# Thread: Maximal Ideals in Integral Domain

1. ## Maximal Ideals in Integral Domain

Let R be an integral domain with quotient field $\displaystyle K = \{ \frac {a}{b} | a,b \in R, b \neq 0 \}$, and let I and J be ideals of R. Let X denote the set of maximal ideals of R.

Let $\displaystyle t \in K$. Prove that if $\displaystyle t \in \bigcap _{M \in X } R_M$, then $\displaystyle t \in R$

pf.

Now, $\displaystyle R_M = \{ \frac {r}{s} | r,s \in R \ , \ s \in R \ not \ M \}$, so $\displaystyle t = \frac {a}{b}, a \in R, b \in (R \ not \ M)$

To be honest, I'm quite lost after this, how should I go on with this problem? thanks.

2. What if $\displaystyle t \in K-R$?

$\displaystyle \exists (a,b) \in R\times (R-\{0\})\ ;\ t=\frac{a}{b}$ , and $\displaystyle b$ is non invertible (or $\displaystyle t=ab^{-1} \in R$) and does not divide $\displaystyle a$.

So there is a maximal ideal $\displaystyle M_{0}$ in $\displaystyle R$ such that $\displaystyle b \in M_{0}$ (Krull's theorem). Thus $\displaystyle t$ isn't an element of $\displaystyle \bigcap _{M \in X } R_M$, because it doesn't belong to $\displaystyle R_{M_{0}}$.

3. Originally Posted by clic-clac
What if $\displaystyle t \in K-R$?

$\displaystyle \exists (a,b) \in R\times (R-\{0\})\ ;\ t=\frac{a}{b}$ , and $\displaystyle b$ is non invertible (or $\displaystyle t=ab^{-1} \in R$) and does not divide $\displaystyle a$.

So there is a maximal ideal $\displaystyle M_{0}$ in $\displaystyle R$ such that $\displaystyle b \in M_{0}$ (Krull's theorem). Thus $\displaystyle t$ isn't an element of $\displaystyle \bigcap _{M \in X } R_M$, because it doesn't belong to $\displaystyle R_{M_{0}}$.
unfortunately the conclusion is not correct! you can have both $\displaystyle b \in M_0$ and $\displaystyle \frac{a}{b} \in R_{M_0}.$ for example $\displaystyle \frac{2}{6} \in \mathbb{Z}_{2\mathbb{Z}}.$ i will show that we actually have $\displaystyle \bigcap_{\mathfrak{m} \in X} R_{\mathfrak{m}}=R$: define $\displaystyle \mathfrak{a}=\{r \in R: \ rt \in R \}.$

obviously $\displaystyle \mathfrak{a}$ is an ideal of $\displaystyle R.$ if $\displaystyle \mathfrak{a}=R,$ then $\displaystyle 1 \in \mathfrak{a},$ and hence $\displaystyle t \in R.$ if $\displaystyle \mathfrak{a} \neq R,$ then there exists a maximal ideal $\displaystyle \mathfrak{m}_0 \supseteq \mathfrak{a}.$ now since $\displaystyle t \in R_{\mathfrak{m}_0},$ there exist $\displaystyle a \in R, \ b \in R-\mathfrak{m}_0$ such that $\displaystyle t=\frac{a}{b}.$

thus $\displaystyle bt=a \in R.$ hence $\displaystyle b \in \mathfrak{a} \subseteq \mathfrak{m}_0,$ which is impossible since $\displaystyle b \in R- \mathfrak{m}_0.$ this proves that $\displaystyle \bigcap_{\mathfrak{m} \in X} R_{\mathfrak{m}} \subseteq R.$ also note that $\displaystyle R \subseteq R_{\mathfrak{m}}$ for all maximal ideals $\displaystyle \mathfrak{m},$ because the map $\displaystyle \varphi_{\mathfrak{m}}: R \longrightarrow R_{\mathfrak{m}}$

defined by $\displaystyle \varphi_{\mathfrak{m}}(r)=\frac{r}{1}$ is injective. thus $\displaystyle R \subseteq \bigcap_{\mathfrak{m} \in X} R_{\mathfrak{m}}.$ hence: $\displaystyle R=\bigcap_{\mathfrak{m} \in X} R_{\mathfrak{m}}. \ \ \Box$

So in this case, we would also have $\displaystyle I = \bigcap _{M \in X } IR_M$ then?

Now one way of this direction should be automatic, as $\displaystyle \bigcap _{M \in X } IR_M$ is contained in I.

But is it true for the other way?

So in this case, we would also have $\displaystyle I = \bigcap _{M \in X } IR_M$ then? yes!
since $\displaystyle I \subseteq IR_{\mathfrak{m}}, \ \forall \mathfrak{m} \in X,$ we have: $\displaystyle I \subseteq \bigcap_{\mathfrak{m} \in X} IR_{\mathfrak{m}}.$ for the other side, the idea is the same as above: let $\displaystyle t \in \bigcap_{\mathfrak{m} \in X} IR_{\mathfrak{m}}.$ define $\displaystyle \mathfrak{a}=\{r \in R: \ rt \in I \}.$ clearly $\displaystyle \mathfrak{a}$ is an ideal of $\displaystyle R.$ now if $\displaystyle \mathfrak{a}=R,$

then $\displaystyle 1 \in \mathfrak{a}$ and hence $\displaystyle t \in I.$ if $\displaystyle \mathfrak{a} \neq R,$ then there exists $\displaystyle \mathfrak{m}_0 \in X: \ \mathfrak{a} \subseteq \mathfrak{m}_0.$ now since $\displaystyle t \in IR_{\mathfrak{m}_0},$ there exist $\displaystyle a \in I, \ b \in R-\mathfrak{m}_0 : \ t=\frac{a}{b}.$ thus $\displaystyle bt=a \in I,$ i.e. $\displaystyle b \in \mathfrak{a} \subseteq \mathfrak{m}_0.$ contradiction! $\displaystyle \Box$

6. Thanks for your massive help!!!

My last question here is: Prove that $\displaystyle I=J$ if and only if $\displaystyle IR_M=JR_M$, for each maximal ideal M of R.

Proof so far.

Suppose that I=J, then $\displaystyle IR_M = I = J = JR_M$, since I and J are ideals.

Conversely, suppose that $\displaystyle IR_M = JR_M$, pick an element in $\displaystyle IR_M$, then it has the form $\displaystyle i \frac {a}{b}, a \in R, b \in R \ not \ M, i \in I$.

Now, I=J implies that there exists an element $\displaystyle j \in J$ such that $\displaystyle j=i$, thus $\displaystyle i \frac {a}{b} = j \frac {a}{b} \in JR_M$, so we have $\displaystyle IR_M \subseteq JR_M$, and the reverse is the same.

Is this correct? My solution look a little bit too easy compare to the other problems...

Thanks!!!

My last question here is: Prove that $\displaystyle I=J$ if and only if $\displaystyle IR_M=JR_M$, for each maximal ideal M of R.

Proof so far.

Suppose that I=J, then $\displaystyle IR_M = I = J = JR_M$, since I and J are ideals.

Conversely, suppose that $\displaystyle IR_M = JR_M$, pick an element in $\displaystyle IR_M$, then it has the form $\displaystyle i \frac {a}{b}, a \in R, b \in R \ not \ M, i \in I$.

Now, I=J implies that there exists an element $\displaystyle j \in J$ such that $\displaystyle j=i$, thus $\displaystyle i \frac {a}{b} = j \frac {a}{b} \in JR_M$, so we have $\displaystyle IR_M \subseteq JR_M$, and the reverse is the same.

Is this correct? My solution look a little bit too easy compare to the other problems...

Thanks!!!
no! your proof is not correct: it's not true that $\displaystyle I=IR_{\mathfrak{m}}.$ if $\displaystyle R$ is an integral domain, then we can only say that $\displaystyle I \subseteq IR_{\mathfrak{m}}.$ but we may have $\displaystyle I \neq IR_{\mathfrak{m}}.$

for example if $\displaystyle R=\mathbb{Z}, \ I=3\mathbb{Z}, \ \mathfrak{m}=2\mathbb{Z},$ then $\displaystyle I \subsetneq IR_{\mathfrak{m}}=\{\frac{3k}{2n+1}: \ k, n \in \mathbb{Z} \}.$ anyway, here's how you should solve the problem:

it's obvious that $\displaystyle I=J$ implies that $\displaystyle IR_{\mathfrak{m}}=JR_{\mathfrak{m}}$ because by definition: $\displaystyle IR_{\mathfrak{m}}=\{\frac{a}{s}: \ a \in I, \ s \in R-\mathfrak{m} \}=\{\frac{a}{s}: \ a \in J, \ s \in R-\mathfrak{m} \}=JR_{\mathfrak{m}}.$

conversely suppose $\displaystyle IR_{\mathfrak{m}}=JR_{\mathfrak{m}},$ for all maximal ideals $\displaystyle \mathfrak{m}$ of $\displaystyle R.$ we'll show that $\displaystyle I \subseteq J$: let $\displaystyle a \in I$ and define $\displaystyle \mathfrak{L}=\{r \in R: \ ra \in J \}.$ see that $\displaystyle \mathfrak{L}$ is an

ideal of $\displaystyle R.$ if $\displaystyle \mathfrak{L}=R,$ then $\displaystyle 1 \in \mathfrak{L}$ and thus $\displaystyle a \in J.$ otherwise, $\displaystyle \mathfrak{L} \subseteq \mathfrak{m},$ for some maximal ideal $\displaystyle \mathfrak{m}.$ now since $\displaystyle a \in I \subseteq IR_{\mathfrak{m}}=JR_{\mathfrak{m}},$ we have $\displaystyle a=\frac{b}{s},$ for some

$\displaystyle b \in J, \ s \in R - \mathfrak{m}.$ but then $\displaystyle sa=b \in J,$ and thus $\displaystyle s \in \mathfrak{L} \subseteq \mathfrak{m},$ which is impossible because $\displaystyle s \in R-\mathfrak{m}.$ this completes the proof of $\displaystyle I \subseteq J.$ an exactly the

same argument shows that $\displaystyle J \subseteq I$ and therefore $\displaystyle I=J. \ \Box$