I am trying to solve the question 27 of Section 10.4 of Dummit and Foote but I am stuck in the first problem: let me state the question and then I will attach the picture of the page of the corresponding book as well:
I am stuck at part c and d
Prove that the map $f: \Bbb C \times \Bbb C \to \Bbb C \times \Bbb C$ by $f(z_1,z_2)=(z_1z_2,z_1\bar{z_2})$ is an $\Bbb R $ bilinear map.
Now I am calculating $f(z_1+z_2, z_3)=((z_1+z_2)z_3,(z_1+z_2)\bar{z_3})=(z_1z_3+z_2z_3,z_1\bar{z_3}+z_2\bar{z_3})$ how is it $f(z_1, z_3)+f(z_2, z_3)$?
Moreover, $f(az_1,z_2)=(az_1z_2,az_1\bar{z_2})$ why is it $af(z_1,z_2)$?
Now part d) is Let $F$ be the $\Bbb R$ module homomorphism from $\Bbb C \otimes \Bbb C$ to $\Bbb C \times \Bbb C$ obtained from $f$. Show that $F$ is $\Bbb C$ linear and deduce $F$ to be surjective.
I have omitted the questions that I have proved. I need help basically on the part that I have mentioned in the question. I am attaching the full question as this will help you.
As a rule of thumb, you should almost never try to prove that a map is (bi-)linear by applying the definition of linearity; every minute spent doing that is a waste of time. There should be a small collection of maps you know are linear (like maps defined by multiplication by a fixed matrix, although if you like you could start with less than that), and you should use that linearity is preserved under linear combination, composition, and combining components into a Cartesian product. Of course you can prove what you want from the the definition of linearity, just expand the definition of $f$ on both ends of what you want to prove, and apply some basic algebra. But I am not going to help with that.
In this question $\def\R{\Bbb R}K=\R$, so $\def\C{\Bbb C}\C$ should be viewed as a real space with basis (for instance) $\def\ii{\mathbf i}\def\B{\mathcal B}\B=[1,\ii]$. Bi-linearity of$~f$ amounts to linearity with respect to the first argument $z_1$ and with respect to the second argument $z_2$; in both cases linearity amounts to linearity of each of the components (defined by $z_1z_2$ and $z_1\overline{z_2}$). So here are four cases to consider, but all are easy. Any map $\C\to\C$ given by multiplication by a fixed complex number $a+b\ii$ is $\R$-linear, as it is given with respect to the basis $\B$ by the matrix $a~-b\choose b~\phantom-a$, and this takes care of three of the four cases. The remaining one, linearity of $z_2\mapsto z_1\overline{z_2}$ involves composition of such a multiplication map with complex conjugation, but the latter is also $\R$-linear since it is given with respect to the basis $\B$ by the matrix $1~\phantom-0\choose0~-1$.
So the linear map $\Phi:\C\otimes_\R\C\to\C\times\C$ of question (d) maps $z_1\otimes z_2\mapsto(z_1z_2,z_1\overline{z_2})$ for pure tensors, and so by linearity sends $\epsilon_1=\frac12(1\otimes1+\ii\otimes\ii)\mapsto\frac12(1-1,1+1)=(0,1)$ and similarly $\epsilon_2=\frac12(1\otimes1-\ii\otimes\ii)\mapsto\frac12(1+1,1-1)=(1,0)$. As for $\C$-linearity, consider the effect of multiplying by $y\in\C$ on pure tensors: it maps $z_1\otimes z_2$ to $(yz_1)\otimes z_2$; then applying $\phi$ gives $(yz_1z_2,yz_1\overline{z_2})$ which indeed is $y(z_1z_2,z_1\overline{z_2})=y\Phi(z_1\otimes z_2)$ (since we already knew $\R$-linearity ,it would have sufficed to do this check for $y=\ii$ only). So $\Phi$ is now $\C$-linear and contains the basis vectors $(1,0)$ and $(0,1)$ of the $\C$-vector space $\C\times\C$, which makes it surjective. Since as an $\R$ vector space, $A$ has dimension $2\times2=4$ it has dimension $4/2=2$ as $\C$-vector space, so $\Phi$ is a $\C$-vector space isomorphism.