $0$. Since the only elements of degree zero are units, we are done.
answered Apr 3 '13 at 20:58
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
answered Apr 3 '13 at 20:58
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
answered Apr 3 '13 at 20:58
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
answered Apr 3 '13 at 20:58
Mariano Suárez-ÁlvarezMariano Suárez-Álvarez
112k7158290
112k7158290
$begingroup$
For OP: If you're wondering how Mariano deduced that $xz - y^2$ is an irreducible polynomial in $k[x,y,z]$, you can apply Eisenstein's criterion with the prime element $x$ to $xz - y^2$ considered as a polynomial in $y$ with coefficients in $k[x,z]$.
$endgroup$
– user38268
Apr 3 '13 at 22:31
add a comment |
$begingroup$
For OP: If you're wondering how Mariano deduced that $xz - y^2$ is an irreducible polynomial in $k[x,y,z]$, you can apply Eisenstein's criterion with the prime element $x$ to $xz - y^2$ considered as a polynomial in $y$ with coefficients in $k[x,z]$.
$endgroup$
– user38268
Apr 3 '13 at 22:31
$begingroup$
For OP: If you're wondering how Mariano deduced that $xz - y^2$ is an irreducible polynomial in $k[x,y,z]$, you can apply Eisenstein's criterion with the prime element $x$ to $xz - y^2$ considered as a polynomial in $y$ with coefficients in $k[x,z]$.
$endgroup$
– user38268
Apr 3 '13 at 22:31
$begingroup$
For OP: If you're wondering how Mariano deduced that $xz - y^2$ is an irreducible polynomial in $k[x,y,z]$, you can apply Eisenstein's criterion with the prime element $x$ to $xz - y^2$ considered as a polynomial in $y$ with coefficients in $k[x,z]$.
$endgroup$
– user38268
Apr 3 '13 at 22:31
add a comment |
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