5
$\begingroup$

Suppose $E$ is a vector space over a field of characteristic $0$. Let $E_1, F_1$ be subspaces of finite codimension and let $E_2, F_2$ be their respective complements, i.e., $E = E_1 \oplus E_2 = F_1 \oplus F_2$.$\DeclareMathOperator{\codim}{codim}$

I know that $\dim E_2 = \codim E_1$ and $\dim F_2 = \codim F_1$ because $E_2 \cong E/E_1$ and $F_2 \cong E/F_1$.

But I don't know how to prove that $\codim (E_1 \cap F_1) \le \dim E_2 + \dim F_2$. I saw a proof that $\codim (E_1 \cap F_1)$ is finite but it used some fancy isomorphism theorem, so I think a more bare hands approach would be helpful. Thanks.

$\endgroup$
1
  • $\begingroup$ Consider the short exact sequence $0 \to E_1 / \left(E_1 \cap F_1\right) \to V / \left(E_1 \cap F_1\right) \to V / E_1 \to 0$. The dimensions of its three middle terms are $\leq \operatorname{codim} F_1$, $\operatorname{codim} \left(E_1 \cap F_1\right)$ and $\operatorname{codim} E_1$, respectively. $\endgroup$ Jul 23, 2017 at 19:50

2 Answers 2

8
$\begingroup$

Maybe there is some way for avoiding the homomorphism theorems, but they're so handy and powerful that it is better trying to understand them.

Consider the linear map $$ f\colon E\to E/E_1\oplus E/F_1,\qquad f(v)=(v+E_1,v+F_1) $$ The kernel of this map is $E_1\cap F_1$, so $f$ induces an injective linear map $$ \tilde{f}\colon E/(E_1\cap F_1)\to E/E_1\oplus E/F_1 $$ In particular, the domain is finite dimensional and $$ \dim E/(E_1\cap F_1)\le\dim(E/E_1\oplus E/F_1)= \dim(E/E_1)+\dim(E/F_1) $$ which is the same as saying that $$\DeclareMathOperator{\codim}{codim} \codim(E_1\cap F_1)\le\codim E_1+\codim F_1 $$

$\endgroup$
2
  • $\begingroup$ In the codomain of $f$, don't you actually mean the cartesian product of the quotient spaces, rather than their linear sum? $\endgroup$
    – Bar Alon
    Nov 22, 2018 at 9:08
  • 1
    $\begingroup$ @BarAlon The coproduct, if you prefer. Since they're not subspaces of some vector space, the $\oplus$ symbol denotes their “external direct sum”. $\endgroup$
    – egreg
    Nov 22, 2018 at 9:31
4
$\begingroup$

Hint:

Consider the diagram \begin{matrix} 0&\!\!\!\longrightarrow \!\!\!&E_1\cap F_1&\!\!\!\longrightarrow \!\!\!&E_1\bigoplus F_1&\!\!\!\longrightarrow \!\!\!&E_1+F_1&\!\!\!\longrightarrow &\!\!\!0\\ & & && \phantom{i\oplus j}\downarrow i\oplus j&& \phantom{k}\downarrow k\\ 0&\!\!\!\longrightarrow \!\!\!&\Delta E\!\!\!&\!\!\!\longrightarrow \!\!\!&E\bigoplus E&\!\!\!\longrightarrow \!\!\!&E&\!\!\!\longrightarrow &\!\!\!0 \end{matrix} where $\Delta E$ is the diagonal of $E\bigoplus E$, $i,j,k$ are the canonical injections, and the rightmost horizontal maps (from the direct sums) are $\; (x, y)\longmapsto x-y $.

You deduce first from this diagram there exists a map $f\colon E_1\cap F_1\longrightarrow \Delta E$ which makes the whole diagram commutative.

Apply the Snake's lemma to show $\operatorname{coker} f$ is (isomorphic to) a submodule of $E/E_1\bigoplus E/F_1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .