TITLE: Show that if $(M_t)_{t \in [0,T]}$ is a non-negative and strict local martingale on [0,T] we have $\mathop{\mathbb{E}}[M_T] < M_0$.
QUESTION [2 upvotes]: I was stuck in proving this statement.
There is a hint: By contradiction, using the tower property and the fact that if a random variable $X \leq 0$ a.s. and $\mathop{\mathbb{E}}[X] = 0$ then $X = 0$ a.s.
As the hint suggests, if we do it by contradiction, then we assuming that $(M_t)_{t \in [0,T]}$ is a non-negative and strict local martingale on $[0,T]$ and that $\mathop{\mathbb{E}}[M_T] \geq M_0$ and we should derive a contradiction from here. But, I don't know how to proceed from here and how the tower property can be applied.
I was wondering if the definition of strict local martingale could be useful? But I just cant put all these things together.
Glad if anyone can help!

REPLY [2 votes]: Lemma If $M_t$ is a non-negative local martingale, then it is a supermartingale.
Proof. This is a simple application of Fatou's lemma. See here. $\blacksquare$
Now, arguing by contradiction, suppose $\mathbb{E}(M_T) = M_0$. Since $M_t$ is a supermartingale, this implies that $\mathbb{E}(M_t) = M_0$ for all $t\in[0,T]$. But then we have that for any $0 \leq s \leq t \leq T$, $M_s - \mathbb{E}(M_t \, | \, \mathcal{F}_s) \geq 0 $ and $\mathbb{E}(M_s - \mathbb{E}(M_t \, | \, \mathcal{F}_s)) = 0$, implying that $M_s - \mathbb{E}(M_t \, | \, \mathcal{F}_s) = 0$. But then, $M_t$ is a true martingale, so it cannot be a strict local martingale.


TITLE: Does codimension equal height in complete local domains?
QUESTION [4 upvotes]: For an ideal $I$ in a commutative ring $R$, define $\operatorname{codim}I=\dim R-\dim R/I$. Does codimension equals height for all ideals in the formal power series ring? Does this hold for complete local domains in general?

REPLY [9 votes]: The standard definition of codimension of $R/I$ is the height of $I$.
Anyway, a first observation is $\dim R - \dim R/I=\operatorname{ht} I$ for all $I$ is equivalent to require the equality for all prime ideals. So let $P$ be a prime ideal in $R$ and you want
$$ \dim R = \dim R/P + \operatorname{ht}P.$$
This is true for any local Cohen-Macaulay ring. If you take a formal power series rings with coefficients in a field and finitely many variables, then it is regular, hence Cohen-Macaulay. Another class of rings for which the egality holds are Noetherian local catenary integral domains (this includes the Noetherian complete local domains).
If $R$ is not a domain, there are easy counterexamples (e.g. if the spectrum of $R$ has two irreducible components and if they have distinct dimensions).