Let $(A,\iota)$ be an [absolutely] simple abelian variety with CM-type $(F,(\phi_i))$, and assume that $A$ is defined over a number field $k$. Let $F^*$ be the reflex field of $(F,(\phi_i))$ and put $K=kF^*$. Then \[\mathrm{rank}(A(K))=[K\colon k]\cdot\mathrm{rank}A(k).\]Why should the rank multiply by the degree $d=[K\colon k]$ in passing from $k$ to $K$? Apart from Honda's rather elaborate argument, the only explanation I could think of was that the Weil restriction $N_{K/k}(A_K)$ of $A_K$ is isogenous to $A^d$. My main theorem (Thm. 3) is that this is in fact true under a hypothesis on the growth of the endomorphism algebra $\mathrm{End}^0(A)$ from $k$ to $K$ that holds in Honda's case.
Of course, knowing that $N_{K/k}(A_K)\sim A^d$ tells you much more about the relation of $A/k$ and $A/K$ than their ranks. In the first section of the article, I worked out the relations between various arithmetic invariants of $A_K$ and $N_{K/k}(A_K)$. These have proved useful to the mathematical community (and resulted in many citations), but it was not very difficult.
At the time I wrote the article the zeta function of a CM abelian variety $A$ over a field $k$ was known (thanks to Shimura and Taniyama) when $A$ has complex multiplication over $k$. In Theorem 4 of the article, I was able to relax this condition a little. Shimura independently obtained somewhat similar results at about the same time.
Consider over $\mathbb{Q}_{2}$ the curves \[ E_{1}:y^{2}=x^{3}+1\qquad E_{2}:y^{2}=x^{3}+3. \] When twisted by the same representation corresponding to $4^{1/3}$, they become \[ E_{1}^{\prime}:y^{2}=x^{3}+16\qquad E_{2}^{\prime}:y^{2}=x^{3}+3.16. \] All curves have additive reduction, except $E_{1}^{\prime}$ has good reduction. In particular the $L$-series of $E_{1}$ and $E_{2}$ coincide, but cease to do so after being twisted by the same representation.