(Hint) Is this set a polynomial ideal?

Question

Definitions and Notation:
Given $\mathbb{P}^n(k)$ for $k$ a(n algebraically closed) field, we can define $n+1$ affine charts $\phi_i$ from $\mathbb{P}^n(k) \to \mathbb{A}^n(k)$ (affine $n$-space), by using the maps $\phi_i: (x_0: \dots: x_n) \mapsto (\frac{x_0}{x_i}, \dots, \frac{x_{i-1}}{x_i}, \frac{x_{i+1}}{x_i}, \dots, \frac{x_n}{x_i})$, $\phi_i^{-1}: (y_1, \dots, y_n) \mapsto (y_1: \dots: y_{i-1}:1:y_{i+1}: \dots:y_n)$, where the domain of each $\phi_i$ (respectively the range of each $\phi_i^{-1}$) is $U_i = \{ x \in \mathbb{P}^n(k): x_i \not= 0 \} \subsetneq \mathbb{P}^n(k)$. Given a projective variety $V \subseteq \mathbb{P}^n(k)$, there exists a corresponding prime, homogeneous ideal $I(V) \subseteq k[x_0, \dots, x_n]$.

Question: The set of all polynomials which vanish on $\phi_i(U_i \cap V)$ is of the form $$\{ f\circ \phi_i^{-1}: f \in I(V) \} \subseteq k[y_1, \dots, y_n] \,.$$ Is this set even an ideal in $k[y_1, \dots, y_n]$, much less a prime ideal? If so, can someone give me a hint for how to start to show this?

Right now I am just staring at the page and telling myself I can't do it and have no idea how.

Background and Motivation: I want to show that for each $i =0, \dots, n$, the set $U_i \cap V \subseteq \mathbb{P}^n(k)$ is isomorphic to an affine variety $\subseteq \mathbb{A}^n(k)$. It seems obvious to me that the only way to do this is to show that $\phi_i(U_i \cap V) \subseteq \mathbb{A}^n(k)$ is an affine variety (irreducible algebraic set).

Since the maps $\phi_i^{-1}$ are polynomial in each component, it seems to follow that the set $$ \phi_i(U_i \cap V) = \{ y \in \mathbb{A}^n(k): \forall f \in I(V), (f \circ \phi_i^{-1})(y) = 0 \} \subseteq \mathbb{A}^n(k) $$ is an algebraic set. However, I also need to show that it is irreducible.

It is unclear to me how to show from the definition that $\phi_i(U_i \cap V)$ is irreducible. So instead I thought to try to utilize the prime ideal - affine variety correspondence. Because the set of polynomials which vanish on $\phi_i(U_i \cap V)$ is $\{ f \circ \phi_i^{-1}: f \in I(V) \}$, my question follows.

Later, I need to show that each $\phi_i(U_i \cap V)$ is an affine variety in $\mathbb{A}^n(k)$ in order to use the definition from my textbook and claim/show that a field of rational functions exists on it. Then I have to show that each such field of rational functions is isomorphic to the rational function field on $V \subseteq \mathbb{P}^n(k)$. This won't be as difficult, since I already know the form of isomorphism. But first I have to show that the range of the isomorphism actually exists and/or is well-defined, hence my question. All of this is just to complete one sub-problem: 5.4.10.(1) from this book.

Answer 1

If I'm not mistaken, composition with $\phi_i^{-1}$ is more or less just the evaluation map at $1$ for the variable $x_i$. Write \begin{align*} (\phi_i^{-1})^*: k[x_0, \ldots, x_n] &\to k[y_1, \ldots, y_n]\\ f(x_0, \ldots, x_i, \ldots, x_n) &\mapsto f(y_1, \ldots, 1, \ldots, y_n) \end{align*}

You can find a proof that this is a ring homomorphism in this thread. Moreover, the evaluation map is surjective: take any polynomial in $f \in k[y_1, \ldots, y_n]$ and include it into $k[x_0, x_1, \ldots, x_n]$ by mapping $y_j \mapsto x_{j-1}$ for $j < i$ and $y_j \mapsto x_j$ for $j > i$. Since the variable $x_i$ does not appear, it is unchanged by the evaluation map, so it maps to itself under $\phi_i^{-1}$.

Now the result follows from the fact that the image of an ideal under a surjective ring homomorphism is an ideal, and in fact primality is also preserved; see this thread.

Answer 2

This is just to expand on the details of Quasicoherent's answer. I.e. to expand the hint (which was precisely what was requested and why Quasicoherent's answer is and will remain the accepted answer) to something closer to a full solution.

By the universal property of polynomial rings (see here), there exists a unique ring homomorphism $$[ev_i(1)]: \left(k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n] \right)[x_i] \to k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n]$$ such that $[ev_i(1)](x_i) = 1$ and $[ev_i(1)](\gamma) = \gamma$ for all $\gamma \in k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n]$.

Moreover, for each $\gamma \in k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n]$, one has that $$\gamma = (\gamma) \cdot (1) = [ev_i(1)](\gamma)\cdot [ev_i(1)](x_i) = [ev_i(1)](\gamma\cdot x_i) = [ev_i(1)](\gamma x_i)$$ (where $\cdot$ denotes ring multiplication, and juxtaposition denotes scalar multiplication, and there is an abuse of notation in identifying constant polynomials with their corresponding scalars) as a result of (1) the definition of $[ev_i(1)]$ (2) the fact that $[ev_i(1)]$ is a ring homomorphism.

Since for each $\gamma \in k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n]$, one has that $\gamma x_i \in \left(k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n]\right)[x_i]$, it follows that every element of $k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n]$ is the image of some element of $\left(k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n]\right)[x_i]$ under $[ev_i(1)]$, and that therefore $[ev_i(1)]$ is not only a ring homomorphism, but a surjective ring homomorphism.

Now observe the existence of the following two equivalences as rings: $$\begin{array}{rcl} k[x_0, \dots, x_n] & \cong & \left(k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n]\right)[x_i] \\ k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n] & \cong & k[y_1, \dots, y_n] \,. \end{array}$$ Denote corresponding ring isomorphisms as follows: $$\begin{array}{rl} \mu: & k[x_0, \dots, x_n] \to \left(k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n]\right)[x_i] \\ \nu: & k[x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_n] \to k[y_1, \dots, y_n] \end{array} $$

Since the composition of ring homomorphisms is again a ring homomorphism, it follows that $$\nu \circ [ev_i(1)] \circ \mu : k[x_0, \dots, x_n] \to k[y_1, \dots, y_n] $$ is a ring homomorphism. (Use associativity of function composition and induction.)

Every ring isomorphism is a bijective function, and every bijective function is a surjective function. Hence $\nu \circ [ev_i(1)] \circ \mu$ is the composition of surjective functions, and therefore surjective.

Hence $\nu \circ [ev_i(1)] \circ \mu: k[x_0, \dots, x_n] \to k[y_1, \dots, y_n]$ is a surjective ring homomorphism.

By definition, (I think), $(\phi_i^{-1})^* = \nu \circ [ev_i(1)] \circ \mu$, hence $(\phi_i^{-1})$ is a surjective ring homomorphism as claimed. (However, I still need to think of a more rigorous argument why $(\phi_i^{-1})^* = \nu \circ [ev_i(1)] \circ \mu$.)

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Ok, so let's recap definitions then. $\phi_i^{-1}$ is a map $\mathbb{A}^n(k) \to \mathbb{P}^n(k)$, $$\phi_i^{-1}: (y_1, \dots, y_n) \mapsto (y_1: \dots: y_{i-1}: 1: y_i: \dots: y_n) \,. $$ But it could also be considered in practice a map from $k^n \to k^{n+1}$, "replacing colons with commas": $$\phi_i^{-1}: (y_1, \dots, y_n) \mapsto (y_1, \dots, y_{i-1}, 1, y_i, \dots, y_n) \,. $$ Then $(\phi_i^{-1})^*: k[x_0, \dots, x_n] \to k[y_1, \dots, y_n]$ is "pre-composition" by $\phi_i^{-1}$ -- yet this is an entirely problematic definition, since pre-composition of functions requires that functions be involved in the first place, and we only want to assume that $k[x_0, \dots, x_n]$ and $k[y_1, \dots, y_n]$ are abstract polynomial rings, not necessarily spaces of functions. (Indeed, the elements of $k[x_0,\dots, x_n]$ in general don't correspond to functions on $\mathbb{P}^n(k)$, so even if we wanted to think in such terms, we can't do so correctly.)