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Since userxxxxx (I don't remember the numbers) deleted his own question which I find interesting, let me repost it:

Let $f,g\in\mathbb Z[X]$ with $\mathrm{gcd}(f,g)=1$. Prove that the ring $\mathbb Z[X]/(f,g)$ is finite.

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    I think this is trivial actually. gcd(f,g) = 1 implies that (f,g) = Z[x] and Z[x]/Z[x] = {1} is clearly finite.2012-12-11
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    gcd$(2,X)=1$ ;)2012-12-11
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    @sunflower: as N.S. alludes, $\mathbb{Z}[x]$ is not a PID, so the result is not as trivial as might be thought.2012-12-11
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    @robjohn, I don't understand his comment. I can't even find a,b such that $2a + bX = 1$..2012-12-11
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    @sunflower Exactly ;) That means that $(2,x)$ cannot be $\mathbb Z[X]$, can it?2012-12-11
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    @N.S., I think it means gcd(2,X) isn't defined but why did you say it's 1?2012-12-11
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    @sunflower not all rings with gcd satisfy Bezout's identity, so you cannot always write the gcd as a linear combination of the elements.2012-12-11
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    @sunflower The GCD is defined, it is one. GCD makes sense in UFDs, but it can only be obtained as linear combination in Euclidina domains. $\mathbb Z[X]$ is UFD but not euclidian.2012-12-11
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    @sunflower: Bezout only holds in a PID. The fact that we cannot find $a$ and $b$ so that $2a+bX=1$ means that the ideal generated by $2$ and $X$ is not principal. However, $\mathrm{GCD}(2,X)=1$.2012-12-11
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    @robjohn Bezout also holds in a Bezout ring :)2012-12-12
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    @YACP: ah, yes. I was only considering finitely generated ideals.2012-12-12

5 Answers 5

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Consider $\mathbb{Z}[x]$ as a subring of $\mathbb{Q}[x]$ which is a PID. Since $\rm{gcd}(f,g) = 1$ in $\mathbb{Z}[x]$ the same is true in $\mathbb{Q}[x]$ so $\mathbb{Q}(f,g) = \mathbb{Q}[x]$ which means that $(f,g)$ contains an element that is invertible inside $\mathbb{Q}[x]$, so it contains some integer. Now we know that the quotient is actually a quotient of $(\mathbb{Z}/n\mathbb{Z})[x]$ which means that unless $(f,g)$ consists of just constants, we get a finite ring, as we get a bounded degree. But if this is the case, then both $f$ and $g$ are constants, and $(f,g) = \mathbb{Z}[x]$ since they have gcd equal to $1$.

Edit: The last part is missing some ingredient as noted in the comment.

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    The last part is where I had troubles: why do we get a bounded degree? Note that the leading degrees of $f,g$ might not be relatively prime, and they could both be non-invertible in $\mathbb Z/n\mathbb Z$...2012-12-11
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    You are right, I had not thought that part properly through.2012-12-11
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    I think that part is more suttle than we expected. BTW: I didn't downvote, I never downvote in questions where I provided answers ;)2012-12-11
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The already-present answers show that $(f,g)$ necessarily contains a non-zero integer $n$, and so the quoitient is equal to $(\mathbb Z/n\mathbb Z)[x]/(f,g).$ So it remains to show that this quotient is finite.

There are various ways to proceed at this point. Here is one.

  • First, suppose $n$ is a prime $p$. Since $f$ and $g$ don't have a non-trivial common factor in $\mathbb Z[x]$, at least one of them is non-zero mod $p$, and the quotient of $\mathbb F_p[x]$ by any non-zero ideal is finite.

  • Now if $n$ is a prime power, say $p^e$, then by what we just proved, the ideal $(f,g)$ has non-zero image in $\mathbb F_p[X]$, hence it contains a polynomial which is monic when reduced mod $p$. An induction proves that some power of this polynomial is actually monic mod $p^e$. Thus $(f,g)$ contains a monic polynomial when reduced mod $p^e$, and so $(\mathbb Z/p^e\mathbb Z)[x]/(f,g)$ is finite.

  • Factor $n = p_1^{e_1}\cdots p_r^{e_r}$. Then by CRT, there is an isomorphism

$$(\mathbb Z/n\mathbb Z)[x]/(f,g) \cong \prod_{i =1}^r (\mathbb Z/p^{e_i}_i \mathbb Z)[x]/(f,g),$$ and so by what we have already proved, $(\mathbb Z/n\mathbb Z)[x]/(f,g)$ is a product of finite rings, hence finite.

Added: Now that some solutions from first principles have been put up, it may be worth describing a way to deduce this statement from general principles of commutative algebra.

First, if $J$ is any ideal in a Noetherian ring $A$, and $I = $rad$(J)$, then since $J \subset I,$ we have that $A/I$ is a quotient of $A/J$. Thus if $A/J$ is finite, so is $A/I$. On the other hand, by Noetherianness (more precisely, by the fact that $I$ is f.g.), we have $I^n \subset J$ for some $n$, and so $A/J$ is a quotient of $A/I^n$, which is filtered by $A\supset I \supset I^2 \supset \cdots \supset I^n.$ Each quotient $I^i/I^{i+1}$ is a f.g. $A/I$-module, and so if $A/I$ is finite, so is each $I^i/I^{i+1}$, hence so is $A/I^n$, and hence so is $A/J$. Thus $A/J$ is finite if and only if $A/I$ is.

Now take $A = \mathbb Z[x]$. This is a UFD of Krull dimension two, and so the prime ideals of $A$ are either $0$, height one and hence principal, or maximal. Let $J = (f,g)$. Since $f$ and $g$ have no non-trivial common divisor, the ideal $(f,g)$ cannot be contained in a principal prime ideal, and so each minimal prime of $J$ must be a maximal ideal of $A$. Thus if $I$ is the radical of $J$, then $I = \mathfrak m_1 \cap \cdots \mathfrak m_r$ for some maximal ideals $\mathfrak m_i$, and so $A/I$ embeds into the product $\prod_{i=1}^r A/\mathfrak m_i$. Thus to show that $A/I$ is finite, it suffices to show that $A/\mathfrak m$ is finite for any maximal ideal of $A$. This last fact follows from the general version of the Nullstellensatz for Jacobson rings applied to $\mathbb Z[x]$ (using that $\mathbb Z$ is Jacobson).

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    I'd be perfectly convinced if you indicated something about the inductive argument about powers of the fixed polynomial in the second step.2012-12-11
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    @MannyReyes: Dear Manny, Suppose that $r$ is a monic, and that $s$ is some polynomial, so that $r \equiv s \bmod p^i$. Then raising to the $p$th power gives that $r^p \equiv s^p \bmod p^{i+1}.$ Noting that $r^p$ is monic since $r$ is, we see that $s^p$ is congruent to a monic polynomial mod $p^{i+1}$. Continuing, and then returning to the context of my answer, we find that if $s \in (f,g)$ is such that $s \bmod p$ is monic, then $s^{p^e}$ is an element of $(f,g)$ which is monic mod $p^e$. Regards,2012-12-11
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    Dear Matt, thanks very much. It was the fact that you wanted to use $p$th powers that I was missing - clever!2012-12-11
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    @Manny: Dear Manny, Thanks! (This is not my cleverness, though! --- raising to the $p$th power like this is a standard way to "improve" a mod $p$ congruence, so you get used to doing it when you work with $p$-adic numbers and such a lot.) Best wishes,2012-12-11
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    @MattE Your add it's nice (but some typos deserve to be corrected) and resumes very well my answer. However I have a comment related to your answer: althought I didn't check it carefully, the use of the Chinese Remainder Theorem in this form $$(\mathbb Z/a\mathbb Z)[x]/(f,g) \cong \prod_{i =1}^r (\mathbb Z/p^{e_i}_i \mathbb Z)[x]/(f,g),$$ by sending the ideal $(f,g)$ on each component of the product, needs some clarifications.2012-12-12
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    @MattE Alternatively, to deduce the finiteness of $A/\mathfrak{m}$ we know (more generally) that $\textrm{Spec}\Bbb{Z}[x]$ consists of $(0)$, ideals generated by prime numbers, $(f)$ for $f$ an irreducible polynomial and the maximal ideals $(p,g)$ where $g$ is irreducible mod $p$. It will now follow that $A/\mathfrak{m}$ is finite.2012-12-12
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    @MattE For those interested in commutative algebra (like me) the last part was extremely insightful! Out of interest, I was trying to do this problem and was wondering if we could reduce to the case when the ring in question is the ring of integers of some finite extension of $\Bbb{Q}$?2012-12-12
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    @BenjaLim I simply wonder if you read my answer, cause the finiteness of $A/m$ is explained there by using exactly what you say.2012-12-12
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    @YACP: Dear YACP, There is an isomorphism of rings $\mathbb Z/a\mathbb Z \cong \prod \mathbb Z/p_i^{e_i} \mathbb Z,$ and hence an isomorphism of rings $(\mathbb Z/a \mathbb Z)[x] \cong \prod (\mathbb Z/p_i^{e_i}\mathbb Z)[x].$ Tensoring both sides with $\mathbb Z[x]/(f,g)$ over $\mathbb Z[x]$ then gives the isomorphism I claim. Regards,2012-12-12
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    @BenjaLim: Dear Benjamin, I'm not sure what you mean in your question about reducing to a ring of algebraic integers. Could you say a little more? Cheers,2012-12-12
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Complete Answer (?)

Note that $\mathbb Z[X]$ is not euclidian, but $\mathbb Q[X]$ is.

Since gcd$(f,g)=1$ there exists $h_1,h_2 \in \mathbb Q[X]$ so that

$$1=h_1f+h_2g$$

Let $a$ be the common denominator of the the coefficients of $h_1,h_2$. Then we get

$$a=(ah_1)f+(ah_2)g \,.$$

Since $ah_1, ah_2 \in \mathbb Z[X]$ we get $a \in (f,g)$.

Now, by the Third Isomorphism Theorem

$$\frac{\mathbb Z[X]}{(f,g)} \sim \frac{\mathbb Z [X]/(a)}{(f,g)/(a)} \sim \frac{\left(\mathbb Z/a\mathbb Z \right)[X]}{(f,g)} $$

Now we know that

$$(\mathbb Z/a\mathbb Z)[x]/(f,g) \cong \prod_{i =1}^r (\mathbb Z/p^{e_i}_i \mathbb Z)[x]/(f,g),$$

We claim that $ (\mathbb Z/p^{e_i}_i\mathbb Z)[x]/(f,g)$ is finite.

Indeed, since gcd$(f,g)=1$ then, one of them has a coefficient not divisible by $p_i$. Lets say this is $f$.

Then, we can write $f=f_1-f_2$ such that all the coefficients of $f_1$ are relatively prime to $p_i$ and all the coefficients of $f_2$ are divisible by $p_i$.

Then, since $f_2^{e_i}=0$ we have

$$f_1^{e^i}=f_1^{e^i}-f_2^{e^i}=(f_1-f_2)(\mbox{junk})=f(\mbox{junk}) \in (f,g)$$

Let $b$ be the largest coefficient of $f_1$, then the largest coefficient of $f_1^{e_i}$ is $b^{e_i}$. Since this is invertible, we get

$$b^{-e_i}f_1^{e^i} \in (f,g)$$

Since we found a polynomial with leading coefficient $1$ in $(f,g)$, it is easy to conclude that $ (\mathbb Z/p^{e_i}_i\mathbb Z)[x]/(f,g)$ is finite.

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    If $a$ is prime, we have a field and finiteness in the last step is clear. Otherwise, one can proceed prime by prime factor until any given polynomial has nonzeor coefficients only at low degrees.2012-12-11
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    @HagenvonEitzen What if $a$ is divisible by some $p^n$? Note that $p$ might divide both the leading coefficients of $f,g$.2012-12-11
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    @N.S. Dear N.S., Since $f$ and $g$ *don't* admit $p$ as a common divisor (by assumption), at least one of them is non-zero mod $p$, and $\mathbb F_p[X]$ mod *any* non-zero ideal is finite. Regards,2012-12-11
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    P.S. As a variant on @HagenvonEitzen 's comment (or perhaps it is what was meant) if you factor your $a$ as a product of primes, then you can find a filtration on your quotient whose successive graded pieces are $\mathbb F_p[X]/(f,g)$ (as $p$ ranges over all primes divising $a$), hence reduce to this this case. (Although there are plenty of variants at this point, not all of which require this filtration/factoring of $a$ argument.)2012-12-11
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    Dear N.S., My filtration argument may be bogus, or at least a bit glib. I am posting an answer below which gives a hopefully correct variant. Cheers,2012-12-11
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    @MattE I wonder if there is any connection between your filtration argument and my argument, found it on the way to school. My idea is to use nilpotency to get rid of the leading terms of $f$ divisible by the prime ;)2012-12-11
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    @N.S. Dear N.S., Indeed, I think that your nilpotency argument is closely related to the induction in my answer below to go from $\bmod p$ to $\bmod p^e$. (If you look at the comment I added in response to a request for more details, it looks a lot like your nilpotency argument!) Cheers,2012-12-11
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    @MattE Yes, that's whart I meant but was too lazy to persue until the end :)2012-12-12
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Proposition. Let $f,g\in\mathbb Z[X]$ with $\mathrm{gcd}(f,g)=1$. Then $\mathbb Z[X]/(f,g)$ is finite.

Proof. If $(f,g)=\mathbb Z[X]$, then there is nothing to prove. Assume $(f,g)\neq\mathbb Z[X]$. Since $\mathrm{gcd}(f,g)=1$, the elements $f,g$ form an $\mathbb Z[X]$-sequence and therefore $\dim\mathbb Z[X]/(f,g)=0$, i.e. the ring $\mathbb Z[X]/(f,g)$ is artinian. In particular, there are only finitely many maximal ideals of $\mathbb Z[X]$ containing $(f,g)$ and their product at some power coincides with $(f,g)$. This shows that $\mathbb Z[X]/(f,g)$ is isomorphic to a finite product of rings of the form $$\mathbb Z[X]/(M^k+(f,g))\simeq\frac{\mathbb Z[X]/M^k}{(\hat{f},\hat{g})},$$ where $M\subset\mathbb Z[X]$ is a maximal ideal. Thus it is enough to show that $\mathbb Z[X]/M^k$ is a finite ring. In order to do this let's recall the form of the maximal ideals of $\mathbb Z[X]$: $M=p\mathbb Z[X]+u\mathbb Z[X]$ where $p\in\mathbb Z$ is a prime number and $u\in\mathbb Z[X]$ is a polynomial irreducible modulo $p$. Obviously $$\mathbb Z[X]/M\simeq\mathbb Z/p\mathbb Z[X]/(\overline{u})$$ is a finite field and $M^{i-1}/M^i$ are $\mathbb Z[X]/M$-vectorspaces of finite dimension. This shows that $\mathbb Z[X]/M^k$ is a finite ring.

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If it's OK I would like to post my rough notes on understanding this brilliant question. I don't have a solution but for beginners to understand the problem itself this may be useful.


Some theory of GCD in $\mathbb Z[X]$ that I was not aware of:

Definition Let $D|P$ and $D|Q$ then $D$ is a common divisor of $P$ and $Q$.

Definition The set of all common divisors of $P$ and $Q$ is a preorder order by the divisibility relation, the greatest element is the greatest common divisor of $P$ and $Q$.

Example $\{-1,1\}$ is the set of common divisors of $2$ and $X$. So $\gcd(2,X) = 1$ (by convention we don't pick $-1$).

Theorem The GCD is unique if it exists. proof: Let $D$,$D'$ be maximal elements of the divisibility preorder order on the set of common divisors of $P$ and $Q$ and suppose for contradiction that they are not associates, then $D \not | D'$ and $D' \not | D$. Since $D|P$ and $D'|P$ let $P=DA=D'A'$ and since $D'|DA$ we find $D|A$ hence $DD'|P$ contradicts the maximality proving that $D$ and $D'$ are associates.

Theorem The GCD exists. proof: factor uniquely into irreducible powers and take the intersection. Any larger element of this in the poset will not be a common divisor so it's maximal.


my idea for a proof and some explicit calculations.

I view $\mathbb Z[X]$ as an infinite dimensional space with axis being integers, X*integers, X^2*integers, ...

To prove that $\mathbb Z[X]/(F,G)$ is finite we need to show that some integer $i \in (F,G)$ (so there are at most $i^\infty$ points in the quotient) and that some $X^r \in (F,G)$ (so the hypercube has finite dimension therefore at most $i^r$ points in the quotient).

Example 1: $(1+X+X^2,3+5X)$. The polynomials have gcd 1. and we can do the following two calculations:

5+5X+5X^2  -3X-5X^2 --------- 5-2X  25-10X  6-10X ------ 31 in an integer in the ideal    3+3X+3X^2 -3-5X ----------   -2X+3X^2    -6X+ 9X^2    6X+10X^2   ---------       19X^2 is in the ideal   and since (31,19)=1 we get X^2 in the ideal. 

Example 2: $(A,B) = (X^3 + X^2 + X + 1, 9X^2 + 5X + 3)$ when we perform the same process as before we get $81A-(9+9X-5)B = 34x + 69$ instead of the integer and $9A - B(3-2X) = 27X^3 - 8X^2$ which is not monic! but.. $\gcd(34X+69,27X^3-8X^2) = 1$ and it does have smaller degree so we can do the process again (induction proof).


It seems like a proof could work along these lines but the hardest bit is showing both processes lead to polynomials which are coprime.

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    The "divides" relation is not quite an ordering. It is only an ordering up to units.2012-12-11
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    Or, in other words, it is a _preorder_.2012-12-11
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    thank you very much I corrected it.2012-12-11