MHB Integral Ring Theory: Proving Integral Extension Property for Quotient Rings

[Fermat1]

Let $S$ be a commutative ring and $R$ a sub-ring. Let $J$ be an ideal of $S$ and $I$ be the intersection of $J$ and $R$. Show that if $S$ is integral over $R$, then $S/J$ is integral over $R/I$.

My attempt: Let $x+J$ be in $S/J$. Then $x$ is integral over $R$ so there is a monic polynomial $m$ in $R[X]$ with $m(x)=0$, say $m(X)=X^n+r_{n-1}x^{n-1}+...+r_{0}$. Then $(m+I)[X]$ is monic and $(m+I)(x+J)$=$x^n+IJ+r_{n-1}x^{n-1}+IJ+...+r_{0}+I$. Why is this zero?

 

[ThePerfectHacker]

Re: integral ring theory

Let $S$ be a commutative ring and $R$ a sub-ring. Let $J$ be an ideal of $S$ and $I$ be the intersection of $J$ and $R$. Show that if $S$ is integral over $R$, then $S/J$ is integral over $R/I$.

My attempt: Let $x+J$ be in $S/J$. Then $x$ is integral over $R$ so there is a monic polynomial $m$ in $R[X]$ with $m(x)=0$, say $m(X)=X^n+r_{n-1}x^{n-1}+...+r_{0}$. Then $(m+I)[X]$ is monic and $(m+I)(x+J)$=$x^n+IJ+r_{n-1}x^{n-1}+IJ+...+r_{0}+I$. Why is this zero?

Small comment, $R/I$ is not a subset of $S/J$, it is therefore not technically a subring. Rather we have an embedding of commutative rings $i:R/I\to S/J$ by $i(r+I) = r+J$. To show that $S/J$ is integral over $R/I$ we therefore need to show that for any $\alpha \in S/J$ there is a monic polynomial $p(X)$ with coeffs in $R/I$ such that $i(P(x))$ has $\alpha$ as a root.

Let $\alpha \in S/J$, so we write $\alpha = s + J$ for some $s\in S$. Since $s\in S$ is integral over $R$ there are $r_j\in R$ such that,
$$ s^n + r_{n-1}s^{n-1} + ... + r_1s + r_0 = 0 $$

Consider the polynomial with coeff in $R/I$:
$$P(X) = X^n + (r_{n-1} + I)X^{n-1}+ ... + (r_1+I)X + (r_0+I) $$
We claim $\alpha$ is root of polynomial $i(P)$, after we substitute,
$$ (s^n +J) + (r_{n-1} + J)(s^{n-1} + J) + ... + (r_1 + J)(s+J) + (r_0+J) = (s^n + r_{n-1}s^{n-1} + ... + r_1s + r_0) + J = J $$
Thus, $p(s+J) = J$, which is the zero element of ring $S/J$.

 

[Fermat1]

Re: integral ring theory

Small comment, $R/I$ is not a subset of $S/J$, it is therefore not technically a subring. Rather we have an embedding of commutative rings $i:R/I\to S/J$ by $i(r+I) = r+J$. To show that $S/J$ is integral over $R/I$ we therefore need to show that for any $\alpha \in S/J$ there is a monic polynomial $p(X)$ with coeffs in $R/I$ such that $i(P(x))$ has $\alpha$ as a root.

Let $\alpha \in S/J$, so we write $\alpha = s + J$ for some $s\in S$. Since $s\in S$ is integral over $R$ there are $r_j\in R$ such that,
$$ s^n + r_{n-1}s^{n-1} + ... + r_1s + r_0 = 0 $$

Consider the polynomial with coeff in $R/I$:
$$P(X) = X^n + (r_{n-1} + I)X^{n-1}+ ... + (r_1+I)X + (r_0+I) $$
We claim $\alpha$ is root of polynomial $i(P)$, after we substitute,
$$ (s^n +J) + (r_{n-1} + J)(s^{n-1} + J) + ... + (r_1 + J)(s+J) + (r_0+J) = (s^n + r_{n-1}s^{n-1} + ... + r_1s + r_0) + J = J $$
Thus, $p(s+J) = J$, which is the zero element of ring $S/J$.

Thanks. Is there a technical meaning to the word embedding?

 

[ThePerfectHacker]

Re: integral ring theory

Thanks. Is there a technical meaning to the word embedding?

Yes. But it depends on the context you are in.

Very often in mathematics you works with objects where one object does not exactly sit in the other object. The simplest example of this is $\mathbb{Z}$ and $\mathbb{Q}$. The integers are not a subset of the rationals. This is because the rationals are equivalence classes of integers. When we write $a/b\in \mathbb{Q}$ we mean the entire class of all fractions that are equal to that. When we write $n\in \mathbb{Z}$ we just mean $n$. Now people often say integers are a subset of rationals because $n=n/1 \in \mathbb{Q}$ but rather what they mean to say is that there is an embedding $i:\mathbb{Z}\to \mathbb{Q}$ defined by $i(n) = n/1$. This map $i$ is injective, so it corresponds $\mathbb{Z}$ with a subset of $\mathbb{Q}$ which is essentially $\mathbb{Z}$, but as sets they are different.

When you are working with rings $R,S$ and want to say $R$ embeds into $S$ all you mean is an injection $i:R\to S$ that preserves the ring structure i.e. a homomorphism of commutative rings. This way $i(R)$ behaves just like $R$ and we can say, abusing terminology, that $R$ is a subset of $S$.

In different contexts embedding can be different. For example, if $X$ is a metric space and $Y$ is another metric space an embedding would be $i:X\to Y$ such that preserves the metric, i.e. an isometry, i.e. $i(d_X(a,b)) = d_Y(i(x),i(y))$.

An embedding is always an injective map that respects the structure you are working with. So in algebra it is usually a homomorphism. In analysis it is usually an isometry. In topology it would be an injective homeomorphism and so forth.

 

[Fermat1]

Thanks. Out of interest, what is the problem with defining the rationals to be a subset of the integers. Just because $b/a$=$bc/ac$ for all non-zero $c$ wouldn't seem to cause any issues. Is it just for the purposes of generalisation?

 

[ThePerfectHacker]

Thanks. Out of interest, what is the problem with defining the rationals to be a subset of the integers.

It is because mathematicians are crazy about precision. Sometimes to the point of being ridiculous about it.

Start with the set $\mathbb{Z}$. Define the set $A = \{ (x,y) \in \mathbb{Z}\times \mathbb{Z} ~ | ~ y \not = 0 \}$. Define a relation $\sim$ on $A$ by saying $(a,b)\sim (c,d)$ iff $ad = bc$. Check that $\sim$ is an equivalence relation on $A$. Given any $(a,b) \in A$ we define,
$$ [(a,b)] = \{ (x,y) \in A ~ | ~ (x,y) \sim (a,b) \} $$
This is what we call the equivalence class of $(a,b)$. Instead of using the complicated notation $[(a,b)]$ we just write $\frac{a}{b}$. Therefore, $ad = bc$ if and only if $(a,b)\sim (c,d)$ if and only if $\frac{a}{b} = \frac{c}{d}$.

Definition: $\mathbb{Q}$ is the set of all equivalence classes of $A$, i.e. $\mathbb{Q}$ is the set of all subsets of $A$ which are of the form $\frac{a}{b}$.

Thus, $\frac{2}{1} = \{ ...,(-6,-3),(-4,-2),(-2,-1),(2,1),(4,2),(6,3),... \}$ as you can see that is not the same thing as $2$.

This defines $\mathbb{Q}$ as a set. It is standard to immediately define binary operations on it that turn it into a ring. This is done by defining,
$$ \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} \text{ and }\frac{a}{b}\cdot \frac{c}{d} = \frac{ac}{bd} $$
Please note the sloppiness. The $+$ sign on the LHS is the addition operation defined on $\mathbb{Q}$ while $+$ sign on RHS is the standard addition operation on $\mathbb{Z}$. So it is better to call the one on LHS by $+'$ to distinguish the different operation. But this is just taking it too far with ridiculous precision. Same thing with multiplication, the one on LHS is the one we are defining on the one on RHS is the standard one for $\mathbb{Z}$.

It is also a standard exercise that prove that since we are choosing representatives for the classes that this type of definition for $\mathbb{Q}$ is well-defined. But I am not going to do that.