# proof of theorem about cyclic subspaces

We first prove the case $r=2$. The $\subseteq $ inclusion is clear, since the right side is a $T$-invariant subspace that contains ${v}_{1}+{v}_{2}$.

For the other inclusion, it is sufficient to show that ${v}_{1},{v}_{2}\in Z({v}_{1}+{v}_{2},T)$. The idea is that the action of $T$ on ${v}_{1}+{v}_{2}$ can ”isolate” the two summands if their annihilator polynomials are coprime^{}. Let’s write ${m}_{i}$ for ${m}_{{v}_{i}}$.

Since $({m}_{1},{m}_{2})=1$, there exist polynomials^{} $p$ and $q$ such that

$$p{m}_{1}+q{m}_{2}=1$$ | (1) |

this is BÃÂ©zout’s lemma (or the Euclidean algorithm^{}, or the fact that $k[X]$ is a principal ideal domain^{}).

Now $p{m}_{1}(T)$ is the projection^{} from $Z({v}_{1},T)\oplus Z({v}_{2},T)$ to $Z({v}_{2},T)$:

$$(p{m}_{1})(T){v}_{1}=p(T){m}_{1}(T){v}_{1}=p(T)0=0$$ | (2) |

(by assumption that ${m}_{1}$ is the annihilator polynomial of ${v}_{1}$) and

$$(p{m}_{1})(T)=1-(q{m}_{2})(T)$$ | (3) |

(by choice of $p$ and $q$), so

$$(p{m}_{1})(T){v}_{2}={v}_{2}-q(T){m}_{2}(T){v}_{2}={v}_{2}-q(T)0={v}_{2}$$ | (4) |

Any subspace^{} that is invariant under $T$ is also invariant under polynomials of $T$. Therefore, the preceding equations show that ${v}_{2}=(p{m}_{1})(T)({v}_{1}+{v}_{2})\in Z({v}_{1}+{v}_{2},T)$. By symmetry, we also get that ${v}_{1}\in Z({v}_{1}+{v}_{2},T)$.

For the last claim, we note that the annihilator polynomial $m$ of $Z({v}_{1},T)\oplus Z({v}_{2},T)$ is the least common multiple^{} of ${m}_{1}$ and ${m}_{2}$ (that $m$ is a multiple of ${m}_{1}$ follows from the fact that $m$ must annihilate ${v}_{1}$, and the set of polynomials that annihilate ${v}_{1}$ is the ideal generated by ${m}_{1}$). Since ${m}_{1}$ and ${m}_{2}$ are coprime, the lcm is just their product.

That concludes the proof for $r=2$. If $r$ is arbitrary, we can simply apply the $r=2$ case inductively. We only have to check that the coprimality condition is preserved under applying the $r=2$ case to $i=1,2$. But it is well-known that if $p,q,r$ (in $k[X]$ or in any principal ideal domain) are pairwise coprime, then $pq$ and $r$ are also coprime.

Title | proof of theorem about cyclic subspaces |
---|---|

Canonical name | ProofOfTheoremAboutCyclicSubspaces |

Date of creation | 2013-03-22 17:32:53 |

Last modified on | 2013-03-22 17:32:53 |

Owner | FunctorSalad (18100) |

Last modified by | FunctorSalad (18100) |

Numerical id | 7 |

Author | FunctorSalad (18100) |

Entry type | Proof |

Classification | msc 15A04 |