0
$\begingroup$

I am trying to understand chapter 4 in John M. Howie's book called "Fields and Galois Theory" (published by Springer). In chapter 4 regarding geometric constructions he gives the following definition:

Let $B_0$be a set of points in the plane. There are two permitted operations on the points of $B_0$:

(1) (Ruler) through any points of $B_0$, draw a straight line;

(2) (Compass) draw a circle whose center is a point in $B_0$, and whose radius is the distance between two points in $B_0$.

Any point which is an intersection of two lines, or two circles, or a line and a circle, obtained by means of the operations (1) and (2), is said to be constructed from $B_0$ in one step. Denote the set of such points $C(B_0)$, and let $B_1 = B_0 \cup C(B_0)$. We can continue this process, definining $B_n = B_{n-1} \cup C(B_{n-1})$ ($n=1,2,3,...)$. A points is said to be constructible from $B_0$ if it belongs to $B_n$ for some $n$. A point that is constructible from $\lbrace O,I \rbrace$ is said to be constructible.

Phew, that was long. Notice, here the points $O$ and $I$ are the origon and unit, and we may choose any two points of our initial set to be these. Now, I have attempted a bisection of an angle $PQR$, which I have illustrated in the picture below.

enter image description here

I tried to convert this bisection to the definition given above. As I see it I need atleast three points in my initial set $B_0$ for this. For example, I can find a bisecting line with $B_0 = \lbrace Q, B, A \rbrace$ ($O$ and $I$ would of course be two of these points). If I, however, try with just two, I can not. For example, suppose $B_0 = \lbrace Q, A \rbrace$. I can construct the red circle which has point $B$ on its exterior. However, to construct $B$ it must be an intersection point of my red circle and some other circle or line, given the definition. But to construct a line passing through $B$, I need atleast one more point.

1) Is the point $C$ in the illustration therefore not constructible, if we follow the definition given by Howie?

Howie later proves that we can not square every circle. I will not state the details on how unless you want me to, but in short, his argument is that you can not construct a square with the radius of a given circle since you can not construct such a square from $B_0 = \lbrace O, I \rbrace$.

2) Why is this a sufficient argument? Certainly, you could square a circle given a larger $B_0$ (for the trivial case, $B_0$ could include $\pi$).

I hope this was not too long to devour. Thank you very much for your patience!

|cite|improve this question
$\endgroup$
1
$\begingroup$

Point C is obtained as the intersection of two circles, the first has center A and passes through Q, the second has center B and passes through Q. Thus the line QC is constructible.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I'm not sure that I understand this, sorry. According to the definition I have provided, for a point to be constructible it must be constructible from the initial set $B_0 = \lbrace O, I \rbrace $. I do not see how we can construct the line QC from $B_0$. To me, atleast, it seems that we require one more point in our initial set $B_0$ to do so. Am I misunderstanding something? $\endgroup$ – kasp9201 Dec 18 '18 at 2:23
  • 1
    $\begingroup$ $B_0=\{Q,B,A\}$ for this construction. $\endgroup$ – i. m. soloveichik Dec 18 '18 at 13:32
  • $\begingroup$ Thank you very much. That lets me deduce an answer to my first question. Perhaps you can also tell me why Howie would say a point is constructible if it is constructible from $B_0 = \lbrace O, I \rbrace$? ($O$ is the origon, $I$ is the unit) He gives this criteria in form of a definition and therefore never states $\textit{why}$ he does so. $\endgroup$ – kasp9201 Dec 18 '18 at 15:47
  • 1
    $\begingroup$ The constructible points are defined so that they only depend on the simplest set $B_0$. This is used later(?) to get a criterion in terms of field theory for a point to be constructible. $\endgroup$ – i. m. soloveichik Dec 19 '18 at 15:06
  • 1
    $\begingroup$ Any two points would work, $B_0$ is a convenient choice. $\endgroup$ – i. m. soloveichik Dec 19 '18 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.