--- title: "Solutions to Worksheet 2 — Affine Algebraic Sets and Ideals" stable_id: br-ak-2025-2026-w02-solutions language: en upstream_solution_count: 9 upstream_solution_map: "authority/wikiversity/unit-02/ORDERED_EXERCISE_MAP.json" license: CC BY-SA 4.0 translation_status: complete --- # Solutions to Worksheet 2 {#br-ak-2025-2026-w02-solutions} This section translates all nine public solutions linked from Worksheet 2 at the source freeze. Their numbering and order follow the source exercises. No additional solutions are supplied for exercises without a public solution at that boundary. ## Solution to Exercise 2.2 {#br-ak-2025-2026-w02-sol-02} A direction vector of the line is $$ \begin{pmatrix}5\\-3\end{pmatrix}. $$ Its equation therefore has the form $$ 3x+5y=c. $$ Substituting either point gives $c=2$. Thus $$ y=\frac{2-3x}{5}. $$ Substituting this into the circle equation $$ x^2+y^2=1 $$ gives $$ x^2+\left(\frac{2-3x}{5}\right)^2=1, $$ or $$ x^2+\frac{4-12x+9x^2}{25}-1 =\frac{34}{25}x^2-\frac{12}{25}x-\frac{21}{25}=0. $$ Normalising the equation gives $$ x^2-\frac6{17}x-\frac{21}{34}=0. $$ Consequently, $$ \begin{aligned} x_{1,2} &=\frac{\frac6{17}\pm \sqrt{\left(\frac6{17}\right)^2+4\cdot\frac{21}{34}}}{2}\\ &=\frac{\frac6{17}\pm \sqrt{\left(\frac6{17}\right)^2+\frac{42}{17}}}{2}\\ &=\frac{6\pm\sqrt{6^2+714}}{34}\\ &=\frac{6\pm\sqrt{750}}{34}\\ &=\frac{6\pm5\sqrt{30}}{34}, \end{aligned} $$ and $$ \begin{aligned} y_{1,2} &=\frac{2-3x_{1,2}}5\\ &=\frac{2-3\left(\frac{6\pm5\sqrt{30}}{34}\right)}5\\ &=\frac{68-3(6\pm5\sqrt{30})}{170}\\ &=\frac{50\mp15\sqrt{30}}{170}\\ &=\frac{10\mp3\sqrt{30}}{34}. \end{aligned} $$ The intersection points are therefore $$ \left(\frac{6+5\sqrt{30}}{34},\frac{10-3\sqrt{30}}{34}\right) \quad\text{and}\quad \left(\frac{6-5\sqrt{30}}{34},\frac{10+3\sqrt{30}}{34}\right). $$ [Back to Exercise 2.2](#br-ak-2025-2026-w02-ex-02). ## Solution to Exercise 2.6 {#br-ak-2025-2026-w02-sol-06} The unit circle is the solution set of $$ x^2+y^2=1, $$ while $K$ is the solution set of $$ (x-1)^2+y^2=x^2-2x+1+y^2=4. $$ Subtracting the first equation from the second gives $$ -2x+1=3, $$ so $x=-1$. The unit circle equation then gives $y=0$. The only intersection point is therefore $(-1,0)$, which indeed satisfies both equations. [Back to Exercise 2.6](#br-ak-2025-2026-w02-ex-06). ## Solution to Exercise 2.7 {#br-ak-2025-2026-w02-sol-07} We seek the solutions of the system $$ x^2+xy+3y^2=3 $$ and $$ 2x^2-xy+y^2=4. $$ Adding the two equations gives $$ 3x^2+4y^2=7, $$ while twice the first equation minus the second gives $$ 3xy+5y^2=2. $$ From the latter equation, $$ x=\frac{2-5y^2}{3y}; $$ there is certainly no solution with $y=0$. Substituting this expression for $x$ into the preceding equation gives $$ 3\left(\frac{2-5y^2}{3y}\right)^2+4y^2=7. $$ Multiplication by $3y^2$ yields $$ \begin{aligned} 0 &=(2-5y^2)^2+12y^4-21y^2\\ &=4-20y^2+25y^4+12y^4-21y^2\\ &=37y^4-41y^2+4. \end{aligned} $$ This is a biquadratic equation. **Edition note:** The frozen public solution stops here, without giving the intersection coordinates. [Back to Exercise 2.7](#br-ak-2025-2026-w02-ex-07). ## Solution to Exercise 2.8 {#br-ak-2025-2026-w02-sol-08} The standard parabola is given by $$ y=x^2, $$ and the unit circle by $$ x^2+y^2=1. $$ The intersection points must satisfy both equations simultaneously. Using the first equation to replace $x^2$ in the second, we obtain $$ y^2+y-1=0. $$ Thus $$ y=\frac{-1\pm\sqrt{1+4}}2=\frac{-1\pm\sqrt5}{2}. $$ The negative sign gives no real value of $x$, so $$ y=\frac{-1+\sqrt5}{2}, \qquad x=\pm\sqrt{\frac{-1+\sqrt5}{2}}. $$ The two intersection points are $$ \left(-\sqrt{\frac{-1+\sqrt5}{2}},\frac{-1+\sqrt5}{2}\right) $$ and $$ \left(\sqrt{\frac{-1+\sqrt5}{2}},\frac{-1+\sqrt5}{2}\right). $$ [Back to Exercise 2.8](#br-ak-2025-2026-w02-ex-08). ## Solution to Exercise 2.9 {#br-ak-2025-2026-w02-sol-09} 1. **Edition note:** The frozen source leaves the sketch item blank. 2. We have $$ \begin{aligned} K &=\{(x,y)\in\mathbb R^2\mid(y-1)^2+x^2=1\}\\ &=\{(x,y)\in\mathbb R^2\mid y^2-2y+1+x^2=1\}\\ &=\{(x,y)\in\mathbb R^2\mid y^2-2y+x^2=0\}. \end{aligned} $$ 3. We seek the common solution set of the two equations $$ y=x^2 $$ and $$ y^2-2y+x^2=0. $$ Replacing $x^2$ by $y$ in the second equation gives $$ 0=y^2-2y+y=y^2-y=y(y-1). $$ Thus $y=0$ or $y=1$. This gives the three intersection points $(0,0)$, $(1,1)$, and $(-1,1)$. 4. The circle equation $$ y^2-2y+x^2=0 $$ is equivalent to $$ y^2-2y=-x^2, $$ and hence to $$ (y-1)^2=1-x^2. $$ Therefore $$ y=1\pm\sqrt{1-x^2}. $$ The lower semicircle is the graph of the function $$ [-1,1]\longrightarrow\mathbb R, \qquad x\longmapsto1-\sqrt{1-x^2}. $$ 5. We claim that on $[-1,1]$ the parabola lies above the lower semicircle. We must show that $$ x^2\ge1-\sqrt{1-x^2}. $$ This is equivalent to $$ \sqrt{1-x^2}\ge1-x^2. $$ Since both sides are non-negative on this interval, this is equivalent to $$ 1-x^2\ge(1-x^2)^2=1+x^4-2x^2. $$ The last inequality is equivalent to $x^4-x^2\le0$, and then to $x^2-1\le0$, which holds because $x\in[-1,1]$. **Edition note:** The source calls both sides positive on $[-1,1]$. They are non-negative and vanish at the endpoints; non-negativity is the condition needed for squaring. [Back to Exercise 2.9](#br-ak-2025-2026-w02-ex-09). ## Solution to Exercise 2.11 {#br-ak-2025-2026-w02-sol-11} 1. Let $$ A=\begin{pmatrix}X_1&X_2\\X_3&X_4\end{pmatrix} \qquad\text{and}\qquad B=\begin{pmatrix}Y_1&Y_2\\Y_3&Y_4\end{pmatrix}. $$ Then $$ AB= \begin{pmatrix}X_1&X_2\\X_3&X_4\end{pmatrix} \begin{pmatrix}Y_1&Y_2\\Y_3&Y_4\end{pmatrix} =\begin{pmatrix} X_1Y_1+X_2Y_3&X_1Y_2+X_2Y_4\\ X_3Y_1+X_4Y_3&X_3Y_2+X_4Y_4 \end{pmatrix} $$ and $$ BA= \begin{pmatrix}Y_1&Y_2\\Y_3&Y_4\end{pmatrix} \begin{pmatrix}X_1&X_2\\X_3&X_4\end{pmatrix} =\begin{pmatrix} X_1Y_1+X_3Y_2&X_2Y_1+X_4Y_2\\ X_1Y_3+X_3Y_4&X_2Y_3+X_4Y_4 \end{pmatrix}. $$ These product matrices are equal precisely when all four corresponding entries agree, that is, when $$ X_1Y_1+X_2Y_3=X_1Y_1+X_3Y_2, $$ $$ X_1Y_2+X_2Y_4=X_2Y_1+X_4Y_2, $$ $$ X_3Y_1+X_4Y_3=X_1Y_3+X_3Y_4, $$ and $$ X_3Y_2+X_4Y_4=X_2Y_3+X_4Y_4. $$ Thus the set is an affine variety. The first and fourth equations are equivalent to each other and to $$ X_2Y_3=X_3Y_2. $$ The commuting matrices are therefore described by the system $$ X_2Y_3=X_3Y_2, $$ $$ X_1Y_2+X_2Y_4=X_2Y_1+X_4Y_2, $$ $$ X_3Y_1+X_4Y_3=X_1Y_3+X_3Y_4. $$ 2. The identity matrix $E_2$ commutes with every matrix. Hence $(A,E_2)$ is a preimage of $A$. 3. We seek the matrices $$ B=\begin{pmatrix}Y_1&Y_2\\Y_3&Y_4\end{pmatrix} $$ that satisfy the preceding system for $$ A=\begin{pmatrix}X_1&X_2\\X_3&X_4\end{pmatrix} =\begin{pmatrix}1&1\\0&0\end{pmatrix}. $$ The conditions become $$ Y_3=0, \qquad Y_2+Y_4=Y_1, \qquad 0=Y_3, $$ and the third condition can be omitted. The inverse image of the given matrix is $$ \left\{ \left( \begin{pmatrix}1&1\\0&0\end{pmatrix}, \begin{pmatrix}Y_2+Y_4&Y_2\\0&Y_4\end{pmatrix} \right) \;\middle|\;Y_2,Y_4\in K \right\}. $$ [Back to Exercise 2.11](#br-ak-2025-2026-w02-ex-11). ## Solution to Exercise 2.15 {#br-ak-2025-2026-w02-sol-15} First we prove $\mathfrak b\subseteq\mathfrak a$ by showing that every generator of $\mathfrak b$ is $0$ in the quotient ring modulo $\mathfrak a$. In that ring, $$ \begin{aligned} X^2+Y^2 &=(2Z^2-1)^2+4Z^2W^2\\ &=(2Z^2-1)^2+4Z^2(1-Z^2)\\ &=4Z^4-4Z^2+1+4Z^2-4Z^4\\ &=1. \end{aligned} $$ Moreover, $$ \begin{aligned} YW &=2ZW^2\\ &=2Z(1-Z^2)\\ &=Z(2-2Z^2)\\ &=Z(1-2Z^2+1)\\ &=Z(1-X), \end{aligned} $$ and $$ \begin{aligned} W(1+X) &=W(2Z^2)\\ &=2WZ^2\\ &=ZY. \end{aligned} $$ The inclusion $\mathfrak c\subseteq\mathfrak b$ is clear, since one generator has been omitted. Finally, we prove $\mathfrak a\subseteq\mathfrak c$ by showing that the generators of $\mathfrak a$ are $0$ in the quotient ring modulo $\mathfrak c$. In this ring, $$ ZX=Z-YW \qquad\text{and}\qquad WX=YZ-W. $$ Consequently, $$ \begin{aligned} X &=X\cdot1\\ &=X(Z^2+W^2)\\ &=Z^2-ZYW+YZW-W^2\\ &=Z^2-W^2\\ &=2Z^2-1, \end{aligned} $$ and $$ \begin{aligned} Y &=Y\cdot1\\ &=Y(Z^2+W^2)\\ &=WXZ+WZ+ZW-ZXW\\ &=2ZW. \end{aligned} $$ [Back to Exercise 2.15](#br-ak-2025-2026-w02-ex-15). ## Solution to Exercise 2.16 {#br-ak-2025-2026-w02-sol-16} 1. Clearly $(1,1)$ is a zero of $F$. 2. We have $$ \begin{aligned} F\cdot(X^2+Y^2+1) ={}&(X^4Y^2+X^2Y^4-3X^2Y^2+1)(X^2+Y^2+1)\\ ={}&X^6Y^2+X^4Y^4-3X^4Y^2+X^2 +X^4Y^4+X^2Y^6-3X^2Y^4+Y^2\\ &+X^4Y^2+X^2Y^4-3X^2Y^2+1\\ ={}&2X^4Y^4+X^6Y^2+X^2Y^6-2X^4Y^2-2X^2Y^4 -3X^2Y^2+X^2+Y^2+1. \end{aligned} $$ For the other side, $$ (X^2Y-Y)^2+(XY^2-X)^2+(X^2Y^2-1)^2 +\frac14(XY^3-X^3Y)^2 +\frac34(XY^3+X^3Y-2XY)^2, $$ we calculate the coefficient of each monomial. Only even degrees occur, and the highest degree is $8$. Only the last three summands contribute at that degree, and the only monomials are $X^6Y^2$, $X^4Y^4$, and $X^2Y^6$: $$ X^6Y^2:\quad\frac14+\frac34=1, $$ $$ X^4Y^4:\quad1+\frac14(-2)+\frac34(2)=2, $$ $$ X^2Y^6:\quad\frac14+\frac34=1. $$ In degree $6$, only $X^4Y^2$ and $X^2Y^4$ occur: $$ X^4Y^2:\quad1+\frac34(-4)=-2, $$ $$ X^2Y^4:\quad1+\frac34(-4)=-2. $$ In degree $4$, only $X^2Y^2$ occurs: $$ X^2Y^2:\quad-2-2-2+\frac34(4)=-3. $$ In degree $2$, the coefficients of $X^2$ and $Y^2$ are both $1$. In degree $0$, the coefficient of $1$ is also $1$. Thus the two sides agree. 3. Dividing the identity in part (2) by $X^2+Y^2+1$ in the field of fractions of $K[X,Y]$, we obtain $$ \begin{aligned} F=\frac{1}{X^2+Y^2+1}\Big(& (X^2Y-Y)^2+(XY^2-X)^2 +(X^2Y^2-1)^2\\ &+\frac14(XY^3-X^3Y)^2 +\frac34(XY^3+X^3Y-2XY)^2\Big). \end{aligned} $$ Since $X^2+Y^2+1$ has no real zero, this identity also holds as an identity of functions $\mathbb R^2\to\mathbb R$. Squares are never negative, and all coefficients of the squares involved are positive. Hence the function is non-negative at every point. **Edition note:** The source's introductory sentence for degree $2$ mentions only $X^2$, but its coefficient list includes both $X^2$ and $Y^2$. The wording above follows that list and the displayed expansion. [Back to Exercise 2.16](#br-ak-2025-2026-w02-ex-16). ## Solution to Exercise 2.20 {#br-ak-2025-2026-w02-sol-20} To prove the inclusion $\subseteq$, take $f\in(I+J)^n$. Since a product of ideals consists of all sums of products, we can write $$ f=f_1+f_2+\cdots+f_k, $$ where $$ f_\ell=c_{\ell1}c_{\ell2}\cdots c_{\ell n}, $$ with $c_{\ell r}\in I+J$. In turn, $$ c_{\ell r}=a_{\ell r}+b_{\ell r} $$ with $a_{\ell r}\in I$ and $b_{\ell r}\in J$. Thus $$ f_\ell=(a_{\ell1}+b_{\ell1})(a_{\ell2}+b_{\ell2}) \cdots(a_{\ell n}+b_{\ell n}). $$ Expanding this product by distributivity gives a sum of products with $n$ factors each: $s$ factors belong to $I$ and $n-s$ to $J$. Each summand therefore belongs to the right-hand side, as do each $f_\ell$ and finally $f$. To prove the inclusion $\supseteq$, it suffices to show $$ I^sJ^{n-s}\subseteq(I+J)^n $$ for every $s$. Since $I,J\subseteq I+J$, we immediately have $$ I^sJ^{n-s} \subseteq(I+J)^s(I+J)^{n-s} =(I+J)^n. $$ [Back to Exercise 2.20](#br-ak-2025-2026-w02-ex-20).