393 reputation
213
bio website math.bme.hu/~pintye
location Budapest, Hungary
age 26
visits member for 2 years, 4 months
seen Nov 27 at 20:07

Whereof one cannot speak, thereof one must be silent.


Sep
11
awarded  Critic
Aug
23
awarded  Necromancer
Aug
18
asked Direct image of the exceptional divisor along a blow-up
Jul
2
awarded  Curious
Jun
12
answered Matrices and rank inequality
Apr
30
awarded  Yearling
Apr
30
revised Why does $T(n) \leq 2 T(\lceil \frac{n}{2} \rceil)+\mathcal{O}(n)$ imply $T(n)=\mathcal{O} (n\log(n))$
corrected spelling
Apr
30
answered Why does $T(n) \leq 2 T(\lceil \frac{n}{2} \rceil)+\mathcal{O}(n)$ imply $T(n)=\mathcal{O} (n\log(n))$
Apr
30
comment Nilpotent Elements and Intersection of Prime Ideals
Where is the problem? The set of nilpotent elements is the radical of $(0)$ (Fact 1), which is the intersection of those prime ideals that contain $(0)$ (Fact 2), that is, the intersection of all prime ideals. At the same time, this immediately implies that the set in question is an ideal. Mind that you need your ring to be commutative with unity.
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
added 100 characters in body