(a) in math block gets treated as list

[madsmh]

I have the following section in an markdown file

(a) Passing to the quotient ring $R/\ideal{q}$, we can assume without loss of 
generality that $\ideal{q}=0$. Let $a,b\in R/\ideal{q}$ such that $a \neq 0$ and $ab=0$
 and consider the ascending chain of inclusions
$$
\ann(b) \subseteq \ann(b^{2})\subseteq \cdots \subseteq\ann(b^{k})\subseteq\cdots \subseteq R
$$
since $R$ it Noetherian there exists an integer $n\geq 1$ such that $\ann(b^{n})=\ann(b^{i})$ for all $i\geq n$. Let $x \in (a) \cap 
(b^{n})$, then there exists $z,y\in R$ such that $x=ya=zb^{n}$, but then 
$$
0=yab=zb^{n+1}
$$
so then $zb^{n+1}=0$ so $z \in \ann(b^{n+1})=\ann(b^{n})$  and hence we must have 
$zb^{n}=0$, which means that $x=0$, so we have that 
$$
   (a) \cap (b^{n})=(0)=\ideal{q}
$$
Assume now that $\ideal{q}$ is irreducible. We assumed that $a\neq 0$, then by the intersection above that $\ann(b^{n})=(0)$ and that 
$b$ is nilpotent and hence $\ideal{q}$ is primary.
(b) Let $X$ be the set of ideals of $R$ that *can not* be written as a finite intersection of irreducible ideals. 
If $X\neq \emptyset$, then we can partially order the elements of $X$ by inclusion. since we   have that $R$ is Noetherian, then every 
ascending chain has an upper bound and then by Zorn's  lemma, we have that $X$ contains a maximal element, denote this element by 
$\ideal{a}$. Note that $\ideal{a}$ is reducible, because if it wasn't, it could be expressed as a finite intersection of one irreducible ideal. 
Hence we have that $\ideal{a} = \ideal{a}_{1} \cap \ideal{a}_{2}$ where $\ideal{a}_{1},\ideal{a}_{2}\notin X$, but then, we have that $
\ideal{a}_{1},\ideal{a}_{2}$  can be  written as an intersection of a finite intersection of irreducible ideals, and hence, so can $\ideal{a}$, 
hence $\ideal{a}\notin X$, which contradicts our assumption that $X\neq \emptyset$. This shows our statement. 

The part

$$
   (a) \cap (b^{n})=(0)=\ideal{q}
$$

is treated as a list and is converted to (note the middle extra \item and $ $):

\begin{enumerate}
    \def\labelenumi{(\alph{enumi})}
    \tightlist
    \item
      Passing to the quotient ring \(R/\ideal{q}\), we can assume
      without loss of generality that \(\ideal{q}=0\). Let
      \(a,b\in R/\ideal{q}\) such that \(a \neq 0\) and \(ab=0\) and
      consider the ascending chain of inclusions \[
      \ann(b) \subseteq \ann(b^{2})\subseteq \cdots \subseteq\ann(b^{k})\subseteq\cdots \subseteq R
      \] since \(R\) it Noetherian there exists an integer \(n\geq 1\)
      such that \(\ann(b^{n})=\ann(b^{i})\) for all \(i\geq n\). Let
      \(x \in (a) \cap (b^{n})\), then there exists \(z,y\in R\) such
      that \(x=ya=zb^{n}\), but then \[
      0=yab=zb^{n+1}
      \] so then \(zb^{n+1}=0\) so \(z \in \ann(b^{n+1})=\ann(b^{n})\)
      and hence we must have \(zb^{n}=0\), which means that \(x=0\), so
      we have that \$\$
    \item
      \cap (b\^{}\{n\})=(0)=\ideal{q} \$\$ Assume now that \(\ideal{q}\)
      is irreducible. We assumed that \(a\neq 0\), then by the
      intersection above that \(\ann(b^{n})=(0)\) and that \(b\) is
      nilpotent and hence \(\ideal{q}\) is primary.
    \item
      Let \(X\) be the set of ideals of \(R\) that \emph{can not} be
      written as a finite intersection of irreducible ideals. If
      \(X\neq \emptyset\), then we can partially order the elements of
      \(X\) by inclusion. since we have that \(R\) is Noetherian, then
      every ascending chain has an upper bound and then by Zorn's lemma,
      we have that \(X\) contains a maximal element, denote this element
      by \(\ideal{a}\). Note that \(\ideal{a}\) is reducible, because if
      it wasn't, it could be expressed as a finite intersection of one
      irreducible ideal. Hence we have that
      \(\ideal{a} = \ideal{a}_{1} \cap \ideal{a}_{2}\) where
      \(\ideal{a}_{1},\ideal{a}_{2}\notin X\), but then, we have that
      \(\ideal{a}_{1},\ideal{a}_{2}\) can be written as an intersection
      of a finite intersection of irreducible ideals, and hence, so can
      \(\ideal{a}\), hence \(\ideal{a}\notin X\), which contradicts our
      assumption that \(X\neq \emptyset\). This shows our statement.
    \end{enumerate}

is there any way to avoid this?

[madsmh]

Adding spaces before the (a) solves the problem.

[jgm]

I assume you're using commonmark or gfm. In these formats, markers of block structure take precedence over markers of inline structure, and math (including display math) is considered inline.
You can work around it in various ways, including the one you found.

[jgm]

The question is how pandoc is being invoked (maybe you don't know, if it's done by obsidian).
With -f commonmark or -f gfm, you will get this behavior. With -f markdown, you probably won't.

[madsmh]

It is invoked by -f markdown+attributes+fenced_divs I am writing a cleanup and conversion script from obsidian -> latex outside of obsidian.

[jgm]

Do you mean header_attributes?

[jgm]

In any case, I guess that markdown behaves the same way as commonmark here -- indicators of block-level structure take priority over indicators of inline-level structure. Perhaps Obsidian's markdown parser behaves differently.

[madsmh]

Obsidian does not really take lists into account...