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Let $k$ be an algebraically closed field, and $f_0,\dots,f_m \in k[x_0,\dots,x_n]$ be homogeneous polynomials of the same degree. Denote by $I\subset k[x_0,\dots,x_m]$ the kernel of the homomorphism sending $x_i$ to $f_i$. Do we have the following statement?

For any $\xi\in k^{m+1}$, $\xi$ is a common zero of $I$ if and only if $\xi=(f_0(a),\dots,f_m(a))$ for some $a\in k^{n+1}$ ---(I)

I think that the direction $\Leftarrow$ is obviously true, so I would want to prove the other direction.

The state if we assume that $f_0,\dots,f_m$ have no common zeros besides $(0,\dots,0)$ since then we have a regular map $f=(f_0,\dots,f_m)$ from $\mathbb{P}^n$ to $\mathbb{P}^m$ and we have

The image of $f$ is a closed set in the Zariski topology of $\mathbb{P}^n$---(II)

let $J$ be the ideal of the image of $f$ and I want to show that $J=I$.

We have $I\subset J$ since every polynomial in $I$ vanishes on $f(\mathbb{P}^n)$. While for any $p\in J,$ we have $(p\circ f)(a)=p(f(a))=0$ for all $a\in \mathbb {P^n}$ and this is possible only when $p\circ f=0$ and thus $p\in I$ hence $J\subset I$.

I don't know how to proceed to the case that $f_0,\dots,f_m$ have a common zero apart from $(0,\dots,0)$. Also, for the case that those $f_i$ do not have a common root in $\mathbb{P^n}$, I want to find a proof not requiring (II) since (I) would imply (II) also.

Any help or hints will be appreciated.

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  • $\begingroup$ (I) and (II) are equivalent. (II) follows from the fact that projective space is proper over the base field: en.wikipedia.org/wiki/Proper_morphism $\endgroup$ – stewbasic Jun 27 '16 at 6:01
  • $\begingroup$ but $f$ may not define a regular map since we may have $f_0(a)=,\cdots, f_m(a)=0$ for some $a \in \mathbb{P^n}$ $\endgroup$ – chan kifung Jun 27 '16 at 6:05
  • $\begingroup$ Ah right, sorry. $\endgroup$ – stewbasic Jun 27 '16 at 6:09
  • $\begingroup$ The issue with (i) is that the image of $f$, defined wherever it may be, doesn't have to be closed. $\endgroup$ – Hoot Jun 27 '16 at 6:37
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Consider the map $f:k[u,v]\rightarrow k[x,y]$ sending $u\mapsto x^2$, $v\mapsto xy$. Note that $f(u)$ and $f(v)$ are homogeneous polynomials of degree $2$. Also $f$ maps the monomials $u^av^b$ to distinct monomials in $k[x,y]$, so $f$ is injective. Thus $I=0$ and every point is a common zero of $I$. However there are no $x,y\in k$ for which $(x^2,xy)=(0,1)$.

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