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I'm just trying to get some basic facts straight,

Given a polynomial map $$F : \mathbb A^n \to \mathbb A^r,x \mapsto (f_1(x), \ldots, f_r(x))$$ with $f_i \in k[x_1,\ldots,x_n]$, I know that the image $Y \subset \mathbb A^r$ of $F$ need not be an algebraic set as in $$F : \mathbb A^2 \to \mathbb A^2, (x,y) \mapsto (x,xy).$$

However, what can I say about the coordinate ring of $Y$ in case it happens to be closed already or when taking its closure? Intuition says that functions on $Y$ should be expressed as functions of $f_1, \ldots, f_r$, so how is the relation to the ring $k[f_1, \ldots, f_r]$? And how I arrive at that ring starting from $k[x_1, \ldots, x_r]/I(Y)$? I don't immediately see equations defining the image.

Thanks for your thoughts and clarifications

Dario
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1 Answers1

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Well, the polynomials that vanish on $F(\mathbf{A}^n)$ are the ones that pull back to the zero function on $\mathbf{A}^n$ under the map $F^*\colon k[y_1, \dots, y_r] \to k[x_1, \dots, x_n]$ sending $y_i$ to $f_i$. So the coordinate ring of the closure of $F(\mathbf{A}^n)$ ought to be $k[y_1, \dots, y_r]/\ker F^* \simeq k[f_1, \dots, f_r] \subseteq k[x_1, \dots, x_n]$.

Hoot
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