Fields - Proof that every multiple of zero equals zero This is one of my first proofs about fields.
Please feed back and criticise in every way (including style and details).
Let $(F, +, \cdot)$ be a field.
Non-trivially, $\textit{associativity}$ implies that any parentheses are meaningless.
Therefore, we will not use parentheses.
Therefore, we will not use $\textit{associativity}$ explicitly.
By $\textit{identity element}$, $F \ne \emptyset$.
Now, let $a \in F$.
It remains to prove that $0a = 0$.
\begin{equation*}
\begin{split}
0a &=  0a +  0      && \quad \text{by }\textit{identity element }(+    ) \\
   &=  0a +  a + -a && \quad \text{by }\textit{inverse element  }(+    ) \\
   &=  0a + 1a + -a && \quad \text{by }\textit{identity element }(\cdot) \\
   &= (0 + 1)a + -a && \quad \text{by }\textit{distributivity   }        \\
   &= (1 + 0)a + -a && \quad \text{by }\textit{commutativity    }(+    ) \\
   &=      1 a + -a && \quad \text{by }\textit{identity element }(+    ) \\
   &=        a + -a && \quad \text{by }\textit{identity element }(\cdot) \\
   &=  0            && \quad \text{by }\textit{inverse element  }(+    )
\end{split}
\end{equation*}
QED
PS: Is "Let $(F, +, \cdot)$ be a field." ok?
Besides, I would not want to call $F$ a field, because $F$ is just a set.
Also, what do you think about using adverbs like "Now"? How would you have said the associativity-thing?
 A: Your proof is fine. As the comments have pointed out, the proof can be shortened. If you still want to do it in a single chain of equalities, then you can do something like this:
\begin{equation*}
\begin{split}
0a &=  0a + 0      && \quad \text{by }\textit{identity element }(+    ) \\
   &=  0a + 0a + (-0a) && \quad \text{by }\textit{inverse element  }(+    ) \\
   &=  (0 + 0)a + (-0a) && \quad \text{by }\textit{distributivity   }        \\
   &= 0a + (-0a) && \quad \text{by }\textit{identity element }(+    ) \\
   &=  0            && \quad \text{by }\textit{inverse element  }(+    )
\end{split}
\end{equation*}
A: A few pointers:


*

*You don't have to use "Now". You could just say "Let $a\in F$."

*Don't say "meaningless". Rather, phrase it like so:

Non-trivially, associativity implies that any parentheses are redundant. Hence, parenthesis will be suppressed and we will thus not explicitly employ associativity.


*It's not wrong to say that $(F,+,\cdot)$ is a field, unless the question goes something like this: "Let $(F,+,\cdot)$ be a field, and let $0\in F$. Then show that every multiple of zero equals zero, i.e., for any $a\in F$, $0a=0$". Considering the way you phrased your proof, I don't think that this is how the question was phrased (correct me if I'm wrong).
Now, your proof is not wrong, but it's not the shortest either. Your proof could go something like this: $0a=(0+0)a=0a+0a$; hence $0a=0$. Alternatively, you could go like so: $0a=(0+0)a=0a+0a$. But $0a=0a+0$. Hence, $0a+0a=0a+0$, implying that $0a=0$. If I was a teacher, I would personally prefer the latter. However, do not take this personally; your proof is also very nice, but a tad bit longer than what I think is the conventional proof of this fact. Mathematicians are lazy; they prefer the shortest proofs (or at least that's what I think. I'm not a professional!)!
