I think this is ok,. Suppose you have $t\longrightarrow (f_1(t),\ldots,f_n(t)).$ Let's call Y the image of that. Then to see if $Y$ is a variety, you can see that $I(Y)$ is prime. But let $\phi: k[X_1, \ldots, X_n] \longrightarrow k[T]$ be given by $X_i \mapsto f_i(T)$. Then $$Ker \phi = {f: \{f: \phi(f) = 00\} = {f: \{f: f(f_1(T), \ldots, f_n(T)) = 00\} = I(Y).$$ So $I(Y)$ is prime because it is the kernel of a map whose image is an integral ring, and then $Y$ is a variety.
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I think this is ok,. Suppose you have $t\longrightarrow (f_1(t),\ldots,f_n(t)).$ Let's call Y the image of that. Then to see if $Y$ is a variety, you can see that $I(Y)$ is prime. But let $\phi: k[X_1, \ldots, X_n] \longrightarrow k[T]$ be given by $X_i \mapsto f_i(T)$. Then $$Ker \phi = {f: \phi(f) = 0} = {f: f(f_1(T), \ldots, f_n(T)) = 0} = I(Y).$$ So $I(Y)$ is prime because it is the kernel of a map whose domain image is an integral ring, and then $Y$ is a variety. |
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I think this is ok,. Suppose you have $t\longrightarrow (f_1(t),\ldots,f_n(t)).$ Let's call Y the image of that. Then to see if $Y$ is a variety, you can see that $I(Y)$ is prime. But let $\phi: k[X_1, \ldots, X_n] \longrightarrow k[T]$ be given by $X_i \mapsto f_i(T)$. Then $$Ker \phi = {f: \phi(f) = 0} = {f: f(f_1(T), \ldots, f_n(T)) = 0} = I(Y).$$ So $I(Y)$ is prime because it is the kernel of a map whose domain is an integral ring, and then $Y$ is a variety. |
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