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Results tagged with real-algebraic-geometry
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user 85592
Real algebraic geometry is the study of real solutions to algebraic equations with real coefficients. Its methods are rather different from classical algebraic geometry, which is typically done over an algebraically closed field (like the complex numbers).
4
votes
Accepted
Is the Segre embedding of two real varieties a real variety?
$\mathrm{Seg}(X\times Y)$ is a real projective variety since the full Segre map is an isomorphism of real algebraic varieties onto its image.
As for your second question, I think the answer is "no". I …
- 1,424
4
votes
Accepted
"Real algebraic varieties" vs finite type separated reduced $\mathbb{R}$-schemes with dense ...
As for your first question, concerning nonaffine R-varieties as you call them, yes, there are nonaffine R-varieties. However, they are considered pathological. Example 12.1.5 on page 301
of Bochnak-Co …
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1
vote
Accepted
Solutions to a system of homogeneous equations (inequalities)
I cannot say much about $\geq0$. For $=0$ one can prove the following statement: if $r\leq n-1$ then then there is an $a\in \mathbf R^n\setminus\{0\}$ such that $f_i(a)=0$ for all $i$.
Indeed, each p …
- 1,424
2
votes
Are continuous rational functions arc-analytic?
If $X$ is all of $\mathbf R^n$ then the answer to the first question is "yes", I think. Indeed, for any continuous rational function $f$ on $\mathbf R^n$ there is a stratification of $\mathbf R^n$ in …
- 1,424
1
vote
Smooth, irreducible surface with real part containing two projective planes
Start with a fibration in conics $Y$ given by the affine equation $$x^2+y^2=-(t+2)(t+1)(t-1)(t-2).$$ It is clear that the set of real points $Y(\mathbf R)$ is a disjoint union of two spheres. Now, blo …
- 1,424
3
votes
Accepted
ideal generated from a truncated "real radical-like" set are still real radical?
I think the answer is no. Let us take $t=3$ and $I_3$ the $1$-dimensional subspace of $\mathbf R[X]_3$ generated by $h=x(x^2+y^2)$. Then $I_3$ satisfies the conditions 1-3. The ideal $I$ in $\mathbf R …
- 1,424
2
votes
Accepted
Decay of real continuous algebraic functions at infinity
Your function $f$ is, in particular, a continuous semi-algebraic function on $\mathbf R^n$, i.e., its graph is a semi-algebraic subset of $\mathbf R^{n+1}$. Such functions are known to have sub-polyno …
- 1,424