# Math Help - Maximal Ideals in Integral Domain

1. ## Maximal Ideals in Integral Domain

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

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

pf.

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

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

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

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

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

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

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

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

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

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

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

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

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

But is it true for the other way?

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

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

6. Thanks for your massive help!!!

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

Proof so far.

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

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

Now, I=J implies that there exists an element $j \in J$ such that $j=i$, thus $i \frac {a}{b} = j \frac {a}{b} \in JR_M$, so we have $IR_M \subseteq JR_M$, 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 $I=J$ if and only if $IR_M=JR_M$, for each maximal ideal M of R.

Proof so far.

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

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

Now, I=J implies that there exists an element $j \in J$ such that $j=i$, thus $i \frac {a}{b} = j \frac {a}{b} \in JR_M$, so we have $IR_M \subseteq JR_M$, 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 $I=IR_{\mathfrak{m}}.$ if $R$ is an integral domain, then we can only say that $I \subseteq IR_{\mathfrak{m}}.$ but we may have $I \neq IR_{\mathfrak{m}}.$

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

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

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

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

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

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