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Suppose $A\subseteq B$ are commutative rings, $B$ integral over $A$. Let $\mathfrak{b}$ be an ideal of $B$, and set $\mathfrak{a}=A\cap\mathfrak{b}$.

Apparently, $B/\mathfrak{b}$ is integral over $A/\mathfrak{a}$.

If $x\in B$, then it satisfies some $$ x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0. $$ Reducing modulo $\mathfrak{b}$ shows $$ (x+\mathfrak{b})^n+(a_{n-1}+\mathfrak{b})(x+\mathfrak{b})^{n-1}+\cdots+(a_1+\mathfrak{b})(x+\mathfrak{b})+(a_0+\mathfrak{b})=0+\mathfrak{b}. $$ This would show that $B/\mathfrak{b}$ is integral over $A/\mathfrak{a}$ if the coefficients are in $A/\mathfrak{a}$. Why is $a+\mathfrak{b}=a+\mathfrak{a}$ for $a\in A$ here? Isn't $a+\mathfrak{b}$ the set of all $y\in B$ such that $y-a\in\mathfrak{b}$? Maybe I've interpreted the image of $a$ incorrectly.

If $y\in a+\mathfrak{b}$, then $y-a\in\mathfrak{b}$, so if $y\in A$ also, then $y-a\in A$, so $y-a\in\mathfrak{a}$. But is there some reason why $y\in A$, if at all, to see explicitly that these cosets are equal? Thanks.

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$a+\mathfrak{b}=a+\mathfrak{a}$ is just false: the two sets are different! (If the two sides of these equality were equal, then adding $-a$ to them would tell you that $\mathfrak{b}=\mathfrak{a}$) – Mariano Suárez-Alvarez Jan 19 '12 at 5:56
I have a feeling that you're going in the right direction. It reminds me of when one wants to show that when given fields $L$, $K$ and $F$, if $L$ is algebraic over $K$ and $K$ is algebraic over $F$ then $L$ is algebraic over $F$. We didn't explicitly write the element $\alpha \in L$ as a root of a polynomial with coefficients in $K$ and then show that the coefficients $K$ were in $F$ ; it was more tricky than that. – Patrick Da Silva Jan 19 '12 at 6:09
@Arturo : $a + \mathfrak b = a + \mathfrak b$ is very true in general. – Patrick Da Silva Jan 19 '12 at 6:10
@Patrick: Sigh; I meant $a+\mathfrak{a}=a+\mathfrak{b}$ is not true in general (what if $a=0$?)... – Arturo Magidin Jan 19 '12 at 17:43
up vote 2 down vote accepted

Recall that we can view $A/\mathfrak{a}$ as a subring of $B/\mathfrak{b}$ because the kernel of the composition $$A\hookrightarrow B\to B/\mathfrak{b}$$ is $A\cap \mathfrak{b}=\mathfrak{a}$, and by the first isomorphism theorem the induced map $A/\mathfrak{a}\to B/\mathfrak{b}$, sending $x+\mathfrak{a}$ to $x+\mathfrak{b}$, is injective. So, when we say that, for some $y\in B$, the coset $y+\mathfrak{b}$ is in $A/\mathfrak{a}$, what we mean is that it lies in the image of this map; i.e., there is some representative $z\in B$ for the coset $y+\mathfrak{b}$ (i.e., an element of $B$ such that $z+\mathfrak{b}=y+\mathfrak{b}$) such that $z\in A$, as this implies that $z+\mathfrak{a}$ will be sent to $z+\mathfrak{b}=y+\mathfrak{b}$ by the "inclusion" of $A/\mathfrak{a}$ into $B/\mathfrak{b}$.

But the coefficients of the equation clearly satisfy this condition, because one representative for the coset $a_i+\mathfrak{b}$ that lies in $A$ is certainly $a_i$ itself.

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Thanks for the explanation. – buzz Jan 19 '12 at 17:30

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