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One of the first things discussed in books on elliptic curves is the change of variables that "puts the curve" in Weierstrass form. To give a concrete example, the curve $6y^2=2x^3+3x^2+x$ (the "cannonball problem" discussed at the beginning of L. Washington's book) can be put in Weierstrass form $v^2=u^3-\frac{9}{4}u$ by the change of variable $(u,v)=(3x+\frac{3}{2},9y)$. In fact, the explicit formulas to define the group law is usually only given for the Weierstrass form, with the understanding that one can always reduce the more general form to it (assuming the characteristic of the field is not 2 or 3). My question is about the precise meaning of "puts the curve" above. The two curves are surely not "isomorphic", as defined for example in http://www.lmfdb.org/knowledge/show/ec.isomorphism

I understand that of course studying and deriving properties of the simpler curve $v^2=u^3-\frac{9}{4}u$ one can easily go back to the original curve $6y^2=2x^3+3x^2+x$ using the simple transformation, but it seems to me that a discussion about the relationship between the two curves would be helpful. But I did not find it in any of the standard references on elliptic curves I know. To put my question in simple minded terms, what happens to the elliptic curve when we make a change of variables? Because one thinks of two curves as being "the same" if they are isomorphic, but apparently the first thing one does when studying an elliptic curve is to change it to another one that is not "the same".

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    $\begingroup$ Why are they "surely not" isomorphic as defined in your reference? On a more basic level; note that the change of variables yields a bijection between the rational points of the two curves. $\endgroup$ – Servaes Jun 13 at 15:46
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    $\begingroup$ The group operation on an elliptic curve also has a geometric interpretation; it is a matter of drawing and intersecting some lines. This is preserved by affine transformations, and in particular by your linear change of variables. $\endgroup$ – Servaes Jun 13 at 22:48
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    $\begingroup$ And I haven't taken up Silverman's book, but it seems that the rational functions of the form $f(x,y)=(u^2x,u^3y)$ are precisely those that preserve Weierstrass form. Your original curve is not in Weierstrass form. $\endgroup$ – Servaes Jun 13 at 23:14
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    $\begingroup$ I don’t see isogenies coming up here. Perhaps if you want an example of an isogeny that isn’t an isomorphism, you could ask a new question. Suffice it to say here that a nonisomorphic isogeny will have a nontrivial kernel, so that $(x,y)\mapsto$ (something with nonconstant functions in denominator). $\endgroup$ – Lubin Jun 14 at 19:40
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    $\begingroup$ Yes, an isogeny with trivial kernel is an isomorphism. $\endgroup$ – Lubin Jun 14 at 19:46

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