How to show $x \otimes 1 = 0 \in R \otimes_{k_0} k$ implies $x=0$?

Question

Let $k_0$ be a field, $k$ a field extension of $k_0$ and $R$ a $k_0$ algebra. I would like to deduce that $x \otimes 1 = 0 \in R \otimes_{k_0} k$ implies $x=0$. I asked a similar question in the past How to show $1 \otimes x = 0 \in A \otimes B$ implies $x=0$? but the situation seems different.. Any explanation is appreciated. Thank you.

Answer 1

Both $R$ and $k$ are vector spaces over $k_0$. That's all we need about $R$ and $k$ thanks to the following claim: if $V$ and $W$ are vector spaces over a field $F$ then for elements $v_0 \not= 0$ in $V$ and $w_0 \not= 0$ in $W$ we have $v_0 \otimes w_0 \not= 0$ in $V \otimes_F W$.

Proof: By Zorn's lemma, $\{v_0\}$ extends to a basis of $V$ and $\{w_0\}$ extends to a basis of $W$ (as $F$-vector spaces). The coordinate function of $v_0$ in the first basis gives us an element $\varphi \in V^*$ such that $\varphi(v_0) = 1$. Similarly, there is an element $\psi \in W^*$ such that $\psi(w_0) = 1$. The $F$-bilinear map $V \times W \rightarrow F$ where $(v,w) \mapsto \varphi(v)\psi(w)$ induces an $F$-linear map $V \otimes_F W \rightarrow F$ where $v \otimes w \mapsto \varphi(v)\psi(w)$ on elementary tensors. In particular, $v_0 \otimes w_0 \mapsto \varphi(v_0)\psi(w_0) = (1)(1) = 1$, so $v_0 \otimes w_0 \not= 0$ (otherwise a linear map has to send $v_0 \otimes w_0$ to the value 0).

Answer 2

This follows since the base change $k_0\to k$ is faithfully flat ($k$ is free over $k_0$).

One elementary way of obtaining your result is then as follows: regarding $R$ just as a $k_0$-vector space, so as $k_0^{(I)}$ for some set $I$, we have $k_0^{(I)}\otimes_{k_0}k\cong k^{(I)}$, so every $k_0$-basis for $R$ induces a $k$-basis for $R\otimes_{k_0}k$. So if $x\neq0$, then $x\otimes1\neq0$.

Alternatively, let $K$ be the kernel of multiplication by $x$. Tensoring with $k$ is flat, so $K\otimes_{k_0}k$ is the kernel of multiplication by $x\otimes1$, so the inclusion $K\subset R$ becomes equality after tensoring. Since $k/k_0$ is faithfully flat, however, we can now conclude that $K\subset R$ is also equality, so $x=0$.