$\ker (T \otimes id_{Z})=\ker(T)\otimes Z$

Question

Does $x\otimes y=0 \implies x=0$ or $y=0$? I don't think so, since its equivalent to $B(x,y)=0$ for some bi-linear form.

But my teacher said: $\ker (T \otimes id_{Z})=\ker(T)\otimes Z$ where $id_{Z}:Z\to Z$ is the identity map.

But $\ker(T \otimes id_{Z})=\{v\otimes z: Tv \otimes z=0\} \subset \ker(T) \otimes Z$, but are they really equal ? In fact $im(T) \otimes \{0\}$ is in the kernel.

Answer 1

This argument should be polished up a bit, but here goes: Changing the notation a bit, let $X, Y, Z$ be vector spaces over some field $k$, and let $T:X \to Z$. Set $T' = T\otimes \text{id}_Y : X\otimes Y \to Z$. Let $v$ be an arbitrary element of $\ker T'$, and write $v = x_1\otimes y_1 + \cdots + x_n\otimes y_n$ with $x_i\in X$, $y_i\in Y$. Assume without loss of generality that the $y_i$ are linearly independent. The vector $T'(v) = T(x_1) \otimes y_1 + \cdots + T(x_n)\otimes y_n = 0$.

Suppose some $Tx_i\not = 0$. Choose a map $\lambda:X \to k$ such that $\lambda(Tx_i)\not =0 $. Then the map $$L = \lambda\otimes \text{id}_Y : X\otimes Y \to k\otimes Y = Y$$ has $L(T'v) \not = 0$, contradicing the assumption that $v'\in \ker T$. It follows that all $T(x_i) = 0$, and thus $(\ker T)\otimes Y\supset \ker T'$. The opposite inclusion is trivial.

Answer 2

There is a more general reason as to why this is true. Every $F$-vector space is free. Thus, $Z$ is flat. In particular, tensoring with $Z$ preserves kernels.