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Let $k[x,y]$ and $k[z_0,z_1,z_2]$ be polynomial algebras over a field, and let $$k[x,y]_{2\bullet}:= k[x^2, xy, y^2]$$ be the subalgebra of polynomials with terms of even degree. Then we clearly have a k-algebra hom $$k[z_0,z_1,z_2]\rightarrow k[x,y], ~~z_i\mapsto x^iy^{2-i}$$ whose kernel contains the ideal $(z_2z_0-z_1^2)$.

Why is this ideal precisely the kernel?

Motivation: this imbeds $\mathbb{P}^1\hookrightarrow \mathbb{P}^3$ as the conic $z_2z_0-z_1^2$.

user26857
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user709090
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  • @Brozovic sorry, i dont understand what you mean by division algo since $k[z_0,z_1,z_2]$ is not a euclidean domain – user709090 Dec 15 '19 at 14:02
  • @Brozoviv i agree it is a ufd. so i take something , p, in the kernel. then you want me to divide p by $z_2z_0-z_1^2$? but i dont know p is divisible by $z_2z_0-z_1^2$ since thats what im trying to prove. im sorry if im being very stupid – user709090 Dec 15 '19 at 14:10
  • You can divide by $z_2z_0-z_1^2$ as a polynomial, in say $z_1$. Then the remainder is of the form $a(z_0,z_2)z_1+b(z_0,z_2)$, with $a,b$ polynomials in $z_0$ and $z_2$ only. So, $az_1+b\mapsto a(x^2,y^2)xy+b(x^2,y^2)=0$. Note that in $b(x^2,y^2)$ all terms have exponents of $x$ that are even, and the same for $y$, while in $a(x^2,y^2)xy$ all exponents are odd. Therefore $a$ and $b$ are both in the kernel. It follows that $a(z_0,z_2)=b(z_0,z_2)=0$. – egorovik Dec 15 '19 at 14:23
  • This division (long division) is a particular case of Buchberger's algorithm with which you can compute kernels of homomorphisms between polynomial rings. – egorovik Dec 15 '19 at 14:29
  • @egorovik it seems like youre assuming there's some kind of "division with remainder" property for $k[z_0,z_1,z_2] =k[z_0,z_2][z_1]$ – user709090 Dec 15 '19 at 14:30
  • There is. Follow the link. Probably it is more often called 'elimination' modulo an ideal, in the general case. But in our case, since the term $z_1^2$ has coefficient $1$, we need nothing more than just one-variable long division. – egorovik Dec 15 '19 at 14:32
  • For this particular case you can prove it yourself. For a given $f(z_0,z_1,z_2)$, gather together in a polynomial $q(z_0,z_1,z_2)z_1^2$ all terms that are multiple of $z_1^2$. Then subtract $f(z_0,z_1,z_2)+q(z_0,z_1,z_2)(z_0z_2-z_1^2)$. The result is the remainder I mentioned above of the form $a(z_0,z_2)z_1+b(z_0,z_2)$. The division/elimination algorithm stops after that one step. – egorovik Dec 15 '19 at 14:37
  • The map, call it $\phi$, is a homomorphism. $0=\phi(f)=\phi(q)\phi(z_0z_2-z_1^2)+\phi(az_1+b)=\phi(az_1+b)$ – egorovik Dec 15 '19 at 14:42
  • wait the $z_1$ degree of $f-qz_1^2 = az_1+b$ is obviously not greater than 1, but you have $f+q(z_0z_2-z_1^2)$ which is $(az_1+b) + qz_0z_2$ and $q$ may have $z_1$ degree greater than 1.... right? – user709090 Dec 15 '19 at 14:48
  • Yes, but you don't care about $q(z_0,z_1,z_2)(z_0z_2-z_1^2)$ anymore. That term is already known to be in the ideal $(z_0z_2-z_1^2)$. That is why the problem reduces to showing that the remainder is also in the ideal. In our case it turns out to be zero. – egorovik Dec 15 '19 at 14:50
  • right but youre claiming $f+q(z_0z_2-z_1^2)$ has $z_1$ degree smaller than 2. i dont see that – user709090 Dec 15 '19 at 14:52
  • Ah, what I said about the division stopping after just one step is wrong. You need to continue further. After you add $q(z_0z_2-z_1^2)$ the degree on $z_1$ has reduced (by at least $2$). So, you repeat again the same idea. Gather the terms that are divisible by $z_1^2$ in a $q_2$ and add $q_2(z_0z_2-z_1^2)$. The degree keeps decreasing, until there are no more terms with degree in $z_1$ larger than $1$. Just like long division. – egorovik Dec 15 '19 at 15:00
  • @egorovik ahh i see. very nice. thank you for your help! – user709090 Dec 15 '19 at 15:07
  • My advice. Find a book about Groebner bases and study Burchberger's algorithm and its applications to compute the basis of an ideal, intersections, products, quotients of ideals, images and kernels or homomorphisms, primary decomposition, etc. All those algorithms are useful in commutative algebra and algebraic geometry. – egorovik Dec 15 '19 at 15:10
  • there is a more general version $$k[z_0,...,z_d]\rightarrow [x,y]{d\bullet},z_i\mapsto x^iy^{d-i}$$ with kernel containing the ideal generated by all $z_iz_j-z{j-1}z_{i+1}$. Do you think a similar approach would work? – user709090 Dec 15 '19 at 15:12
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    It should work. It doesn't mean that it is the most convenient, though. One would need to write a convenient form of the remainder after division(elimination) by that ideal. One needs to fix a monomial order. For example $z_0<z_1<...<z_d$. I think that the same argument on the remainders of the degrees of $x$ and $y$ after division by $d$ of the image of the remainder will tell that the remainder is zero. See here – egorovik Dec 15 '19 at 16:28
  • See it also here. – egorovik Dec 15 '19 at 16:32
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    Or here – user26857 Dec 15 '19 at 16:33

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