183 reputation
11
bio website math.bme.hu/~pintye
location Budapest, Hungary
age 26
visits member for 1 year, 8 months
seen 2 days ago

Whereof one cannot speak, thereof one must be silent.


Mar
22
comment Rational roots of polynomials
$D(f)$ is the complement of $V(f)$. It is the set of those points where $f$ does not vanish.
Mar
22
awarded  Revival
Mar
21
awarded  Informed
Mar
19
awarded  Cleanup
Mar
19
revised Rational roots of polynomials
rolled back to a previous revision
Mar
18
revised Rational roots of polynomials
workaround added
Mar
15
comment Rational roots of polynomials
I managed to prove that for $n=2$, there is no non-empty open subset of $Y$ contained in $\varphi(X)$. Therefore, the method above does not work.
Mar
11
comment Open set in the image of a dominant morphism of affine spaces
Indeed. However, we have $\dim(Y\setminus U)\leq 1$ now. Is it possible (though highly unlikely) that we have a bound independently of $n$ in general?
Mar
10
comment How to show $\phi(X)$ contains a non-empty open set of its closure $\overline{\phi(X)}$?
'This proof doesn't need $k$ algebraically closed'. It does. Take for example $k=\mathbb{Q}$, $X=Y=\mathbb{A}^1_k$ and $\varphi(x)=x^2$. Then $\varphi$ is dominant, yet its image does not contain any non-empty open.
Mar
10
revised Rational roots of polynomials
major mistake found
Mar
10
revised Open set in the image of a dominant morphism of affine spaces
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Mar
10
comment Open set in the image of a dominant morphism of affine spaces
I have just noticed that $k$ has to be algebraically closed, otherwise we might not have any non-empty open set contained in the image. Take for example $k=\mathbb{Q}$, $n=1$ and $\varphi(x)=x^2$. On the other hand, the case $n=1$ is trivial, for then $\varphi$ is surjective.
Mar
10
awarded  Commentator
Mar
10
comment Polynomials with rational zeros
I suspect that the polynomials in question has distinct roots (that is, the discriminant does not vanish). If that is the case, using the rational root test, one can check whether there are enough rational roots at all. As an indication, it seems there are no solutions for $n\geq 4$, and only a few for $n=1,2,3$.
Mar
10
asked Open set in the image of a dominant morphism of affine spaces
Mar
9
revised Rational roots of polynomials
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Mar
9
revised Rational roots of polynomials
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Mar
9
revised Rational roots of polynomials
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Mar
9
answered Rational roots of polynomials
Mar
4
revised Let $\phi : G \rightarrow H$ be a homomorphism from group $G$ to group $H$. Let $g \in G$ and $h \in H$ such that $\phi(g)=h$…?
added 157 characters in body