Why is zero mod zero undefined? (Edit: This is not a duplicate of the question about n % 0. I'm specifically asking about 0 % 0.)
Why is zero mod zero undefined?
To me, the answer must be zero, because $0 \times N + M = 0$ has only one solution for $M$, zero.
(Assuming $M$ and $N$ are integers.)
However, today I found out that it is undefined. Why?
 A: In general, when taking the remainder of $a$ divided by $b$, we use the division algorithm to find nonnegative integers $q,r$ such that $$a=bq+r$$ but also that $r<b$. If we have $b=0$, then we can't find a remainder when dividing by $b$ in the conventional sense.
A: To say $a=b \mod N$ is exactly, literally, to say that $a-b\in N\cdot \mathbb Z$ (for integers...). It is true that this would normally only be asked in situations where (as the other answers suggest) something is actually happening. Here. $0\cdot \mathbb Z=\{0\}$, so for integers $a,b$ the condition $a=b\mod 0$ literally is that $a-b\in\{0\}$, which is $a-b=0$.
As suggested by the other answers, there's little point to talking this way for a "modulus" $0$, and, more confusingly, the things that usually matter, that usually help understanding or provide alternatives, fail in various regards.
Ok, those failures are irrelevant in real situations, happily. And, if we cautiously revert to the minimal formal characterization in terms of equality in the quotient ring $\mathbb Z/(N\cdot \mathbb Z)$, we find that it's just asking about equality in $\mathbb Z/\{0\}\cong \mathbb Z$, which is the usual equality in $\mathbb Z$.
That is, the corollaries mess up for modulus $0$, but the "fancier/higher-level" notion is perfectly fine... if somewhat boring/pointless.
A: As you noted it only makes any sort of sense when applied to zero. By your definition the statement $$ a \equiv  b\mod 0$$
Only makes sense when $a=b=0$. For any other numbers you would need a remainder after division by $0$ which you can't do. So it seems that there is no natural way of extending the concept of a congruence  to include modding by $0$. 
Incidentally it is an $ideal$ since $I = \lbrace 0 \rbrace$ is a trivial subgroup of $\mathbb{Z}$ with respect to addition and $i \in I \quad n \in \mathbb{Z} \Rightarrow in \in I$. Ideals are related to a generalization of modular arithmetic so it may be possible to come up with a consistent notion of modding by $0$.
