$f(x)$ is irreducible polynomial on field $k$, $f(\alpha)=0$. $k'$ is field extension of $k$, then $k(\alpha)\otimes_kk'\cong k'[x]/(f(x))$

Question

$k$ is a field, $f(x)$ is irreducible polynomial on $k$, $\alpha$ is a root of $f(x)$.

If $k'$ is field extension of $k$, then $k(\alpha)\otimes_kk'\cong k'[x]/(f(x))$.

My idea:

Since $f(\alpha)=0$, $f(x)$ is irreducible on $k$, $k(\alpha)\cong k[x]/(f(x))$.

It's sufficient to prove $k[x]/(f(x))_k \otimes_kk'\cong k'[x]/(f(x))_k'$

$(f(x))_k$ means ideal in $k[x]$ generated by $f(x)$, same meaning for $(f(x))_k'$.

Since left side occurs $\otimes$, I think Base Change for tensor product might help.

Then how to proceed? Any help would be appreciated. Thanks in advance :)

Answer 1

Hint:

Consider the short exact sequence $$ 0\longrightarrow k[x]\xrightarrow{\times f(x)} k[x]\longrightarrow k[x]/(f(x))\longrightarrow 0, $$ and tensor by $k'$ (which is flat over $k$).

Answer 2

Hint $1$:

$1$. From property of field extension, $k(\alpha)\cong k[x]/(f(x))$.

$2$. $k[x]\otimes_kk'\cong k'[x]$.

$3$. From Base Change Formula, $\cdot \otimes_k k' \cong \cdot \otimes_{k[x]}(k[x]\otimes k')$.

Hint $2$:

$\varphi:k[x]/(f(x))\times k' \to k'[x]/(f(x)), (\bar x,a)\mapsto a\bar x$ induces $\bar \varphi:k[x]/(f(x))\otimes_k k' \to k'[x]/(f(x))$.

"Inverse" of $\bar \varphi$ is given by $\Sigma a_i \bar x^i \mapsto \Sigma \bar x^i \otimes_k a_i.$ Do some check.