Exercise 2.2.6

Question

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Here is an example of polynomials which are not algebraically independent. Consider $x_1^2,x_1x_2,x_2^2 \in F[x_1,x_2]$, and let $\phi : F[u_1,u_2,u_3] 	o F[x_1,x_2]$ be defined by
$$\phi(u_1) = x_1^2,\phi(u_2)= x_1x_2, \phi(u_3) = x_2^2.$$
Show that $\phi$ is not one-to-one by finding a nonzero polynomial $h \in F[u_1,u_2,u_3]$ such that $\phi(h)=0$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

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ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $h = u_1u_3 -u_2^2$.

Then the unique algebra morphism $\phi$ such that 
$$\phi(u_1) = x_1^2, \phi(u_2) = x_1x_2, \phi(u_3) = x_2^2$$
verifies
$$\phi(h) = \phi(u_1)\phi(u_3) - (\phi(u_2))^2 = x_1^2 x_2^2 - (x_1x_2)^2 = 0.$$
So $h 
eq 0$ is in the kernel of $\phi$, and $\phi$ is not one-to-one. Thus $x_1^2,x_1x_2,x_2^2$ are not algebraically independent.
\end{proof}

Exercise 2.2.6

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Exercise 2.2.6

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