MATH 128A: Homework 6 Due date: Thursday, May 14

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Solution:

(a) Label the \(9\) points as follows:

\[ \begin{array}{ccc} 0 & 1 & 2\\ 3 & 4 & 5\\ 6 & 7 & 8 \end{array} \]

One affine plane of order \(3\) is obtained by taking the following \(12\) lines:

\[ \begin{array}{lll} \ell _{012}=\{0,1,2\}, & \ell _{345}=\{3,4,5\}, & \ell _{678}=\{6,7,8\},\\[2mm] \ell _{036}=\{0,3,6\}, & \ell _{147}=\{1,4,7\}, & \ell _{258}=\{2,5,8\},\\[2mm] \ell _{048}=\{0,4,8\}, & \ell _{156}=\{1,5,6\}, & \ell _{237}=\{2,3,7\},\\[2mm] \ell _{246}=\{2,4,6\}, & \ell _{138}=\{1,3,8\}, & \ell _{057}=\{0,5,7\}. \end{array} \]

Each line contains exactly \(3\) points.

(b) The parallel classes are

\[ \{\ell _{012},\ell _{345},\ell _{678}\}, \]

\[ \{\ell _{036},\ell _{147},\ell _{258}\}, \]

\[ \{\ell _{048},\ell _{156},\ell _{237}\}, \]

and

\[ \{\ell _{246},\ell _{138},\ell _{057}\}. \]

Thus there are \(4\) parallel classes, each containing \(3\) lines.

(c) To construct the projective completion, we add one point at infinity for each parallel class. We name these points by choosing one line from each class:

\[ [\ell _{012}],\qquad [\ell _{036}],\qquad [\ell _{048}],\qquad [\ell _{246}]. \]

Here

\[ [\ell _{012}]=[\ell _{345}]=[\ell _{678}], \]

\[ [\ell _{036}]=[\ell _{147}]=[\ell _{258}], \]

\[ [\ell _{048}]=[\ell _{156}]=[\ell _{237}], \]

and

\[ [\ell _{246}]=[\ell _{138}]=[\ell _{057}]. \]

Then we add the line at infinity

\[ \ell _\infty = \{[\ell _{012}],[\ell _{036}],[\ell _{048}],[\ell _{246}]\}. \]

In the picture below, the colors match the parallel classes above: black for \(\{\ell _{012},\ell _{345},\ell _{678}\}\), magenta for \(\{\ell _{036},\ell _{147},\ell _{258}\}\), blue for \(\{\ell _{048},\ell _{156},\ell _{237}\}\), and green for \(\{\ell _{246},\ell _{138},\ell _{057}\}\).

Therefore the projective completion has \(9+4=13\) points and \(12+1=13\) lines.

Solution:

Let \(r_1,\dots ,r_k\) be the distinct lines incident with \(p\).

(a) Let \(q\in \mathcal P\setminus \{p\}\). By axiom I1, there is a unique line through \(p\) and \(q\). Since \(r_1,\dots ,r_k\) are all the lines incident with \(p\), this line must be one of the \(r_i\). Hence \(q\) lies on one of the lines \(r_1,\dots ,r_k\).

(b) Suppose \(i\neq j\), and suppose for contradiction that

\[ q\in \bigl (R(r_i)\setminus \{p\}\bigr )\cap \bigl (R(r_j)\setminus \{p\}\bigr ). \]

Then \(q\neq p\), and both \(p\) and \(q\) are incident with \(r_i\) and with \(r_j\). By axiom I1, there is a unique line through the two distinct points \(p\) and \(q\). Therefore \(r_i=r_j\), contradicting \(i\neq j\). Hence

\[ \bigl (R(r_i)\setminus \{p\}\bigr )\cap \bigl (R(r_j)\setminus \{p\}\bigr )=\varnothing . \]

(c) By part (a), every point of \(\mathcal P\setminus \{p\}\) lies in one of the sets

\[ R(r_1)\setminus \{p\},\dots ,R(r_k)\setminus \{p\}. \]

By part (b), these sets are pairwise disjoint. Therefore they form a partition of \(\mathcal P\setminus \{p\}\).

(d) Since the affine plane has order \(n\), every line has exactly \(n\) points. Thus each \(R(r_i)\) has \(n\) points. Since \(p\in R(r_i)\), removing \(p\) leaves exactly \(n-1\) points:

\[ \left |R(r_i)\setminus \{p\}\right |=n-1. \]

(e) Since \(|\mathcal P|=n^2\), we have

\[ |\mathcal P\setminus \{p\}|=n^2-1. \]

By part (c), \(\mathcal P\setminus \{p\}\) is partitioned into \(k\) disjoint sets, each of size \(n-1\). Therefore

\[ n^2-1=k(n-1). \]

(f) Since

\[ n^2-1=(n-1)(n+1), \]

we get

\[ k(n-1)=(n-1)(n+1). \]

Since \(n\geq 2\), we have \(n-1\neq 0\), so we can divide by \(n-1\). Hence

\[ k=n+1. \]

Therefore exactly \(n+1\) lines are incident with \(p\).

Solution:

Recall that in the projective completion \(\alpha ^\ast \), each old affine line \(\ell \in \mathcal L\) is extended by adding the point at infinity \([\ell ]\) corresponding to its parallel class. The new line \(\ell _\infty \) is incident exactly with the points at infinity.

(a) Let \(\ell \in \mathcal L\) be an old affine line. By axiom I2 in the affine plane, \(\ell \) is incident with at least two distinct ordinary points. In the projective completion, \(\ell \) is also incident with the point at infinity \([\ell ]\). This point is not an ordinary point, so \(\ell \) is incident with at least three distinct points of \(\mathcal P^\ast \).

(b) By axiom I3 in the affine plane, there exist three ordinary points \(A,B,C\in \mathcal P\) which are not all incident with one common affine line.

Let \(\ell \) be the unique affine line incident with \(A\) and \(B\), let \(m\) be the unique affine line incident with \(A\) and \(C\), and let \(n\) be the unique affine line incident with \(B\) and \(C\).

We first claim that \(\ell ,m,n\) belong to three distinct parallel classes. Indeed, \(\ell \) and \(m\) both contain \(A\). If \(\ell =m\), then this one line would contain \(A\), \(B\), and \(C\), contradicting the choice of \(A,B,C\). Hence \(\ell \neq m\). Since \(\ell \) and \(m\) are distinct lines meeting at \(A\), they are not parallel, and therefore they cannot belong to the same parallel class.

Similarly, \(\ell \) and \(n\) are distinct lines meeting at \(B\), so they are not parallel and cannot belong to the same parallel class. Also, \(m\) and \(n\) are distinct lines meeting at \(C\), so they are not parallel and cannot belong to the same parallel class.

Thus \(\ell ,m,n\) determine three distinct parallel classes. Therefore the corresponding points at infinity

\[ [\ell ],\qquad [m],\qquad [n] \]

are three distinct points of \(\mathcal P^\ast \). All points at infinity are incident with the line at infinity \(\ell _\infty \), so \(\ell _\infty \) is incident with at least three distinct points.

(c) Every line of \(\alpha ^\ast \) is either an old affine line or the line at infinity. By parts (a) and (b), every such line is incident with at least three distinct points.

(d) By axiom I3 in the affine plane, there exist three ordinary points \(A,B,C\in \mathcal P\) which are not all incident with one common affine line. These same three points are also points of \(\mathcal P^\ast \).

They are not all incident with \(\ell _\infty \), because \(\ell _\infty \) contains only points at infinity. They are also not all incident with any old line, because that would mean they were collinear in the original affine plane. Therefore \(A,B,C\) are three points of \(\mathcal P^\ast \) which are not all incident with one common line. Hence \(\alpha ^\ast \) satisfies axiom I3.

(e) Let \(\ell ,m\in \mathcal L\) be two distinct old affine lines. If \(\ell \) and \(m\) are not parallel in the affine plane, then they meet in an ordinary point, so they also meet in \(\alpha ^\ast \).

If \(\ell \) and \(m\) are parallel in the affine plane, then they belong to the same parallel class. Hence

\[ [\ell ]=[m]. \]

In the projective completion, both \(\ell \) and \(m\) are incident with this common point at infinity. Therefore they meet in \(\alpha ^\ast \).

Thus any two distinct old lines meet in \(\alpha ^\ast \).

(f) Let \(\ell \in \mathcal L\) be an old affine line. In the projective completion, \(\ell \) is incident with the point at infinity \([\ell ]\). By definition, \([\ell ]\) is also incident with the line at infinity \(\ell _\infty \). Therefore \(\ell \) meets \(\ell _\infty \).

(g) Let \(L,M\in \mathcal L^\ast \) be two distinct lines. There are three cases.

If both \(L\) and \(M\) are old affine lines, then they meet by part (e). If one of them is an old affine line and the other is \(\ell _\infty \), then they meet by part (f). The remaining case, where both are \(\ell _\infty \), cannot occur because \(L\) and \(M\) are distinct.

Therefore any two distinct lines in \(\mathcal L^\ast \) meet. Hence \(\alpha ^\ast \) satisfies the elliptic parallel property.