> [!NOTE] Theorem (Product With Ring Negative) > Let $(R,+,\times)$ be a [[Rings|ring]]. Then $\forall x,y \in R: x \times(-y)=(-x)\times y=-(x\times y).$ *Proof*. We have $\begin{align} (-a)b + ab &= (-a+a)\times b & \text{Distributivity of $\times$ over $+$ } \\ &= 0 \times b &\text{Definition of additive inverse} \\ &= 0 \end{align}$by [[Ring Product With Zero is Zero|ring product with zero is zero]]. Adding $-(ab)$ to both sides gives $(-a)b=-(ab).$ Similarly $a(-b)=-(ab).$ # Applications Corollary: it follows that to test for ideals we only need to check for closure under addition and product and theres no need to check for inverses directly since $-1_{R}\times x=-x$.