In a ring $a*a=0$ then $a+a=0$ I am stuck in a question for Algebraic Structures:
I need to prove: If given any ring $R$, for any $a$ in $R$ $a\cdot a=0$ then $a+a=0$.
(of course $+$, $*$ and $0$ regarding $R$)
I didn't find any question regarding this question.
 A: I am assuming that the ring has a unit and that all elements of the ring square to 0.
In particular $1=1^2=0$ so the unit and zero must coincide. Thus for all $a\in R$ $$0=(a+1)^2=a^2+a+a+1=a+a,$$ proving that $a+a=0$ as desired.

Here is a counterexample if the ring is non-unital. Consider the matrix ring
$$
R=\left\{\begin{pmatrix}0&n\\0&0\end{pmatrix}\colon n\in\mathbb Z\right\},
$$
with usual matrix multiplication and addition as the ring operations. Then this ring has no unit, every element squares to zero, yet adding a non-zero element to itself results in a non-zero element of the ring.
A: For a more non-trivial example of a non-unital ring which satisfies the hypothesis but not the conclusion, consider the set of matrices $$\begin{pmatrix}0 & 0 & 0 & 0 \\ s & 0 & 0 & 0 \\ t & 0 & 0 & 0 \\ u & -t & s & 0\end{pmatrix},$$ with $s, t, u \in \mathbb Z$. All such matrices square to zero, but adding such a matrix to itself does not give zero. This example contrasts from the ones given in the other answers by the fact that multiplying two of these matrices does not necessarily always give zero.
Technical explanation:
Let $R$ be the non-unital associative ring generated by elements $a, b$ modulo the relation that every element squares to zero. In fact, it is sufficient to enforce $a^2=0$, $b^2=0$, and $ab+ba=0$; that last relation is equivalent to $(a+b)^2=0$ in the presence of the first two relations. The resulting ring has a $\mathbb Z$-basis given by $a, b, ab$. We embed $R$ into a unital ring $R_1$ with basis $1, a, b, ab$, and consider the left regular representation of $R_1$ with respect to that basis. This gives the above faithful representation of $R \hookrightarrow R_1$.
Even more technical overview of how one might attempt to simplify my example:
One might ask if there is a lower-dimensional faithful representation for $R$, which would give smaller matrices above. I'm pretty sure the answer is no: If we extend scalars to a field (say, we denote $R^F := F \otimes_{\mathbb Z} R$ and $R_1^F := F \otimes_{\mathbb Z} R_1$, for $F$ a field), then the socle of this representation is $1$-dimensional, spanned by $ab$. The quotient by this socle yields a representation in which $ab$ is associated to the zero map. The radical of the regular representation is $3$-dimensional, spanned by $a$, $b$, and $ab$. This sub-representation also associates $ab$ to the zero map.
So as far as I can tell, the above $4$-dimensional representation is minimal dimension with respect to the constraint that $ab$ does not act by zero. More precisely, what the above argument establishes is that the only subquotient of the left regular representation of $R_1^F$ which doesn't kill $ab$ is the full left regular representation itself. It doesn't eliminate the possibility of a more clever way of taking a subquotient of a direct sum of a few copies of the left regular representation.
Another note: After extending scalars to a field as before, a representation of $R^F$ which doesn't kill $ab$ (i.e., one which doesn't render this counter-example trivial) is precisely the same thing as a faithful representation of $R^F$, precisely because of the socle condition mentioned above - every nontrivial ideal of $R^F$ contains $ab$. We have that $R_1^F$ is a local ring (non-units form an ideal) with Jacobson radical equal to $R^F$. So a faithful representation for $R^F$ automatically becomes a faithful representation for $R_1^F$, which is a unital embedding of $R_1^F$ into a matrix ring. An embedding into a $2 \times 2$ matrix ring would be an isomorphism, but matrix rings are not local rings. So that leaves the possibility of an embedding into a $3 \times 3$ matrix ring, a possibility which I don't think I've rigorously excluded.
A: If the ring has a unit then follow pre-kidney's answer. If not, the conclusion no longer follows. Take for instance $\mathbb Z$ with regular addition and a new multiplication defined by $x*y:=0$ for $x,y\in\mathbb Z$. (It is trivial to check that the ring axioms hold.) Clearly $x*x=0$ for all $x$ but $x+x\neq0$ unless $x=0$.
