Question: If I have a non-zero polynomial $P(x_1,\ldots,x_n)$ defined on $\mathbb R^n$, is it true that the zero set $$Z(P) = \{x \in \mathbb R^n: P(x_1,\ldots,x_n) = 0\}$$ can be written as finite union of smooth submanifolds of $\mathbb R^n$?  This is probably a very basic thing of real algebraic geometry, but I'm totally ignorant in the area. If the response is affirmative, can anyone provide a reference?
123516
Can we write every semi-algebraic set $A \subset \mathbb{R}^m$ as the disjoint union of finitely many Nash submanifolds of $\mathbb{R}^m$? By Nash submanifolds are meant those manifolds that are smooth and semi-algebraic.
53199
Suppose that $P$ is a monic polynomial of degree $n$ in one variable with real coefficients and $K$ is a real number. Then which of the following statements are necessarily correct ?         If $n$ is even and $K&gt;0$ then there exists $x_0\in \mathbb R$ such that $P(x_0)=Ke^{x_0}$.   If $n$ is odd and $K&lt;0$ , then there exists $x_0 \in \mathbb{R}$ such that $P(x_0)=Ke^{x_0}$.   For any natural number $n$, and $0&lt;K&lt;1$ then there exists $x_0 \in \mathbb{R}$ such that $P(x_0)=Ke^{x_0}$.   If $n$ is odd and $K \in \mathbb R$ , then there exists $x_0 \in \mathbb{R}$ such that $P(x_0)=Ke^{x_0}$.      For (3), consider, $P(x)=x^2+5$. Then , there does not exist $x_0\in \mathbb R$ such that $0&lt;e^{-x_0}P(x_0)&lt;1$. So it is FALSE.  But what about the others ?
9240
I am reading Gathmann's free online notes on Algebraic Geometry. One exercise asks to show that     "Every affine variety in $\mathbb A^n$ consisting of finitely many points can be written as the zero locus of $n$ polynomials".    There is a hint says "interpolation". I don't know how to start with the hint.   If $n=2$, we can use interpolation to get 1 polynomial for finitely many points. But we need to show 2 polynomials instead. I am also not sure how to apply interpolation for higher dimensions. Anyone can help? Thank you!
24054
Trying to solve a problem, I got stuck on the following question.     If $P\in\mathbb R[X]$ is an irreducible polynomial and $\operatorname {deg} P=2$, then is it true that $$\mathbb{R}[X]/(P)\cong\mathbb C?$$   I believe it must be true from the bits I've been hearing throughout my studies, but of which I haven't seen proofs. I've been unable to find anything on the internet other than . It tells me that there is something called Weierstrass' theorem which should be relevant, but can't find even the precise statement of the theorem, let alone the proof.  Here's what I have tried.   First of all, I can assume that $P$ is monic, because for any polynomial of degree $2$ there's a monic polynomial generating the same ideal in $\mathbb R[X].$ So I have  $$P(x)=x^2+bx+c$$  for some real numbers $b,c$ such that $b^2&lt;4c$.   I want to assume there is an isomorphism $\phi:\mathbb R[X]/(P)\to\mathbb C$ and work out what it should look like. I will denote by $[S]$ the coset of $S\in \mathbb R[X]$ in $\mathbb R[X]/(P).$ I'd like to find $S$ such that $\phi([S])=i.$  I have   $$ [x^2+bx+c-1]\mapsto -1, $$  So the coset that is mapped to $-1$ has to be a square root of $[x^2+bx+c-1]$ in $\mathbb R[X]/(P).$ Such a coset must be represented by a linear polynomial $kx+l$. Thus $$(kx+l)^2-(x^2+bx+c-1)\in (P),$$  that is $$ (kx+l)^2-(x^2+bx+c-1)\equiv a(x^2+bx+c) $$  for some real number $a$, and further  $$ (k^2-1-a)x^2+(2kl-b-ab)x+l^2-c-ac+1\equiv 0. $$  This gives   $$ \begin{cases} k^2=1+a\\ l^2=(1+a)c+1\\ 2kl=(1+a)b \end{cases} $$  The problem is that solving this gives formulas that are very difficult to handle. And I need to handle them if I want to prove that what I'm constructing is indeed an isomorphism. Is there a better way of doing it?
120339
This question is related to this:  We have $P_3=P_2\cdot P_1$,for three non-zero polynomials. The degree of each polynomial is at least 1.  Question: Does $P_1$ overlap $P_3$ over more than one point?    By overlap I do not mean intersect. Instead, I mean a part of the polynomials can lay on each other over more than one point. edit: $P_1\neq P_2$
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Does there exist a polynomial $p(x) \in \mathbb C[x]$ such that $p(x) \notin \mathbb R[x]$ and $p(x)p(-x)=p(x^2)$ ?  I have noticed that if $a_n$ is the leading coefficient of $p(x)$ then $a_n=(-1)^n $, so the leading coefficient cannot be imaginary; also $p(0)^2=p(0)$ so the constant term also is real. Please help. Thanks in advance.
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Let $V = P_{2}$ be the vector space of polynomials of degree less than or equal to $2$ with real coefficients, and let $W$ be the subset of polynomials $p(x)$ in $P_{2}$ such that:  $$\int_{-2}^{0}p(x)\,dx = 4\int_{0}^{2}p(x)\,dx.$$  $b)$ Find a basis for $W$, and compute $\dim(W)$.  For $b)$, I know after plugging in an example p(x)=a and p(x)=a+bx+c^2 I get that a = 0, where do I go from here?
98862
I'm totally stuck with the following question, I even don't know how to start:  For each natural number $n$ we consider a space $X_n$ that is obtained by removing $n$ distinct points from $\mathbb{R}^2$. We consider the 1-point compactification $X_n^+$ and we denote by $\infty_n\in X_n^+$ the point at infinity (so that $X_n^+=X_n\cup \{\infty_n\}$). Show that   $X_n^+$ can be embedded in $\mathbb{R}^3$. I do not have to write down explicit formulas for the embedding, but I have to explain my reasoning using pictures and mention what result(s) I use in order to reach my final conclusion If $X_n$ and $X_m$ are homeomorphic, then $n=m$.    Both questions I find very hard to answer. I really hope someone can help me!
86871
Let $p(z) = c_0 + c_1z + c_2z^2 + \dots + c_nz^n$ where $0 \le c_0 \le c_1 \le \dots \le c_n$.  I would like to show that all zeroes of this polynomial lie inside the unit disk by applying Rouche's theorem to the polynomial $(1-z)p(z)$.  I'm not completely sure how to do this. Using the given, information I can deduce that  $$|(1 - z)p(z)| \le |c_nz^{n+1}|$$  on the unit circle, but this doesn't really match the assumptions of Rouche's theorem.   Help would be appreciated.
68274
why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$  in $n$?  I know this is well-known. But how to prove it rigorously? Even mathematical induction does not seem so straight-forward.  Thanks.
21449
I can't prove this statement, can anybody show me how to prove it?    $$f:\mathbb{C}\rightarrow \mathbb{C} \in \mathcal{O}(\mathbb{C}), \exists n\in \mathbb{N}, R &gt;0 , M&gt;0 : |f(z)| \le M|z|^{n} \ \ \forall |z|&gt;R \Rightarrow \deg(f)\le n $$  To show is that if there exists such an $M$, that then $f$ is a polynomial of max degree $n$. I started like this:   $$f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n$$ So if I put this into the inequality: $$|f(z)| = \left| \sum_{n=0}^{\infty} a_n (z-z_0)^n \right| \le M |z|^n .$$
100033
Let, $p$ be a polynomial in $1$-complex variable. Suppose all zeros of $p$ are in the upper half plane $H=\{z\in \mathbb C|\Im(z)&gt;0\}. $ Then , which are corrct ?   $\Im\left\{\frac{p'(z)}{p(z)}\right\}&gt;0$ for $z\in \mathbb R$. $\Re \left\{i\frac{p'(z)}{p(z)}\right\}&lt;0$ for $z\in \mathbb R$. $\Im\left\{\frac{p'(z)}{p(z)}\right\}&gt;0$ for $z\in \mathbb C$ with $\Im (z)&lt;0$. $\Im\left\{\frac{p'(z)}{p(z)}\right\}&gt;0$ for $z\in \mathbb C$ with $\Im (z)&gt;0$.   I know that , since all roots of $p(z)$ lie in the upper half plane so, all roots of $p'(z)$ lie in the same upper half plane , from Luca's theorem.  But from there how we can say about real part or imaginary part of $\frac{p'(z)}{p(z)}$ ?
119842
Let $n$ be a positive integer, $x_0, x_1,\ldots,x_n$ be $(n+1)$ pairwise distinct real numbers, and $y_0, y_1,\ldots,y_n$ be $(n+1)$ real numbers. We have know that there exists a unique polynomial $P(x)$ where its degree is not greater than $n$.  What is the necessary and sufficient condition guaranteeing $P(x)$ has a degree of $n$.  Thank you for all comments and helping.
57453
Let $p(t) =  a_0+\dots +a_nt^n, a_n\neq 0,$ a polynomial with degree $n&gt;0$. Prove that $p:\mathbb{R}\rightarrow \mathbb{R}$ is not bounded, but for any bounded subset $X\subset \mathbb{R}$ the restriction $p\vert X$ is bounded.   I want to prove this in very elementary terms. I mean, without using the notion of limit or even continuity. I can only show this with "strong" results. Is it possible with elementary terms?
33444
To prove that: The $n$-dimensional projective space can be presented as:  $\mathbb{RP}^n=\bigcup_{i=0}^{n} A^{i}$.  (The intersection of each two $A^{i}$ is empty)، $A^{i}$ is isomorphic to $\mathbb R^{i}$ ($A^{0}$ includes only one point). The definition of $\mathbb{RP}^n$ is: $\mathbb{RP}^n= (\mathbb R^{n+1} \setminus \{0\})/ \sim$ , $\sim$ is defined as: $x \sim \lambda x$for all $\lambda\in \mathbb R^\times$ and $x\in \mathbb R^{n+1} \setminus \{0\}$. $\mathbb R^\times $ is the productive group of scalars $\ne0$ in the field $\mathbb R$. I'm just a beginner in this field so any help would be appreciated.
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If $[x:y]$ are coordinates of $\mathbb{P}^1$ and $[X:Y:Z]$ are coordinates of $\mathbb{P}^2$, what do the following varieties look like?   $V(x^2X+y^2Y+xyZ)\subset \mathbb{P}^2\times \mathbb{P}^1$ $V(x^2X^2+y^2Y^2+xyZ^2)\subset \mathbb{P}^2\times \mathbb{P}^1$   As of now, I don't quite know how to visualize these varieties. A detailed geometric picture would be especially welcome.
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I'm trying to prove that $A$ a UFD implies that $A[T]$ is a UFD.   The only thing I am sure I could try to use is Gauss's lemma.  Also, how can we deduce that the polynomial rings $\mathbb{Z}[x_1,\ldots,x_n]$ and $k[x_1,\ldots,x_n]$ are UFDs?