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Zero set of non-constant polynomial , in more than one variable, over uncountable algebraically closed field
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Let $n>1$, $K$ be an uncountable algebraically closed field and $f(X_1,...,X_n)in K[X_1,...,X_n] $ be a non-constant polynomial. Then it is known that $Z(f):=bar a in K^n : f(bar a)=0$ is infinite. My question is, can $Z(f)$ be countable ?
If $K=mathbb C$, it is known that $Z(f)$ contains a connected (in the Euclidean topology of $mathbb C^n$) subset (corresponding to an irreducible factor of $f$), hence is always uncountable.
algebraic-geometry polynomials commutative-algebra field-theory model-theory
asked Jul 28 at 23:38
user521337
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Let $n>1$, $K$ be an uncountable algebraically closed field and $f(X_1,...,X_n)in K[X_1,...,X_n] $ be a non-constant polynomial. Then it is known that $Z(f):=bar a in K^n : f(bar a)=0$ is infinite. My question is, can $Z(f)$ be countable ?
If $K=mathbb C$, it is known that $Z(f)$ contains a connected (in the Euclidean topology of $mathbb C^n$) subset (corresponding to an irreducible factor of $f$), hence is always uncountable.
algebraic-geometry polynomials commutative-algebra field-theory model-theory
asked Jul 28 at 23:38
user521337
606
if you vary $n-1$ of the coordinates, you get an uncountable family of polynomials in the last coordinate, all of the same degree. probably no two are scalar multiples of eachother except in some degenerate cases.... just some thoughts that may have already occured to you
– Tim kinsella
Jul 28 at 23:53
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $n>1$, $K$ be an uncountable algebraically closed field and $f(X_1,...,X_n)in K[X_1,...,X_n] $ be a non-constant polynomial. Then it is known that $Z(f):=bar a in K^n : f(bar a)=0$ is infinite. My question is, can $Z(f)$ be countable ?
If $K=mathbb C$, it is known that $Z(f)$ contains a connected (in the Euclidean topology of $mathbb C^n$) subset (corresponding to an irreducible factor of $f$), hence is always uncountable.
algebraic-geometry polynomials commutative-algebra field-theory model-theory
asked Jul 28 at 23:38
user521337
606
Let $n>1$, $K$ be an uncountable algebraically closed field and $f(X_1,...,X_n)in K[X_1,...,X_n] $ be a non-constant polynomial. Then it is known that $Z(f):=bar a in K^n : f(bar a)=0$ is infinite. My question is, can $Z(f)$ be countable ?
If $K=mathbb C$, it is known that $Z(f)$ contains a connected (in the Euclidean topology of $mathbb C^n$) subset (corresponding to an irreducible factor of $f$), hence is always uncountable.
algebraic-geometry polynomials commutative-algebra field-theory model-theory
asked Jul 28 at 23:38
user521337
606
asked Jul 28 at 23:38
user521337
606
asked Jul 28 at 23:38
user521337
606
asked Jul 28 at 23:38
user521337
606
606
if you vary $n-1$ of the coordinates, you get an uncountable family of polynomials in the last coordinate, all of the same degree. probably no two are scalar multiples of eachother except in some degenerate cases.... just some thoughts that may have already occured to you
– Tim kinsella
Jul 28 at 23:53
add a comment |Â
if you vary $n-1$ of the coordinates, you get an uncountable family of polynomials in the last coordinate, all of the same degree. probably no two are scalar multiples of eachother except in some degenerate cases.... just some thoughts that may have already occured to you
– Tim kinsella
Jul 28 at 23:53
if you vary $n-1$ of the coordinates, you get an uncountable family of polynomials in the last coordinate, all of the same degree. probably no two are scalar multiples of eachother except in some degenerate cases.... just some thoughts that may have already occured to you
– Tim kinsella
Jul 28 at 23:53
if you vary $n-1$ of the coordinates, you get an uncountable family of polynomials in the last coordinate, all of the same degree. probably no two are scalar multiples of eachother except in some degenerate cases.... just some thoughts that may have already occured to you
– Tim kinsella
Jul 28 at 23:53
add a comment |Â
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There are tons and tons of ways to prove this. Here's probably the lowest-tech way.
Since $f$ is nonconstant, some variable appears in $f$; suppose without loss of generality that it is $X_1$. Consider $f$ as a polynomial in $X_1,dots,X_n-1$ with coefficients in $K[X_n]$. Since $X_1$ appears in $f$, this polynomial with coefficients in $K[X_n]$ is nonconstant. For all but finitely many values of $ain K$, then, $f(X_1,dots,X_n-1,a)$ will still be nonconstant (since $a$ just needs to not be a simultaneous root of all the coefficients besides the constant term). Since such a nonconstant polynomial has a root, this gives uncountably many roots of $f$.
(Or, if you don't know that a nonconstant polynomial has a root, you can continue this argument by induction on $n$ to prove it.)
answered Jul 29 at 0:13
Eric Wofsey
162k12188298
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
There are tons and tons of ways to prove this. Here's probably the lowest-tech way.
Since $f$ is nonconstant, some variable appears in $f$; suppose without loss of generality that it is $X_1$. Consider $f$ as a polynomial in $X_1,dots,X_n-1$ with coefficients in $K[X_n]$. Since $X_1$ appears in $f$, this polynomial with coefficients in $K[X_n]$ is nonconstant. For all but finitely many values of $ain K$, then, $f(X_1,dots,X_n-1,a)$ will still be nonconstant (since $a$ just needs to not be a simultaneous root of all the coefficients besides the constant term). Since such a nonconstant polynomial has a root, this gives uncountably many roots of $f$.
(Or, if you don't know that a nonconstant polynomial has a root, you can continue this argument by induction on $n$ to prove it.)
answered Jul 29 at 0:13
Eric Wofsey
162k12188298
add a comment |Â
up vote
1
down vote
There are tons and tons of ways to prove this. Here's probably the lowest-tech way.
Since $f$ is nonconstant, some variable appears in $f$; suppose without loss of generality that it is $X_1$. Consider $f$ as a polynomial in $X_1,dots,X_n-1$ with coefficients in $K[X_n]$. Since $X_1$ appears in $f$, this polynomial with coefficients in $K[X_n]$ is nonconstant. For all but finitely many values of $ain K$, then, $f(X_1,dots,X_n-1,a)$ will still be nonconstant (since $a$ just needs to not be a simultaneous root of all the coefficients besides the constant term). Since such a nonconstant polynomial has a root, this gives uncountably many roots of $f$.
(Or, if you don't know that a nonconstant polynomial has a root, you can continue this argument by induction on $n$ to prove it.)
answered Jul 29 at 0:13
Eric Wofsey
162k12188298
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There are tons and tons of ways to prove this. Here's probably the lowest-tech way.
Since $f$ is nonconstant, some variable appears in $f$; suppose without loss of generality that it is $X_1$. Consider $f$ as a polynomial in $X_1,dots,X_n-1$ with coefficients in $K[X_n]$. Since $X_1$ appears in $f$, this polynomial with coefficients in $K[X_n]$ is nonconstant. For all but finitely many values of $ain K$, then, $f(X_1,dots,X_n-1,a)$ will still be nonconstant (since $a$ just needs to not be a simultaneous root of all the coefficients besides the constant term). Since such a nonconstant polynomial has a root, this gives uncountably many roots of $f$.
(Or, if you don't know that a nonconstant polynomial has a root, you can continue this argument by induction on $n$ to prove it.)
answered Jul 29 at 0:13
Eric Wofsey
162k12188298
There are tons and tons of ways to prove this. Here's probably the lowest-tech way.
Since $f$ is nonconstant, some variable appears in $f$; suppose without loss of generality that it is $X_1$. Consider $f$ as a polynomial in $X_1,dots,X_n-1$ with coefficients in $K[X_n]$. Since $X_1$ appears in $f$, this polynomial with coefficients in $K[X_n]$ is nonconstant. For all but finitely many values of $ain K$, then, $f(X_1,dots,X_n-1,a)$ will still be nonconstant (since $a$ just needs to not be a simultaneous root of all the coefficients besides the constant term). Since such a nonconstant polynomial has a root, this gives uncountably many roots of $f$.
(Or, if you don't know that a nonconstant polynomial has a root, you can continue this argument by induction on $n$ to prove it.)
answered Jul 29 at 0:13
Eric Wofsey
162k12188298
answered Jul 29 at 0:13
Eric Wofsey
162k12188298
answered Jul 29 at 0:13
Eric Wofsey
162k12188298
answered Jul 29 at 0:13
Eric Wofsey
162k12188298
162k12188298
add a comment |Â
add a comment |Â
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if you vary $n-1$ of the coordinates, you get an uncountable family of polynomials in the last coordinate, all of the same degree. probably no two are scalar multiples of eachother except in some degenerate cases.... just some thoughts that may have already occured to you
– Tim kinsella
Jul 28 at 23:53