The Stacks project

Proof. Assume the contrary and take $x \in I \setminus \mathfrak p$ and $y \in J \setminus \mathfrak p$. Their product is an element of $IJ \subset \mathfrak p$, which contradicts the assumption that $\mathfrak p$ was prime. $\square$

Proof. The result is true for $r = 1$. If $r = 2$, then let $x, y \in J$ with $x \not\in I_1$ and $y \not\in I_2$. We are done unless $x \in I_2$ and $y \in I_1$. Then the element $x + y$ cannot be in $I_1$ (since that would mean $x + y - y \in I_1$) and it also cannot be in $I_2$.

For $r \geq 3$, assume the result holds for $r - 1$. After renumbering we may assume that $I_ r$ is prime. We may also assume there are no inclusions among the $I_ i$. Pick $x \in J$, $x \not\in I_ i$ for all $i = 1, \ldots , r - 1$. If $x \not\in I_ r$ we are done. So assume $x \in I_ r$. If $J I_1 \ldots I_{r - 1} \subset I_ r$ then $J \subset I_ r$ (by Lemma 10.15.1) a contradiction. Pick $y \in J I_1 \ldots I_{r - 1}$, $y \not\in I_ r$. Then $x + y$ works. $\square$

Proof. We may assume there are no inclusions among the $\mathfrak p_ i$. After reordering we may assume $x \not\in \mathfrak p_ i$ for $i < s$ and $x \in \mathfrak p_ i$ for $i \geq s$. If $s = r + 1$ then we are done. If not, then we can find $y \in I$ with $y \not\in \mathfrak p_ s$. Choose $f \in \bigcap _{i < s} \mathfrak p_ i$ with $f \not\in \mathfrak p_ s$. Then $x + fy$ is not contained in $\mathfrak p_1, \ldots , \mathfrak p_ s$. Thus we win by induction on $s$. $\square$

Proof. Let us first prove $I_1 \cap \ldots \cap I_ r = I_1 \ldots I_ r$ as this will also imply the injectivity of the induced ring homomorphism $R/(I_1 \ldots I_ r) \rightarrow R/I_1 \times \ldots \times R/I_ r$. The inclusion $I_1 \cap \ldots \cap I_ r \supset I_1 \ldots I_ r$ is always fulfilled since ideals are closed under multiplication with arbitrary ring elements. To prove the other inclusion, we claim that the ideals

generate the ring $R$. We prove this by induction on $r$. It holds when $r = 2$. If $r > 2$, then we see that $R$ is the sum of the ideals $I_1 \ldots \hat I_ i \ldots I_{r - 1}$, $i = 1, \ldots , r - 1$. Hence $I_ r$ is the sum of the ideals $I_1 \ldots \hat I_ i \ldots I_ r$, $i = 1, \ldots , r - 1$. Applying the same argument with the reverse ordering on the ideals we see that $I_1$ is the sum of the ideals $I_1 \ldots \hat I_ i \ldots I_ r$, $i = 2, \ldots , r$. Since $R = I_1 + I_ r$ by assumption we see that $R$ is the sum of the ideals displayed above. Therefore we can find elements $a_ i \in I_1 \ldots \hat I_ i \ldots I_ r$ such that their sum is one. Multiplying this equation by an element of $I_1 \cap \ldots \cap I_ r$ gives the other inclusion. It remains to show that the canonical map $R/(I_1 \ldots I_ r) \rightarrow R/I_1 \times \ldots \times R/I_ r$ is surjective. For this, consider its action on the equation $1 = \sum _{i=1}^ r a_ i$ we derived above. On the one hand, a ring morphism sends 1 to 1 and on the other hand, the image of any $a_ i$ is zero in $R/I_ j$ for $j \neq i$. Therefore, the image of $a_ i$ in $R/I_ i$ is the identity. So given any element $(\bar{b_1}, \ldots , \bar{b_ r}) \in R/I_1 \times \ldots \times R/I_ r$, the element $\sum _{i=1}^ r a_ i \cdot b_ i$ is an inverse image in $R$.

To see (2), by the very definition of being distinct maximal ideals, we have $\mathfrak {m}_ a + \mathfrak {m}_ b = R$ for $a \neq b$ and so the above applies. $\square$

Proof. For $I \subset \{ 1, \ldots , n\}$ with $|I| = m$, we denote by $E_ I$ the $m \times n$ matrix of the projection

and set $A_ I = E_ I A$, i.e., $A_ I$ is the $m \times m$ matrix whose rows are the rows of $A$ with indices in $I$. Let $B_ I$ be the adjugate (transpose of cofactor) matrix to $A_ I$, i.e., such that $A_ I B_ I = B_ I A_ I = \det (A_ I) 1_{m \times m}$. The $m \times m$ minors of $A$ are the determinants $\det A_ I$ for all the $I \subset \{ 1, \ldots , n\}$ with $|I| = m$. If $f \in J$ then we can write $f = \sum c_ I \det (A_ I)$ for some $c_ I \in R$. Set $B = \sum c_ I B_ I E_ I$ to see that (1) holds.

If $f 1_{m \times m} = BA$ then by the Cauchy-Binet formula (72) we have $f^ m = \sum b_ I \det (A_ I)$ where $b_ I$ is the determinant of the $m \times m$ matrix whose columns are the columns of $B$ with indices in $I$. $\square$

Proof. Since the adjugate has the property $A_1B = B A_1 = f$ the first block of the expression for $AB$ is correct. Note that

where $A_1^{ij}$ means $A_1$ with the $j$th row and $k$th column removed. This last expression is the row expansion of the determinant of the matrix in the statement of the lemma. $\square$

Proof. Choose a surjection $R^{\oplus n - 1} \to M$. We may lift the map $f$ to a map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ (Lemma 10.5.2). It suffices to prove $f'$ has a nonzero kernel. The map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ is given by a matrix $A = (a_{ij})$. If one of the $a_{ij}$ is not nilpotent, say $a = a_{ij}$ is not, then we can replace $R$ by the localization $R_ a$ and we may assume $a_{ij}$ is a unit. Since if we find a nonzero kernel after localization then there was a nonzero kernel to start with as localization is exact, see Proposition 10.9.12. In this case we can do a base change on both $R^{\oplus n}$ and $R^{\oplus n - 1}$ and reduce to the case where

Hence in this case we win by induction on $n$. If not then each $a_{ij}$ is nilpotent. Set $I = (a_{ij}) \subset R$. Note that $I^{m + 1} = 0$ for some $m \geq 0$. Let $m$ be the largest integer such that $I^ m \not= 0$. Then we see that $(I^ m)^{\oplus n}$ is contained in the kernel of the map and we win. $\square$

Proof. Immediate from Lemma 10.15.7. $\square$