Original problem

Suppose $K$ is a field, $f,g\in K[x]$ and $\gcd(f,g)=1$. $V$ is a linear space based on $K$ and $\phi\in\operatorname{End}(V)$. Notation $f_\phi$ and $g_\phi$ denote linear transformations $f(\phi)$ and $g(\phi)$. Try to show that $$\ker(f_\phi g_\phi)=\ker(f_\phi)\oplus\ker(g_\phi)$$

The sketch of the proof

Notice that sub-ring $K[\phi]\subseteq\operatorname{End}(V)$ is commutative, and $f_\phi,g_\phi\in K[\phi]$, so it's easy to show that $\ker(f_\phi),\ker (g_\phi)\subseteq\ker(f_\phi g_\phi)$.

$\because\gcd(f,g)=1$, $\therefore\exists u,v\in K[x]: uf+vg=1\implies u_\phi f_\phi+v_\phi g_\phi=I_V$, where $I_V$ is the identity operator of $V$.

$\forall x\in\ker(f_\phi g_\phi)$, plug $x$ into the preceding equation, we have $x=u_\phi f_\phi(x)+v_\phi g_\phi(x)$, and it's not quite difficult to show that $u_\phi f_\phi(x)\in\ker(g_\phi)$ and $v_\phi g_\phi(x)\in\ker(f_\phi)$, so $x\in\ker(f_\phi)+\ker(g_\phi)$.

$\forall x\in\ker(f_\phi)\cap\ker(g_\phi)$, plug $x$ into that equation and we have $x=0$, so $\ker(f_\phi)\cap\ker(g_\phi)=0$, Q.E.D


It seems that the proposition is quite similar to Chinese Remainder Theorem, say: $$\mathbb Z/mn\mathbb Z\cong\mathbb Z/m\mathbb Z\oplus\mathbb Z/n\mathbb Z$$ if $\gcd(m,n)=1$.

Can we find the deeper relationship between these two propositions? Or just an unique aspect?


  • 2
    $\begingroup$ A first thing to observe is that your first decomposition is one of vector spaces, and the second one of rings. What common ground do you propose for the two statements? $\endgroup$ – Marc van Leeuwen Jan 10 '13 at 15:06
  • $\begingroup$ @MarcvanLeeuwen I don't know. I'm not very familiar but I doubt there's some connections. $\endgroup$ – Yai0Phah Jan 10 '13 at 15:14
  • $\begingroup$ One might propose that both statements are holding as modules, for exmaple. One for a $K$ module, one for a $\Bbb Z$ module. The fact that one is also a ring homomorphism and the other is definitely not is conspicuous, though. $\endgroup$ – rschwieb Jan 10 '13 at 16:24
  • $\begingroup$ I've edited the title, the old one made it hard to guess what the question is about $\endgroup$ – Marc van Leeuwen Jan 13 '13 at 5:16

I think one analogy is being crossed-up with another.

For one thing, your work shows that $f_\phi$ and $g_\phi$ are coprime in the ring $K[\phi]$, and so $K[\phi]/(f_\phi g_\phi)\cong K[\phi]/(f_\phi)\oplus K[\phi]/(g_\phi)$ would be a better analogy with the Chinese remainder theorem comment on $\Bbb Z$ that you made, both of these being ring decompositions.

Since the kernel equation ceases to talk about $f_\phi$ and $g_\phi$ as elements of a ring of transformations $V\rightarrow V$, and instead switches to their kernels, the analogy with $\Bbb Z$ is lost. It's not clear what the elements of $\Bbb Z$ should be operating on to produce kernels.

However, if we allowed elements of $\Bbb Z$ to act as maps from $\Bbb Z\rightarrow \Bbb Z/(k)$ for different $k$'s, then the kernels produced would be subgroups of $\Bbb Z$, which are all cyclic of course. And it is true that the cyclic group equation $C_n\times C_m\cong C_{mn}$ when $m$ and $n$ are coprime. This might be the closest relative of your kernel equation.

A final difference to note is that the kernel equation is taking place in a vector space (a $K$ module), and we know that vector spaces decompose very nicely. On the other hand, $\Bbb Z$ modules do not have these nice decomposition properties.

  • $\begingroup$ In fact, I'm learning linear algebra so I don't know the notations of the deep theories, but there's a theorem about primary decomposition in the next chapter, normal forms and space decompositions, of the linear algebra textbook: Suppose $R$ is PID and $p_1,\ldots,p_n\in R$ are pairwise coprime, $m=p_1\cdots p_n$ and $M$ is $R$-module, then $\ker m=\ker p_1\oplus\cdots\oplus\ker p_n$, where $\ker p=\left\{\,x\in M\,\big|\,px=0\,\right\}$ is submodule of $M$. $\endgroup$ – Yai0Phah Jan 11 '13 at 6:35
  • $\begingroup$ I finally discover that it's a general case of Chinese Remainder Theorem, for $\mathbb Z/m\mathbb Z\cong\ker(n+mn\mathbb Z)$. $\endgroup$ – Yai0Phah Jan 11 '13 at 6:39
  • $\begingroup$ @FrankScience Yes, it's true that all the rings discussed are principal ideal rings, and that is something in common. $\endgroup$ – rschwieb Jan 11 '13 at 13:54

There is a connection between these results, but it is harder to formulate than one might think. Maybe there is something better than what I propose here by making the setting (even) more general, but this is what I found (when I thought about this connection some time ago).

I'll call the result you started out with the kernel decomposition theorem; I'm not sure this is the correct English term or even if the result carries a name at all, but in France it is called (in a version with any number of pairwise relatively prime polynomials) "Théorème de décomposition des noyaux" (and it is a cornerstone in any advanced linear algebra course). The result is very general, and does not even require any finite dimensionality, as the given proof shows. So I'll try to deduce from it some form of the Chinese remainder theorem. It would seem difficult to do this for the original form of that theorem, which does not involve any vector spaces, so instead I'll go for the form that there is a $K$-algebra decomposition $$ K[X]/(P_1\ldots P_k)=K[X]/(P_1)\oplus\cdots\oplus K[X]/(P_k) $$ whenever $P_1,\ldots,P_k$ are pairwise relatively prime polynomials; this is the natural "$K$-linear version" of the Chinese remainder theorem following the usual analogy between $\def\Z{\Bbb Z}\Z$ and $K[X]$.

Although the hypotheses on the polynomials match, this "Chinese remainder" statement is not that of the kernel decomposition theorem; the main problem is that we are decomposing a quotient of $K[X]$ into other such quotients, rather than kernels, which are subspaces of a given space. To remedy this, I need to see the CR theorem as one of a decomposition of $K[X]$-modules, and in particular of submodules of one big module that contains all quotients of $K[X]$ in a unique way as submodule. That submodule is the analogue of the $\Z$-module $\Bbb Q/\Z$ (in which every finite cyclic group occurs exactly once as submodule): it is the $K[X]$-module $M=K(X)/K[X]$ of rational functions in $X$ (over $K$), modulo polynomials in $X$.

For any monic polynomial $P$ the quotient module $K[X]/(P)$ is realised inside $M$ as the elements with a representative rational function having $P$ as denominator: for such rational functions the numerator is effectively an element of $K[X]/(P)$. This submodule is also the kernel of the endomorphism $P_\phi$ of $M$ of multiplication by $P$ (this follows the notation in the question, with $\phi:M\to M$ being multiplication by $X$). Under these identifications, the kernel decomposition theorem gives the stated version of the Chinese remainder theorem, but as an isomorphism of $K[X]$-modules. However, since the occurring $K[X]$-modules are all cyclic modules, they can be equipped by a $K[X]$-algebra structure compatible with the module structure by the choice of any generator $u$ of the module as the unit element of the algebra (the map from $K[X]$ sending $Q\mapsto Qu$ becomes a quotient map). Then any $K[X]$-module morphism becomes a $K$-algabra morphism provided it sends unit element to unit element, and the decomposition of $K[X]$-modules obtained by the kernel decomposition theorem will become one of $K[X]$-algebras, provided that the unit elements in the summands of the right hand side are chosen to be the projections of the unit element on the left.

The transition from $K[X]$-modules to $K[X]$-algebras is not entirely innocent, as is witnessed by the fact that the most obvious set of choices for a generator in the kernel of each $P_\phi$, namely taking (the image in $M$ of) $\frac1P$, is not compatible with the decomposition: putting $P=P_1\ldots P_k$, the rational function $\frac1P$ is not congruent, modulo polynomials, to the sum $\frac1{P_1}+\cdots+\frac1{P_k}$. One either needs to choose the strange generator $\frac1{P_1}+\cdots+\frac1{P_k}$ in $\ker(P_\phi)$, or if one chooses the generator $\frac1P$ there, then the generator in $\ker((P_i)_\phi)$ must be the "Bezout term" $C_i\tilde P_i$, where $\tilde P_i=P/P_i$ is the product of all $P_j$ for $j\neq i$ and $C_i\in K[X]$ is the Bezout coefficient such that $C_1\tilde P_1+\cdots+C_k\tilde P_k=1$. Either way the projection $K[X]/(P)\to K[X]\to(P_i)$ is not realised by multiplying the denominator by $\tilde P_i$, but rather by $C_i\tilde P_i$.

All this has a (simpler) equivalent for $\Z$ in place of $K[X]$. Instead of the kernel decomposition theorem there is a decomposition theorem for Abelian groups, namely that if $m\in\Z$ is a product of pairwise relatively prime factors $q_i$, then the torsion subgroup annihilated by multiplication by $m$ decomposes as direct sum of the subgroups annihilated by multiplication by each of the $q_i$. This gives the decomposition of the cyclic group $\Z/m\Z$ as direct sum of cyclic groups $\Z/q_i\Z$, which can be realised inside the group $\Bbb Q/\Z$. To make this into a decomposition of rings one needs to choose generators in the cyclic groups, and the most obvious choosing $\frac1n+\Z$ as generator for each cyclic subgroup $\Z/n\Z$ inside $\Bbb Q/\Z$ will not work here either.

Note that if one states (as is usual) the Chinese remainder theorem as a decomposition of $\Z/m\Z$ as a product of rings, then the projections $\Z/m\Z\to\Z/q_i\Z$ are painless (just mod out further by $q_i$), but it is the other direction (lifting a tuple of elements in the various $\Z/q_i\Z$ to an element of $\Z/m\Z$) that requires using the Bezout terms. The situation is in this respect quite opposite to that of submodules of $\Bbb Q/\Z$!

  • $\begingroup$ As I've commented under the answer of rschwieb's, it might be a decomposition theorem of torsion module. Well, in fact, soon after I've asked the problem here, I read the textbook randomly and found that theorem. $\endgroup$ – Yai0Phah Jan 11 '13 at 6:52

Since my previous answer, I have realised there is a very simple way to link the Chinese remainder theorem for $K[X]$, which is a result for rings, to the kernel decomposition theorem for vector spaces.

The Chinese remainder theorem for a principal ideal domain $R$ says that if $a_1,\ldots,a_k\in R$ are pairwise relatively prime, then $R/a_1\ldots a_kR\cong(R/a_1R)\times\cdots\times(R/a_kR)$ via the map that on component$~i$ of the product is given by reduction modulo$~a_i$ as morphism $R/a_1\ldots a_kR\to R/a_iR$. If $\widehat a_i=a_1\ldots a_k/a_i$ denotes the product of all factors except $a_i$, it essentially states the existence of Bezout coefficients $c_1,\ldots,c_k\in R$ such that $c_1\widehat a_1+\cdots+c_k\widehat a_k=1$, since this requires the class of $c_i\widehat a_i$ to project to the element $e_i=(0,\ldots,0,1,0,\ldots,0)$ of the product ring (the $1$ being in position$~i$), and once these classes are given one can just choose representatives, adapting the last choice so as to make the equality to$~1$ exact (rather than just a congruence modulo $a_1\ldots a_k$).

Applying this for $R=K[X]$, one gets that if $P_1,\ldots,P_k\in K[X]$ are pairwise relatively prime polynomials, then $K[X]/(P_1\ldots P_k)\cong(K[X]/(P_1))\times\cdots\times(K[X]/(P_k))$. Also for any endomorphism$~\phi$, the kernel $\ker(P_1\ldots P_k[\phi])$ is naturally a module over that ring. The final step that I was missing before is the following simple result (it surprises me that I never saw it stated explicitly)

Lemma. Any module $M$ over a product $R=R_1\times\cdots\times R_k$ of (unitary) rings canonically decomposes as direct sum $M=M_1\oplus\cdots\oplus M_k$, where $M_i$ is a module over$~R_i$ (the other factors acting as $0$ on it).

The proof is almost trivial: if $e_i\in R$ is the element with coordinate$~1$ in the factor $R_i$ and$~0$ in the the factors, one has $M_i=e_iM$, and the direct sum decomposition is a consequence of $e_1+\cdots+e_k=1$ together with the relations $e_i^2=e_1$ and $e_ie_j=0$ for $i\neq j$. Not even commutativity of$~R$ is required.

Back to linear algebra, this says that $\ker(P_1\ldots P_k[\phi])$ canonically decomposes as a direct sum of submodules, summand $M_i$ being obtained as image of (multiplication by) the polynomial $C_i[\phi]$ in$~\phi$, where $C_i$ is the Bezout coefficient satisfying $C_1\widehat{P_1}+\cdots+C_k\widehat{P_k}=1$ as above. This submodule can be seen to be equal to the kernel of $P_i[\phi]$, whence the kernel decomposition theorem.


Assuming I haven't overlooked anything....

Your proof doesn't make use of the fact that $f$ and $g$ are polynomials or that $V$ was actually a vector space: it would work for any commutative subring of $End(A)$ for an Abelian group $A$. (although you'd have to replace "$\gcd(f,g)=1$" with "$f$ and $g$ are coprime")

In particular, in your Chinese Remainder Theorem example, consider the two endomorphisms "multiplication by $m$" and "multiplication by $n$" on applied to $\mathbb{Z}/mn\mathbb{Z}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.