Find the number of the ideals in this ring.

Question

Question) Let the integer set, $Z$

The ring, $R = Z/ \langle 300 \rangle$

For ideal $I$ of $R$ $s.t.$ $I = \langle a \rangle / \langle 300 \rangle$

Then How many number of the $I$ which is commutative ring with unity?

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My trail)

Since $R$ has a element that $1+ \langle 300 \rangle$, $R$ is a commutative ring with unity. Hence $I$ is a commutative.

(1) Plus, If the ideal $I $ have a unity $s.t. I \subset R$, then $R=I$

(2) By third isomorphic theorem, $R/I \simeq (Z/<300>) / (<a> / 300) \simeq Z/<a>$

By (1) and (2), All we have to do is just find the $a \vert 300$ $s.t.$ $\vert Z/<a> \vert = 1$

Hence $a = \pm 1$. (The number of the the $I$ is $1$).

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But the answer sheet said there are $a = 1,3,4,12,25,72,100,300$. Therefore there are $8$ ideals.

There aren't any solution in my text book. They give me only answer. :(

I don't know why I'm not correct and why should be $a = 1,3,4,12,25,75,100,300$?

Any help would be thanksful.

Answer 1

I believe you are looking for those ideals $I$ of $R=\Bbb{Z}/300\Bbb{Z}$ such that $R/I$ is a commutative ring with identity. If not, then I am not sure what you are asking and this whole answer is meaningless.

By the correspondence theorem (basically using the projection map $\pi:\Bbb{Z} \longrightarrow \Bbb{Z}/300\Bbb{Z}$) the ideals of $\Bbb{Z}/300\Bbb{Z}$ are in bijective correspondence to ideals of $\Bbb{Z}$ that contain $300\Bbb{Z}$. The ideals that contain $300\Bbb{Z}$ are of the form $d\Bbb{Z}$, where $d \mid 300$.

Moreover by the third isomorphism theorem $$R/I=\left(\Bbb{Z}/300\Bbb{Z}\right)/\left(d\Bbb{Z}/300\Bbb{Z}\right) \cong \Bbb{Z}/d\Bbb{Z}.$$ Now (if) we want $R/I$ to be commutative ring with identity, we want an (identity) element $d\alpha \in [d]_{300}$ such that $$d(d\alpha) \equiv d \pmod{300} \implies d\alpha \equiv 1 \pmod{\frac{300}{d}}.$$ This means we want those $d$ such that

$$\gcd\left(d, \frac{300}{d}\right)=1.$$

Now you can get the values of $d$ that are you looking for. Essentially your $d=2^{a} \,3^{b} \,5^{c}$, and based on the condition derived above, the choices for $a,b,c$ are $a=0,2, \,\, b=0,1, \,\, c=0,2$, a totality of $8$ possibilities.

NOTE: one of the values in your answer is a typo. Instead of $72$ it should be $75$.