Prime ideals in a polynomial ring in infinitely many indeterminates Suppose we have a polynomial ring $R = \mathbb{Q}[x_i, y_{i,j}, z_{i,j} : i, j \in \mathbb{N}]$ over $\mathbb{Q}$ in infinitely many indeterminates $x_i, y_{i,j}, z_{i,j}$, as $i, j$ range over $\mathbb{N}$.

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*Consider the ideal $I = (x_i - y_{i,j}z_{i,j} : i,j \in \mathbb{N})$ in $R$. Is $I$ prime?


*In general, how does one show a given ideal of, say, $\mathbb{Q}[x_0, x_1, \ldots]$ is prime (i.e. what techniques are there)?
 A: Let $k$ be a field. I claim that
$$A(k) := k[y_{i,j},z_{i,j},x_i : i,j \in \mathbb{N}] / (x_i = y_{i,j}z_{i,j} : i,j \in \mathbb{N})$$
is an integral domain, so that the ideal we quotient out is a prime ideal. My aim is to prove this with almost no element calculations.
Since $A(k) \to A(k) \otimes_k \overline{k}$ is injective (linear algebra) and $A(k) \otimes_k \overline{k} \cong A(\overline{k})$, we may assume that $k$ is algebraically closed. Now observe that there is a tensor product (i.e. coproduct) decomposition
$$A(k) \cong \bigotimes_{i \in \mathbb{N}} A_i(k)$$
where
$$A_i(k) := k[y_{i,j},z_{i,j},x_i : j \in \mathbb{N}] / (x_i = y_{i,j}z_{i,j} : j \in \mathbb{N}).$$
Since a tensor product of two integral domains over $k$ is again an integral domain (here we use that $k$ is algebraically closed), the same holds for any finite number, and hence for any number, since arbitrary tensor products of algebras are filtered colimits of finite tensor products of algebras.
Thus, we only need to show that
$$k[y_j,z_j,x : j \in \mathbb{N}]/(x = y_j z_j : j \in \mathbb{N})$$
is an integral domain. It is the filtered colimit of the algebras where we replace $\mathbb{N}$ by $\mathbb{N}_{\leq n}$ for some $n$. By induction, it therefore suffices to prove the following:
Lemma 1. Let $R$ be an integral domain, $r \in R \setminus \{0\}$. Then the $R$-algebra
$$B(R,r) := R[y,z]/(r = yz)$$
is an integral domain as well, and $R \to B(R,r)$ is injective.
Notice that, for example, $B(R,1) = R[y,y^{-1}]$ is the algebra of Laurent polynomials over $R$, and you can see $B(R,r)$ as a generalization of this construction, where we replace $yz=1$ by $yz=r$. This example also motivates the basis below. Also notice that for every homomorphism $\varphi : R \to S$ we have
$$B(S,\varphi(r)) \cong S \otimes_R B(R,r),$$
which shows that $B$ is a "natural" construction. We now need
Lemma 2. The set
$$\{1,y,y^2,\dotsc,z,z^2,\dotsc\}$$
is an $R$-module basis of $B(R,r)$.
This even holds when $r=0$. For a proof, consider the homomorphism $\mathbb{Z}[x] \to R$ defined by $x \mapsto r$. Because of $B(R,r) \cong R \otimes_{\mathbb{Z}[x]} B(\mathbb{Z}[x],x)$, it suffices to show the claim for the special (universal) case $R=\mathbb{Z}[x]$ and $r=x$. Observe that $B(\mathbb{Z}[x],x) \cong \mathbb{Z}[y,z]$, and $\mathbb{Z}[x] \to \mathbb{Z}[y,z]$ is isomorphic to the subring $\mathbb{Z}[yz] \hookrightarrow \mathbb{Z}[y,z]$. Thus, we have to show that the set above is a $\mathbb{Z}[yz]$-basis of $\mathbb{Z}[y,z]$. It is clear that it is a generating set. For linear independence, we have to do a calculation. Assume that $p_0,p_1,\dotsc,q_1,q_2,\dotsc$ are polynomials in $\mathbb{Z}[yz]$ with
$$\sum_{n \geq 0} p_n y^n + \sum_{m > 0} q_m z^m = 0$$
in $\mathbb{Z}[y,z]$. The first sum only consists of terms $t$ with $\deg_y(t) \geq \deg_z(t)$, and for the second sum we have $\deg_y(t) < \deg_z(t)$. It follows that both of the sums are zero. Let us look for example at the first one (the second one is similar), so $0 = \sum_{n \geq 0} p_n y^n$. Write $p_n = \sum_{j \geq 0} a_{n,j} y^j z^j$ with $a_{n,i} \in \mathbb{Z}$. Then we get $0 = \sum_{n,j \geq 0} a_{n,j} y^{n+j} z^j$. Since the pairs $(n+j,j)$ are pairwise distinct, it follows $a_{n,j}=0$ for all $n,j$, so that $p_n = 0$ for all $n$. This finishes the proof of Lemma 2.
The basis implies immediately that $R \to B(R,r)$ is injective. It also implies that $B(R,r)$ is a flat $R$-module. Therefore, we get an injection $$B(R,r) \to Q(R) \otimes_R B(R,r) \cong B(Q(R),r).$$
But since $Q(R)$ is a field and $r \in Q(R)^{\times}$, the substitution $y = r y'$ gives an isomorphism $B(Q(R),r) \cong B(Q(R),1) \cong Q(R)[y,y^{-1}]$, which is an integral domain. This finishes the proof of Lemma 1.
