MHB Understanding a problem in ring theory

[kalish1]

Can someone here help me fill in the gaps of my understanding for this problem? I would appreciate it.

**Problem:** If $f(x) \in \mathbb{C}[x]$ is a nonzero polynomial of degree $n$, prove that the ring $R=\frac{\mathbb{C}[x]}{(f)}$ has finitely many distinct ideals. How many distinct ideals does it have?

**Teacher's explanation:** $$R=\frac{\mathbb{C}[x]}{(f)} \longleftarrow \frac{J}{(f)}$$ means that $J \supseteq (f).$

*She mentioned that $(u \cdot g) = (g)$, but I don't see how.*

Now, $\frac{(g)}{(f)}$ where $(g) \supseteq (f)$, implies that $f \in (g)$. Thus $f$ is a multiple of $g$ and $g|f$. Thus $g$ is a monic divisor of $f$. Note that $f=c(x-x_1)(x-x_2)\cdots(x-x_n)$ by the Linear Factorization Theorem. Now we count the monic divisors of $f$, which turns out to be:
$$ 1, (x-x_1), \ldots, (x-x_n)$$ and all products of the linear factors. Thus there are $2^n$ distinct ideals.

**My questions:**
What does $R=\frac{\mathbb{C}[x]}{(f)} \longleftarrow \frac{J}{(f)}$ mean? Why is $(u \cdot g) = (g)?$ How does counting the monic divisors of $f$ correlate with the number of distinct ideals?

Thanks.

This question has been crossposted here: abstract algebra - Understanding a problem in ring theory - Mathematics Stack Exchange

 

[Fallen Angel]

Hi,

I haven't seen before that notation with the arrow, I guess is just saying $J$ being an ideal on the quotient implies $J$ is an ideal of $C[X]$ which contain $(f)$.

$(u\cdot g)=(g)$ I suppose $u\in \mathbb{C}$, and that means we can always consider a monic generator for any ideal.

$\mathbb{C}[X]$ is an Euclidean domain, so also a $PID$, and $(f)\subset (g)$ means that there exist some polynomial $h$ such that $f=gh$ so $g$ is a divisor of $f$, and monic for the reason above.

 

[kalish1]

Hi,

I haven't seen before that notation with the arrow, I guess is just saying $J$ being an ideal on the quotient implies $J$ is an ideal of $C[X]$ which contain $(f)$.

$(u\cdot g)=(g)$ I suppose $u\in \mathbb{C}$, and that means we can always consider a monic generator for any ideal.

$\mathbb{C}[X]$ is an Euclidean domain, so also a $PID$, and $(f)\subset (g)$ means that there exist some polynomial $h$ such that $f=gh$ so $g$ is a divisor of $f$, and monic for the reason above.

We haven't learned about domains or PIDs yet, and I'm still unsure of how $(u\cdot g)=(g).$ Also, where does the Correspondence Theorem come into play? Could you please elaborate with some examples?

 

[Fallen Angel]

Hi Kalish,

I will state a couple of theorems that I was assuming you know, let me know if you have heard about it or have some trouble proving it.

Theorem 1
If $K$ is a field, then $K[X]$ is an euclidean domain, i.e. for any given polynomials $f(x),g(x)\in K[X]$ there exists two polynomials $q(x),r(x)\in K[X]$ such that
$\bullet$ $f(x)=g(x)q(x)+r(x)$
$\bullet$ $deg(r)\leq deg(g)-1$
Furthermore, this polynomials are unique.Theorem 2 (Bezout's identity restricted to $K[X]$)
Let $K[X]$ be the ring of polynomials over a field and let $f_{1},f_{2},\ldots f_{n}\in K[X]$ be a finite set of polynomials such that $gcd(f_{1},f_{2},\ldots ,f_{n})=h$, then there exists $g_{1},g_{2},\ldots g_{n}\in K[X]$ such that
$h=\displaystyle\sum_{i=1}^{n}f_{i}g_{i}$

Theorem 3
Let $R$ be a ring and $I\subset R$ an ideal, if $a\in R^{*}$ then $I=aI$, where $R^{*}$ is the set of all elements in $R$ that have an inverse with respect to the multiplication.

Theorem 4
Let $R$ be a ring and $I\subset R$ an ideal, and $\pi_{I}: R \longrightarrow R/I$.
$\hspace{8cm} r \ \ \mapsto \ r+I$
Then $J\subseteq R/I$ is an ideal of the quotient ring if and only if it there exists an ideal $J'\subseteq R$ containing $I$ such that $\pi_{I}(J')=J$.

 

[kalish1]

Thanks for that. Do you think $2^n$ is correct for the number of distinct ideals?

Hi Kalish,

I will state a couple of theorems that I was assuming you know, let me know if you have heard about it or have some trouble proving it.

Theorem 1
If $K$ is a field, then $K[X]$ is an euclidean domain, i.e. for any given polynomials $f(x),g(x)\in K[X]$ there exists two polynomials $q(x),r(x)\in K[X]$ such that
$\bullet$ $f(x)=g(x)q(x)+r(x)$
$\bullet$ $deg(r)\leq deg(g)-1$
Furthermore, this polynomials are unique.Theorem 2 (Bezout's identity restricted to $K[X]$)
Let $K[X]$ be the ring of polynomials over a field and let $f_{1},f_{2},\ldots f_{n}\in K[X]$ be a finite set of polynomials such that $gcd(f_{1},f_{2},\ldots ,f_{n})=h$, then there exists $g_{1},g_{2},\ldots g_{n}\in K[X]$ such that
$h=\displaystyle\sum_{i=1}^{n}f_{i}g_{i}$

Theorem 3
Let $R$ be a ring and $I\subset R$ an ideal, if $a\in R^{*}$ then $I=aI$, where $R^{*}$ is the set of all elements in $R$ that have an inverse with respect to the multiplication.

Theorem 4
Let $R$ be a ring and $I\subset R$ an ideal, and $\pi_{I}: R \longrightarrow R/I$.
$\hspace{8cm} r \ \ \mapsto \ r+I$
Then $J\subseteq R/I$ is an ideal of the quotient ring if and only if it there exists an ideal $J'\subseteq R$ containing $I$ such that $\pi_{I}(J')=J$.

 

[Fallen Angel]

No, it makes no sense.

The number of ideals will be the number of monic divisors of the polinomyal, so it can be just 2 if the polyomial is irreducible or whatever if not.

The four theorems above are the way to prove it. (They are the hidden facts in you first prove)