# Using Quotient Spaces to “Kill” Elements

I've a doubt about one kind of construction in linear algebra. First, I understand the definition of quotient space. Given the vector spaces $V$ and $W\subset V$ for all $x,y\in V$ we can say that $x\sim y$ if $x-y \in W$. With this we create the set of all equivalence classes denoted by $V/W$ and then we simply give it the structure of a vector space with convenient operations.

All of that is fine, however, I've seen already the use of this construction to "kill" elements. Like in the definition of tensor product, where we use this to grant linearity to the tensor product. I've heard that this construction is very common, and that it's one of the main uses of the quotient space.

How's that done ? Why does this construction kill the elements ? I've thought a little but I couldn't interpret it from the definition.

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The phrasing "killing elements" maybe has been chosen to indicated that all elements in the quotient are considered equivalent to zero. – Vahid Shirbisheh Mar 15 '13 at 17:56

Whenever you have $x,y \in V$ with $x-y \in W$, $y+W= y+(x-y)+W = x+W$. In a sense, you can ignore that y and just use x and vice versa. In a special case, if $y=0$, then $x-y \in W$ means $x\in W$ and for every $x \in W$, $x+W = 0+W =W$. So, you "kill" those x by seeing them as $0$.
Hi Ustun, I think I've got. Since, $x \in V$ is equivalent to $y \in V$ if $x-y \in W$, if we pick $x \in W$, then $x - 0 \in W$ and hence $x$ is equivalent to zero. So when we consider $V/W$ all elements of $W$ are "killed" because of that right ? Thanks for your answer! – user1620696 Mar 16 '13 at 21:41
For example, let $\rm P = X^2 + 1$. This polynomial has no roots in $\mathbb R$. But if you call $\mathbb C = \mathbb R[X]/\rm P$, then in the quotient, the polynomial $\rm P$ is the same as the element 0. So in this quotient $\rm X^2 + 1 = 0$, and your element $\rm X$ is a square root of $-1$.