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This is an independent English edition of the complete classical 30-unit sequence of Holger Brenner’s Algebraische Kurven. Its scope is thirty lectures and thirty worksheets, together with every public solution available at the frozen source boundary. The source material comes from two distinct course boundaries. Units 1–23 follow the frozen revisions of Algebraische Kurven (Osnabrück 2025–2026). At the frozen boundary, that course has no Unit 24 or later units. Units 24–30 therefore follow the official lectures and worksheets of the earlier complete course, Algebraische Kurven (Osnabrück 2012). They are not an official continuation of the 2025–2026 edition, and the two courses are not treated as one source edition.

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Lecture 1: Plane Algebraic Curves

Plane algebraic curves

What is an algebraic curve? For example, the objects shown in the following beautiful pictures.

Of course, we can draw all sorts of things. The following curves are beautiful too, but they are not algebraic curves.

The word “algebraic” in algebraic curve comes from the requirement that only algebraic operations may be used in its definition: addition and multiplication, but not analytic processes such as taking limits, infinite sums, approximation, differentiation or integration. The maps allowed in our context are given by polynomials in several variables. The pictures above show plane algebraic curves defined by a polynomial in two variables. The first two pictures are graphs of a polynomial function in one variable; they are described by

$$Y=P(X).$$

In the first picture, $P(X)=X$ (so the polynomial is linear), whereas in the second the polynomial has a form such as

$$P(X)=a_4X^4+a_3X^3+a_2X^2+a_1X+a_0,$$

with coefficients $a_i$ in a field $K$. In algebraic geometry we fix a base field $K$. Important fields for us are the real numbers (the pictures are primarily to be understood in this sense) and the complex numbers $\mathbb C$. Such a graph is a simple object in that each value of $X$ has exactly one corresponding value of $Y$, namely the function value, which is also easy to calculate if we can calculate in the given field. In a certain sense, the graph is a “curved” copy of the base line, the $X$-axis.

Now consider the third picture. It is the graph of a rational function: we take two polynomials $P,Q$ in the variable $X$ and consider their quotient $P(X)/Q(X)$. This expression makes sense only where the denominator is nonzero. The rational function is undefined at the zeros of the denominator polynomial. If numerator and denominator vanish at the same point, cancellation sometimes gives the quotient a meaning there as well. If the denominator vanishes but the numerator does not, the undefined point is a pole: the real graph tends to $+\infty$ or $-\infty$. It is tempting to say that the rational function takes the value “infinity” at these points; in projective geometry this idea really does make sense, as we shall see later.

Because of these undefined points, however, the “graph equation” $Y=P(X)/Q(X)$ is not an ideal description of the curve. Multiplying by the denominator instead gives the condition, or equation,

$$YQ(X)=P(X), \qquad\text{or, more precisely,}\qquad \{(x,y)\in K^2\mid yQ(x)=P(x)\},$$

whose two sides are well-defined polynomials. The set satisfying the equation (or solution set) is uniquely defined. For an $x$ with $Q(x)=0$, the left-hand side is zero. If $P(x)\ne0$, there is no solution at this $x$, as in the picture; if $P(x)=0$, every value of $Y$ is allowed. In the latter case, the object therefore contains the line through $(x,0)$ perpendicular to the $X$-axis.

Example: the hyperbola

A typical and important example of a rational function is $Y=1/X$. Its graph is called a hyperbola $H$. Written without a denominator, the equation becomes

$$XY=1, \qquad\text{or}\qquad H=\{(x,y)\mid xy=1\}.$$

On $K^\times=K\setminus\{0\}$ this rational function is an ordinary function, with graph $H$, and it gives a “natural” bijection

$$K^\times\longrightarrow H, \qquad x\longmapsto\left(x,\frac1x\right).$$

Thus $K^\times$ and $H$ are “equivalent” or “isomorphic” in a sense that will be made precise later.

Both descriptions have advantages. The description as $K^\times\subset K$ takes place on a line (if we think of $K=\mathbb R$), but the point $0$, which is a limit point of $K^\times$, does not belong to $K^\times$. In other words, $K^\times$ is not closed. The hyperbola, by contrast, is closed in $\mathbb R^2$; realising the object as a closed set therefore requires moving to a higher dimension. The question of what constitutes a good description of an algebraic-geometric object will recur throughout the course.

In the real case, $K=\mathbb R$, the set $\mathbb R^\times$ (and likewise $H_{\mathbb R}$) consists of two disjoint “branches”, so it is not connected. In the complex case, $K=\mathbb C$, the set $\mathbb C^\times$ (and likewise $H_{\mathbb C}$) is a punctured real plane and is therefore connected. This is a typical phenomenon in algebraic geometry: important properties may depend on the base field. Nevertheless, properties that depend only on the defining equations and hold for their solution sets over every field have particular significance.

The fourth picture shows a circle, with equation

$$K=\{(x,y)\mid x^2+y^2=r^2\},$$

where $r$ denotes its radius. The picture already shows that this object cannot be the graph of a function, since on a graph each $x$-value is paired with exactly one $y$-value. There is no function $y=\varphi(x)$ satisfying

$$K=\{(x,\varphi(x))\mid x\in\mathbb R\}.$$

The question of whether an algebraic solution set can be realised as a graph is equivalent to asking whether its defining equation can be “solved” for $y$. In this example we can write

$$y^2=r^2-x^2, \qquad y=\sqrt{r^2-x^2}=\sqrt{(r-x)(r+x)}.$$

Is the circle a graph after all? There are two interpretations.

1. If we restrict ourselves to real numbers and positive square roots, the last step is not an equivalent transformation: we have “added” information that was not in the original equation. Taking the positive square root means restricting to the upper semicircle. Adding information or conditions makes the solution set smaller.

2. If instead $\sqrt{\phantom{x}}$ is understood to include all solutions—in the real case, both the positive and the negative square root, often written $\pm\sqrt{\phantom{x}}$—we have added no information, but we have not solved for a function either; we have only obtained what is sometimes called a “multivalued function”.

Both viewpoints are useful. The attempt to describe part of a geometric object, such as the upper arc, simply as a graph reappears in the implicit function theorem, power-series methods, parametrisations and local theory.

Equations of the form $Y^2=G(X)$

A circle equation can be viewed as an equation of the form

$$Y^2=G(X),$$

where $G$ is a polynomial in the single variable $X$; for the circle, $G=-X^2+1$. This is not a graph, but the “square root” of a graph. More generally, allow $G(X)$ to be more complicated. The zero set (or zero locus) represents the square root $\sqrt{G(X)}$. For any chosen value $x$ of $X$, there are three possibilities for the corresponding real solutions $y$.

1. If $G(x)<0$, there is no solution.
2. If $G(x)=0$, there is exactly one solution, $y=0$.
3. If $G(x)>0$, there are two solutions, $y=\pm\sqrt{G(x)}$.

This also suggests how to visualise the real picture: for each $x$, calculate $G(x)$ and, if the radicand is nonnegative, mark the points $(x,\pm\sqrt{G(x)})$.

Over the complex numbers, we need distinguish only $G(x)=0$ from $G(x)\ne0$. If $G$ has degree two, the resulting curve is a conic section, a subject studied since antiquity (see Lecture 7).

Isaac Newton studied intensively the case in which $G(X)$ is a real cubic polynomial, that is, a polynomial of degree three. Even this collection of examples is already very rich.

Consider the case $G(X)=X^3$, the object described by

$$\{(x,y)\mid y^2=x^3\}.$$

This object is called Neil’s parabola. A new phenomenon appears here: the origin is different from all the other points. We call it a singularity; the other points, by contrast, are called smooth or nonsingular. Giving a precise definition is part of this course. As a first, imprecise formulation, a curve near a smooth point looks, in suitable coordinates, like the possibly rotated graph of a differentiable function. The singularity of Neil’s parabola is called a cusp—the source uses the German words Spitze and Kuspe, both meaning a pointed tip. By contrast, the singularity in the eighth picture is a crossing point or double point.

In the seventh picture at the beginning, and in the picture above, we also see zero loci of the form $Y^2=G(X)$ with $G(X)$ of degree three. What must $G(X)$ look like to produce such a curve? The last examples also show that the presence of a singularity depends on the precise form of $G(X)$.

Let us stay with Neil’s parabola $C$. If $t$ is any real or complex number, the point with coordinates

$$(x,y)=(t^2,t^3)$$

always lies on Neil’s parabola, since $(t^2)^3=t^6=(t^3)^2$. Conversely, one can show (see Exercise 1.6) that every point of Neil’s parabola has this form: for each $(x,y)$ satisfying $y^2=x^3$, there is exactly one $t$ with $(x,y)=(t^2,t^3)$. The map

$$\mathbb R\longrightarrow C, \qquad t\longmapsto(t^2,t^3),$$

is called a (bijective polynomial) parametrisation of Neil’s parabola. Determining which algebraic curves admit a polynomial parametrisation is a nontrivial question. A smooth curve of the form $Y^2=G(X)$ with $\deg G=3$ has no such parametrisation. In elementary number theory, we learn that all Pythagorean triples can be written in a simple uniform form; see Theorem 10.6 (Number Theory, Osnabrück 2025). An equivalent statement is the existence of a rational parametrisation of the rational unit circle; see Theorem 10.4. We shall discuss this in greater generality in Theorem 7.6.

We now come to the first general definition.

Definition: affine plane algebraic curve

Let $K$ be a field. An affine plane algebraic curve over $K$ is the zero locus $V(F)\subseteq K^2$ of a nonconstant polynomial $F$ in two variables, that is,

$$F=\sum_{0\le i,j\le m}a_{ij}X^iY^j \qquad (a_{ij}\in K).$$

In other words,

$$V(F)=\left\{(x,y)\in K^2\;\middle|\; F(x,y)=\sum_{0\le i,j\le m}a_{ij}x^iy^j=0\right\}.$$

The following favourite polynomials in the variables $X$ and $Y$ were suggested in class:

1. $X^2-Y^2$;
2. $2X+4Y+3$;
3. $X^2+Y^2-3$;
4. $X^2+Y^2$;
5. $5X^2+12Y^2-26$;
6. $3X^2-15Y^2-3$;
7. $X^3+Y^3+XY$;
8. $X^3-4Y^2-XY$;
9. $X^4$;
10. $X^2Y^2-X^2$.

The corresponding zero loci $V(F)$ vary in how difficult they are to understand. By the difference-of-squares identity,

$$X^2-Y^2=(X+Y)(X-Y),$$

and a product of two field elements is zero exactly when one factor is zero. This zero locus is therefore simply the union of the two diagonals: a union of two affine lines. The zero locus of $2X+4Y+3$ is the solution set of this linear equation, an affine line. The real zero locus of $X^2+Y^2-3$ is a circle centred at the origin with radius $\sqrt3$. By contrast, the real zero locus of $X^2+Y^2$ consists only of the origin $(0,0)$; over $\mathbb C$ the situation is different. The zero locus of $5X^2+12Y^2-26$ is an ellipse with axes parallel to the coordinate axes, whereas the zero locus of $3X^2-15Y^2-3$ is a compressed hyperbola.

We shall discuss and classify the zero loci of quadratic polynomials in detail in Lecture 7. The two polynomials $X^3+Y^3+XY$ and $X^3-4Y^2-XY$ have degree $3$ and are much harder to understand. The first question is whether their curves are smooth or have singularities. The zero locus $V(X^4)$ equals $V(X)$ and is therefore the $y$-axis. The last polynomial factors as

$$X^2Y^2-X^2=X^2(Y^2-1)=X^2(Y-1)(Y+1),$$

so it is easy to understand. Its zero locus is the union of three lines: the $y$-axis and two lines parallel to the $x$-axis.

We shall prove a lemma that immediately shows why the four nonalgebraic curves pictured above are not algebraic.

Lemma: intersection with a line

Let $C$ be an affine plane algebraic curve and $L$ a line in $K^2$.

Then $C\cap L$ is either the whole line $L$ or a finite set of points.

Proof

By definition, a plane algebraic curve $C=V(F)$ is always the zero locus of a polynomial $F$ in two variables. Suppose the line $L$ is given by

$$aX+bY+c=0.$$

Without loss of generality, assume $a\ne0$. Solving for $X$ gives $X=\alpha Y+\beta$. An intersection point $P\in C\cap L$ must satisfy both $F(P)=0$ and the line equation. Using the line equation, replace $X$ in $F$ by $\alpha Y+\beta$. This turns $F$ into a polynomial in the single variable $Y$, which we call $\widetilde F$.

Now $P\in C\cap L$ is equivalent to $P\in L$ and $\widetilde F(P)=0$. Thus the intersection is described by $\widetilde F$. If $\widetilde F=0$, the whole line is the intersection. If $\widetilde F\ne0$, Corollary 19.9 (Linear Algebra, Osnabrück 2024–2025) says that it has only finitely many zeros. $\square$

For the four nonalgebraic examples above, there are lines meeting the curves in infinitely many points. The curves are therefore not algebraic.

Polynomial rings

After these introductory examples, we fix some terminology that is probably already familiar.

Definition: polynomial ring in one variable

The polynomial ring over a commutative ring $R$ consists of all polynomials

$$P=a_0+a_1X+a_2X^2+\cdots+a_nX^n,$$

with $a_i\in R$ for $i=0,\ldots,n$ and $n\in\mathbb N$, equipped with componentwise addition and multiplication defined by extending the following rule distributively:

$$X^n\cdot X^m:=X^{n+m}.$$ From this definition we can also define polynomial rings in several variables. Set

$$K[X,Y]:=(K[X])[Y], \qquad K[X,Y,Z]:=(K[X,Y])[Z],$$

and so on. A polynomial in $n$ variables has the form

$$F=\sum_{(\nu_1,\ldots,\nu_n)} a_{(\nu_1,\ldots,\nu_n)}X_1^{\nu_1}\cdots X_n^{\nu_n}.$$

The sum runs over a finite family of exponent tuples $(\nu_1,\ldots,\nu_n)$. Expressions of the form $X_1^{\nu_1}\cdots X_n^{\nu_n}$ are also called monomials. A polynomial is usually abbreviated as $F=\sum_\nu a_\nu X^\nu$. Multiplying two monomials means adding their exponent tuples:

$$\left(X_1^{\nu_1}\cdots X_n^{\nu_n}\right) \left(X_1^{\mu_1}\cdots X_n^{\mu_n}\right) :=X_1^{\nu_1+\mu_1}\cdots X_n^{\nu_n+\mu_n}.$$

In algebraic geometry, the case of greatest interest to us is when the base ring $R$ is a field. Algebraic geometry studies the shape of zero loci of polynomials in several variables. We shall see later that the relationship between algebraic and geometric properties is particularly strong when the base field is algebraically closed.

Definition: algebraically closed field

A field $K$ is called algebraically closed if every nonconstant polynomial $F\in K[X]$ has a zero in $K$.

The fundamental theorem of algebra was first proved by Gauss.

Theorem: fundamental theorem of algebra

The field of complex numbers $\mathbb C$ is algebraically closed.

Proof

We shall not prove this theorem here. Its proofs use topological or analytic methods. $\square$

Source navigation: course · Lecture 2 · worksheet for this lecture

Worksheet 1

Practice exercises

Exercise 1.1

Linear algebra often deals with systems of linear equations. What nonlinear equations or systems of equations arise in linear algebra?

Exercise 1.2

Sketch the solution sets in $\mathbb R^2$ of the following equations.

1. $x^2-y^2-1=0$;
2. $x^2+xy+y^2=0$;
3. $x^2+y^2+1=0$;
4. $x^2+y^2=0$;
5. $x^2+y^3=0$;
6. $x^3-y^5=0$;
7. $x^2-x^3=0$;
8. $x^3+y^3=1$;
9. $x^4+y^4=1$;
10. $-5+3x+4x^2+x^3-y^2=1$.

Exercise 1.3

Calculate the intersection of each curve in Exercise 1.2 with each of the following lines.

1. $x=0$;
2. $y=0$;
3. $x=1$;
4. $y=-2$;
5. $x=y$;
6. $x=-y$;
7. $2x-3y+4=0$.

Exercise 1.4 ★

1. Find an integer solution $(x,y)\in\mathbb Z\times\mathbb Z$ of

   $$x^2-y^3+2=0.$$

2. Show that

   $$\left(\frac{383}{1000},\frac{129}{100}\right)$$

   is a solution of

   $$x^2-y^3+2=0.$$

Exercise 1.5 ★

Find a point on the plane algebraic curve

$$V(X^3-Y^3+4X^2-2XY+Y+3)\subset\mathbb C^2.$$

Exercise 1.6

Let $K$ be a field. The image of the curve defined by

$$K\longrightarrow K^2, \qquad t\longmapsto(t^2,t^3),$$

is called Neil’s parabola. Show that a point $(x,y)\in K^2$ belongs to this image if and only if it satisfies $x^3=y^2$.

Exercise 1.7

Let $C\subseteq\mathbb R^2$ be the image of the polynomial map

$$\mathbb R\longrightarrow\mathbb R^2, \qquad t\longmapsto(t^2-t+2,t^2-3).$$

Find a nonzero polynomial $F$ in two variables such that $C$ lies in the zero locus of $F$.

Exercise 1.8

Let $C\subseteq\mathbb R^2$ be the image of the polynomial map

$$\mathbb R\longrightarrow\mathbb R^2, \qquad t\longmapsto(t^3-1,t^2-1).$$

Find a nonzero polynomial $F$ in two variables such that $C$ lies in the zero locus of $F$.

Exercise 1.9

Consider the curve

$$\mathbb R\longrightarrow\mathbb R^2, \qquad t\longmapsto(t^2-1,t^3-t).$$

1. Show that the image points $(x,y)$ of this curve satisfy

   $$y^2=x^2+x^3.$$

2. Show that every point $(x,y)\in\mathbb R^2$ satisfying $y^2=x^2+x^3$ belongs to the image of the curve.

3. Show that there are exactly two parameter values $t_1$ and $t_2$ with the same image point, and that the map is otherwise injective.

Exercise 1.10

Discuss the relationship between plane algebraic curves and the implicit function theorem.

Exercise 1.11

Let $K=\mathbb Z/(7)$. Find all points in $K^2=K\times K$ lying on the curve defined by

$$X^2Y+2Y^3+3Y^2=0.$$

How many solutions are there?

Exercise 1.12 ★

Find a line $G\subseteq\mathbb A_{\mathbb C}^2$ meeting the curve

$$C=V(X^3+Y^3+1)\subseteq\mathbb A_{\mathbb C}^2$$

in exactly one point.

Exercise 1.13 ★

Show that Neil’s parabola

$$C=V(Y^2-X^3)\subseteq\mathbb A_{\mathbb C}^2$$

meets every line through $P=(1,1)\in C$ in at least one further point.

Exercise 1.14 ★

Give an analytic argument showing that the unit circle $V(x^2+y^2-1)$ and Neil’s parabola $V(y^2-x^3)$ have a real intersection point. Determine numerically the real $x$-coordinate of such a point with error $\le 0.1$.

Exercise 1.15

Consider equations of the form

$$y^2=G(x), \qquad G(x)=x^3+ax^2+bx+c,$$

over $\mathbb R$. Sketch the different solution sets for coefficients $a,b,c\in\{1,-1,0\}$.

The statement in the next exercise also follows from Theorem 6.1 using Lemma 1.3.

Exercise 1.16

Let $K$ be a field and let

$$K\longrightarrow K^2, \qquad t\longmapsto(P(t),Q(t))$$

be a map given by two polynomials $P(t),Q(t)\in K[t]$. Let $B$ be its image and $G\subseteq K^2$ a line. Show that either $B\subseteq G$ or the intersection $B\cap G$ is finite.

Exercise 1.17

Multiply the following two polynomials in $\mathbb Z[x,y,z]$:

$$x^5+3x^2y^2-xyz^3 \qquad\text{and}\qquad 2x^3yz+z^2+5xy^2z-x^2y.$$

Exercise 1.18

Multiply the following two polynomials in $(\mathbb Z/(5))[x,y]$:

$$x^4+2x^2y^2-xy^3+2y^3 \qquad\text{and}\qquad x^4y+4x^2y+3xy^2-x^2y^2+2y^2.$$

Exercise 1.19

Let $R$ be an integral domain. Show that the polynomial ring $R[X]$ is also an integral domain.

Exercise 1.20 ★

Let $R$ be an integral domain and $R[X]$ the polynomial ring over $R$. Show that the units of $R[X]$ are precisely the units of $R$.

Exercise 1.21 ★

Let $K$ be a field. Show that the following two properties are equivalent.

1. $K$ is algebraically closed.
2. Every nonconstant polynomial $F\in K[X]$ factors into linear factors.

Exercise 1.22

Let $K$ be an algebraically closed field. Determine all irreducible polynomials in $K[X]$.

Exercise 1.23

Let $K$ be an algebraically closed field. Show that $K$ cannot be finite.

Exercises to hand in

Exercise 1.24 — 2 points

Carry out the following polynomial division in $(\mathbb Z/(7))[X]$:

$$X^4+5X^2+3 \qquad\text{divided by}\qquad 2X^2+X+6.$$

Exercise 1.25 — 5 points

Find all monic irreducible polynomials of degree $4$ in the polynomial ring $\mathbb F_3[X]$.

Exercise 1.26 — 3 points

Find all solutions of the circle equation

$$x^2+y^2=1$$

for the fields $K=\mathbb Z/(2)$, $\mathbb Z/(5)$ and $\mathbb Z/(11)$.

Exercise 1.27 — 5 points

Let $C\subseteq\mathbb C^2$ be the image of the polynomial map

$$\mathbb C\longrightarrow\mathbb C^2, \qquad t\longmapsto(t^3-t^2+4t+3,-t^2+5t-1).$$

Find a nonzero polynomial $F$ in two variables such that $C$ lies in the zero locus of $F$.

Exercise 1.28 — 4 points

Consider the map

$$f:\mathbb R\longrightarrow S^1\subseteq\mathbb R^2$$

assigning to each $t\in\mathbb R$ the unique intersection point other than $(0,-1)$ of the line $G_t$ through $(t,1)$ and $(0,-1)$ with the unit circle

$$S^1=\{(x,y)\in\mathbb R^2\mid x^2+y^2=1\}.$$

Show that this map is well defined and find the formulas describing it. Show that $f$ is differentiable. Is $f$ injective? Is $f$ surjective?

Source navigation: course · Worksheet 2 · Lecture 1

Solutions to Worksheet 1

This section translates the seven public solutions linked from Worksheet 1. Their numbering and order follow the source exercises. No additional solutions are supplied for exercises without a public solution at the frozen source revision.

Solution to Exercise 1.4

1. $(5,3)$ is an integer solution.

2. We have

   $$\begin{aligned} \left(\frac{383}{1000}\right)^2 -\left(\frac{129}{100}\right)^3+2 &=\frac{146\,689}{1\,000\,000} -\frac{2\,146\,689}{1\,000\,000}+2\\ &=\frac{146\,689-2\,146\,689}{1\,000\,000}+2\\ &=\frac{-2\,000\,000}{1\,000\,000}+2\\ &=-2+2\\ &=0. \end{aligned}$$

Back to Exercise 1.4.

Solution to Exercise 1.5

Consider the intersection of the curve with the line $V(X-Y)$, imposing the additional condition $X=Y$. Substituting $Y=X$ into the curve equation gives

$$X^3-X^3+4X^2-2X^2+X+3=2X^2+X+3=0.$$

The quadratic formula gives

$$X=-\frac14\pm\sqrt{-\frac{23}{16}} =-\frac14\pm\frac{\sqrt{23}}4i.$$

Therefore

$$\left(\frac{-1+\sqrt{23}\,i}{4}, \frac{-1+\sqrt{23}\,i}{4}\right)$$

is a point on the curve.

Back to Exercise 1.5.

Solution to Exercise 1.12

Choose the line

$$G=V(Y+1).$$

To calculate $C\cap G$, substitute the equation $Y=-1$, which holds on $G$, into the curve equation. This gives

$$0=X^3+(-1)^3+1=X^3.$$

The only solution is $X=0$, so $(0,-1)$ is the unique intersection point of $G$ and $C$.

Back to Exercise 1.12.

Solution to Exercise 1.13

Every line in the plane has an equation

$$ax+by=c,$$

where $a$ and $b$ are not both zero. If the line passes through $(1,1)$, then $a+b=c$.

If $b=0$, the line is $x=1$ and has the further intersection point $(1,-1)$ with the curve. Hence assume $b\ne0$. Solving the line equation for $y$ gives

$$y=rx+s, \qquad s=1-r.$$

On this line, the curve equation becomes

$$\begin{aligned} 0 &=y^2-x^3\\ &=(rx+(1-r))^2-x^3\\ &=-x^3+r^2x^2+2r(1-r)x+(1-r)^2. \end{aligned}$$

Since $x=1$ is a zero, we can factor out $x-1$:

$$-x^3+r^2x^2+2r(1-r)x+(1-r)^2 =(x-1)\bigl(-x^2-(1-r^2)x-(1-r)^2\bigr).$$

After multiplication by $-1$, the quadratic factor on the right is monic of degree $2$, so it has roots in $\mathbb C$. We must show that at least one additional root differs from $1$. Evaluating the displayed quadratic factor at $x=1$ gives

$$-1-(1-r^2)-(1-r)^2=-3+r^2+2r-r^2=2r-3.$$

If $r\ne\frac32$, then $1$ is not a root of that quadratic factor. It remains to consider $r=\frac32$. In this case,

$$x^2+(1-r^2)x+(1-r)^2 =x^2-\frac54x+\frac14 =(x-1)\left(x-\frac14\right),$$

so there is another root, $x=\frac14$. Thus every line through $(1,1)$ meets Neil’s parabola in at least one further point.

Edition note: The source calls the displayed quadratic factor monic, although its leading coefficient is $-1$. The clarification above makes the multiplication by $-1$ explicit; its roots are unchanged.

Back to Exercise 1.13.

Solution to Exercise 1.14

Substitute $y^2=x^3$ into the circle equation to obtain

$$x^3+x^2-1=0.$$

At $x=1$ the polynomial has value $1$, whereas at $x=0.5$ it has a negative value. By the intermediate value theorem it has a root $x_0$ in $[0.5,1]$. Since $x_0^3$ is positive, its real square root

$$y_0=\sqrt{x_0^3}$$

exists, and $(x_0,y_0)$ is a real intersection point.

To approximate $x_0$ numerically, calculate

$$(0.7)^3+(0.7)^2-1 <0.49+0.49-1<0$$

and

$$\begin{aligned} (0.8)^3+(0.8)^2-1 &=0.64\cdot0.8+0.64-1\\ &>0.48+0.64-1>0. \end{aligned}$$

Thus an intersection point exists whose $x$-coordinate lies in $[0.7,0.8]$.

Edition note: The source ends the second estimate with $0.48+0.64-1=0$. This edition uses the correct inequality $0.48+0.64-1>0$.

Back to Exercise 1.14.

Solution to Exercise 1.20

Let $a\in R$ be a unit. There is a $b\in R$ with $ab=1$, and the same identity holds in the polynomial ring. Hence

$$R^\times\subseteq R[X]^\times.$$

Conversely, let

$$P=\sum_{i=0}^d a_iX^i, \qquad a_d\ne0,$$

be a unit in $R[X]$. Then there is a polynomial

$$Q=\sum_{j=0}^e b_jX^j, \qquad b_e\ne0,$$

with $PQ=1$. Since $R$ is an integral domain, $a_db_e\ne0$, and the product has the form

$$a_db_eX^{d+e}+\text{terms of lower degree}.$$

Because $PQ=1$, we must have $d+e=0$ and $a_db_e=1$. Thus $P$ is a constant unit. Consequently the units of $R[X]$ are precisely the units of $R$.

Back to Exercise 1.20.

Solution to Exercise 1.21

Suppose $K$ is algebraically closed and $F\in K[X]$ is nonconstant. We prove by induction on $n=\deg F$ that $F$ factors into linear factors.

For $n=1$, we have $F=a_1X+a_0$, so $F$ is already a single linear factor. Assume every polynomial $G\in K[X]$ of degree $n-1$ factors into linear factors. Since $K$ is algebraically closed, $F$ has a root $x_0$. We can therefore write

$$F=G\cdot(X-x_0)$$

for a polynomial $G\in K[X]$ of degree $n-1$. This degree assertion follows directly from the fact that a field is also an integral domain and a slight adaptation of the proof in Exercise 8. By the induction hypothesis, $G$ factors into linear factors, so $G\cdot(X-x_0)$ does too. Hence every nonconstant polynomial $F\in K[X]$ factors into linear factors.

Conversely, if every nonconstant polynomial factors into linear factors, each has a root represented by one of its linear factors. Thus $K$ is algebraically closed.

Edition note: “Exercise 8” is an unlinked reference in the frozen source and does not identify Exercise 1.8 of this worksheet. The degree assertion used here is $\deg(G(X)(X-x_0))=\deg G+1$ over a field; no renumbered source reference is inferred.

Back to Exercise 1.21.

Lecture 2: Affine Algebraic Sets

Affine algebraic sets

Definition: affine space

Let $K$ be a field. The space

$$\mathbb A_K^n=K^n$$

is called affine space of dimension $n$ over $K$.

Thus, to begin with, affine space is simply a set of points. A point in affine space is an $n$-tuple $(a_1,\ldots,a_n)$ with coordinates in $K$. Why, then, introduce a new term? The term “affine space” indicates that we wish to regard $K^n$ as an object of algebraic geometry. In other words, we regard $n$-dimensional affine space as the natural geometric object on which polynomials in $n$ variables act as functions. We shall gradually equip affine space with further structures—the Zariski topology and the structure sheaf—which make it clear that it is “more” than “just” $K^n$. For $n=1$ we speak of the affine line, and for $n=2$ of the affine plane.

A polynomial $F\in K[X_1,\ldots,X_n]$ can naturally be regarded as a function on affine space. To a point

$$P=(a_1,\ldots,a_n)\in\mathbb A_K^n,$$

we assign the value

$$F(P)=F(a_1,\ldots,a_n)$$

by replacing each variable $X_i$ by $a_i$ and carrying out all the operations in $K$. Given a polynomial $F\in K[X_1,\ldots,X_n]$, we can ask whether $F(P)=0$. One object of particular interest associated with $F$ is therefore the zero locus it defines,

$$V(F)=\{P\in\mathbb A_K^n\mid F(P)=0\}.$$

We encountered several examples in the first lecture. It is also useful, however, to study the common, or simultaneous, zero locus of several polynomials. This is the intersection of their individual zero loci—for example, in the case of conic sections, where a cone in three-dimensional space is intersected with various planes.

We therefore make the following general definition.

Definition: the zero locus of a family of polynomials

Let $K$ be a field, and let

$$F_j\in K[X_1,\ldots,X_n],\qquad j\in J,$$

be a family of polynomials in $n$ variables. The set

$$\{P\in\mathbb A_K^n\mid F_j(P)=0\text{ for every }j\in J\}$$

is called the zero locus (or zero set) defined by the family. It is denoted by $V(F_j,j\in J)$.

Those subsets of affine space that arise as zero sets deserve a name of their own.

Definition: affine algebraic set

Let $K$ be a field and $K[X_1,\ldots,X_n]$ the polynomial ring in $n$ variables. A subset $V\subseteq\mathbb A_K^n$ is called an affine algebraic set if it is the zero set of a family of polynomials $F_j$, $j\in J$, with $F_j\in K[X_1,\ldots,X_n]$; that is, if

$$V=V(F_j,j\in J).$$

The simplest examples are finite sets of points on the affine line $\mathbb A_K^1$, each given by a single polynomial, and affine linear subspaces of $\mathbb A_K^n$, which are the solution sets of inhomogeneous systems of linear equations over $K$.

Example: the axes and the origin

Consider the affine plane $\mathbb A_K^2$ and some affine algebraic subsets defined by the variables $X$ and $Y$.

• The zero locus $V(X,Y)$ consists only of the origin $(0,0)$, since both variables must be zero.

• The set $V(X)$ is the $Y$-axis: all points of the form $(0,y)$.

• The set $V(Y)$ is the $X$-axis.

• The set $V(X+Y)$ consists of all points $(x,y)$ with $y=-x$: it is the antidiagonal.

• The set $V(XY)$ consists of the points $(x,y)$ with $xy=0$. Since $K$ is a field, a product can be zero only if one of its factors is zero (Lemma 3.10 of Linear Algebra, Osnabrück 2024–2025). Thus

  $$V(XY)=V(X)\cup V(Y),$$

  the union of the two coordinate axes.

Points in affine space or on an affine algebraic set are often interpreted as representing more complicated mathematical objects. Properties of those objects are then reflected in whether their representing points satisfy certain algebraic equations or, equivalently, lie on certain affine algebraic sets. The following example illustrates this idea.

Example: matrices as points of affine space

A $2\times2$ matrix

$$\begin{pmatrix} a_{11}&a_{21}\\ a_{12}&a_{22} \end{pmatrix}$$

is uniquely determined by the four numbers $a_{11},a_{21},a_{12},a_{22}\in K$. It can therefore be identified with a point of $\mathbb A_K^4$. In this interpretation, it is natural to denote the variables by $X_{11},X_{21},X_{12},X_{22}$. We can now ask which properties of matrices can be described by algebraic equations. We discuss several typical properties.

A matrix is upper triangular precisely when $a_{12}=0$. Thus the set of upper triangular matrices is the zero locus of $X_{12}$.

A matrix is invertible when

$$a_{11}a_{22}-a_{12}a_{21}\ne0.$$

Consequently, the set of non-invertible matrices is described by the algebraic determinant condition

$$X_{11}X_{22}-X_{12}X_{21}=0.$$

A matrix describes multiplication by a scalar if it is diagonal with equal diagonal entries. This set is described by the three equations

$$X_{12}=0,\qquad X_{21}=0,\qquad X_{11}-X_{22}=0.$$

An element $\lambda\in K$ is an eigenvalue of a matrix precisely when it is a root of its characteristic polynomial (Theorem 23.2 of Linear Algebra, Osnabrück 2024–2025), that is, when

$$\det\begin{pmatrix} \lambda-a_{11}&-a_{21}\\ -a_{12}&\lambda-a_{22} \end{pmatrix} =\lambda^2-\lambda(a_{11}+a_{22})+a_{11}a_{22}-a_{12}a_{21}=0.$$

In linear algebra, the matrix is usually given and we seek roots $\lambda$ of this polynomial in one variable. We can also reverse the viewpoint: fix $\lambda$ and study the zero locus

$$\lambda^2-\lambda(X_{11}+X_{22}) +X_{11}X_{22}-X_{12}X_{21}=0$$

in four variables. This equation describes all matrices having $\lambda$ as an eigenvalue.

Similarly, a matrix has the two distinct eigenvalues $\lambda\ne\delta$ precisely when

$$\lambda^2-\lambda(X_{11}+X_{22}) +X_{11}X_{22}-X_{12}X_{21}=0$$

and

$$\delta^2-\delta(X_{11}+X_{22}) +X_{11}X_{22}-X_{12}X_{21}=0.$$

Subtracting the two equations gives

$$\lambda^2-\delta^2-(\lambda-\delta)(X_{11}+X_{22})=0,$$

which such a matrix must also satisfy. Since $\lambda\ne\delta$, we can write this as

$$X_{11}+X_{22}=\lambda+\delta.$$

The sum of a matrix’s diagonal entries is called its trace. The last equation therefore says that the trace of a matrix with eigenvalues $\lambda\ne\delta$ must equal their sum.

The characteristic polynomial of a matrix can also be written as

$$\lambda^2-\lambda\operatorname{Trace}(M)+\det(M),$$

where

$$\operatorname{Trace}(M)=X_{11}+X_{22},\qquad \det(M)=X_{11}X_{22}-X_{12}X_{21}.$$

Thus two matrices have the same characteristic polynomial precisely when they have the same trace and determinant. The set of matrices with a prescribed characteristic polynomial can therefore be regarded as a fibre of the map

$$\mathbb A_K^4\longrightarrow\mathbb A_K^2, \qquad M\longmapsto(\operatorname{Trace}(M),\det(M)).$$

This map is given by simple polynomial expressions. Is it surjective? Do its fibres always look alike—that is, does the set of matrices with a prescribed trace and determinant always have the same structure—or are there differences?

Fix $s$ and $d$. We must study the solution set of the system

$$X_{11}+X_{22}=s, \qquad X_{11}X_{22}-X_{12}X_{21}=d.$$

The variable $X_{11}$ is uniquely determined by $X_{22}$, and conversely. We can therefore eliminate one variable by putting $X_{22}=s-X_{11}$. This gives an “equivalent” system in the three variables $X_{11},X_{12},X_{21}$ with the single equation

$$X_{11}(s-X_{11})-X_{12}X_{21}=d,$$

or

$$X_{11}^2-sX_{11}+X_{12}X_{21}+d=0.$$

Here “equivalent” means that the two solution sets are in bijection through maps given by polynomials. The last form shows that a solution always exists: we may choose any value of $X_{11}$ and obtain an equation of the form $X_{12}X_{21}=a$, which has solutions.

A linear change of variables simplifies the equation further. Suppose that $2$ is invertible in $K$, so the characteristic of $K$ is not $2$. With

$$X=X_{11}-\frac{s}{2},\qquad Y=X_{12},\qquad Z=X_{21},$$

we obtain

$$X^2+YZ+c=0, \qquad c=-\frac{s^2}{4}+d.$$

Thus the shape of the set of matrices with a prescribed trace and determinant depends only on $-s^2/4+d$. Indeed, the zero locus differs according to whether this expression is zero or nonzero. In the first case it has a singularity; in the second it does not, as we shall see later.

Ideals and zero loci

Since for now we allow arbitrary families of polynomials to define zero loci and hence affine algebraic sets, these objects initially seem difficult to get a handle on. Three important statements nevertheless hold, which we shall prove in stages.

1. The zero locus of a family of polynomials equals the zero locus of the ideal generated by that family.
2. Every ideal has a finite set of generators. Thus every zero locus can be described by finitely many polynomials (Hilbert’s basis theorem).
3. Over an algebraically closed field, zero loci correspond bijectively to radical ideals, a special class of ideals (Hilbert’s Nullstellensatz).

We can prove the first statement immediately. The other two require some algebraic preparation, which we shall develop in the following lectures.

Lemma: a family of polynomials and the ideal it generates

Let $K$ be a field and $F_j\in K[X_1,\ldots,X_n]$, $j\in J$, a family of polynomials in $n$ variables. Let $\mathfrak a$ be the ideal of $K[X_1,\ldots,X_n]$ generated by all the $F_j$. Then

$$V(F_j,j\in J)=V(\mathfrak a).$$

Proof

The ideal $\mathfrak a$ consists of all finite linear combinations of the polynomials $F_j$ and in particular contains every $F_j$. The inclusion

$$V(F_j,j\in J)\supseteq V(\mathfrak a)$$

is therefore clear. For the reverse inclusion, take $P\in V(F_j,j\in J)$ and $H\in\mathfrak a$. There are polynomials $A_i\in K[X_1,\ldots,X_n]$ and indices $j_1,\ldots,j_k$ such that

$$H=\sum_{i=1}^k A_iF_{j_i}.$$

Then

$$H(P)=\sum_{i=1}^k A_i(P)F_{j_i}(P)=0.$$

Thus every element of the ideal vanishes at $P$, so $P\in V(\mathfrak a)$. $\square$

Henceforth, then, we may assume that every zero set is given by an ideal.

Lemma: inclusion of ideals reverses inclusion of zero loci

For ideals $\mathfrak a\subseteq\mathfrak b$ in $K[X_1,\ldots,X_n]$, the corresponding zero loci satisfy

$$V(\mathfrak a)\supseteq V(\mathfrak b).$$

Proof

Take $P\in V(\mathfrak b)$. This means that $F(P)=0$ for every $F\in\mathfrak b$. Since $\mathfrak a\subseteq\mathfrak b$, it follows in particular that $F(P)=0$ for every $F\in\mathfrak a$. Thus $P\in V(\mathfrak a)$. $\square$

Affine algebraic subsets of affine space have the following important structural properties.

Proposition: unions and intersections of affine algebraic sets

Let $K$ be a field, $K[X_1,\ldots,X_n]$ the polynomial ring in $n$ variables, and $\mathbb A_K^n$ the corresponding affine space. The following properties hold.

1. $V(0)=\mathbb A_K^n$: the whole affine space is an affine algebraic set.

2. $V(1)=\varnothing$: the empty set is an affine algebraic set.

3. If $V_1,\ldots,V_k$ are affine algebraic sets with $V_i=V(\mathfrak a_i)$, then

   $$V_1\cup V_2\cup\cdots\cup V_k =V(\mathfrak a_1\mathfrak a_2\cdots\mathfrak a_k).$$

   In particular, a finite union of affine algebraic sets is again an affine algebraic set.

4. If $V_i$, $i\in I$, are affine algebraic sets with $V_i=V(\mathfrak a_i)$, then

   $$\bigcap_{i\in I}V_i=V\left(\sum_{i\in I}\mathfrak a_i\right).$$

   In particular, an arbitrary intersection of affine algebraic sets is again an affine algebraic set.

Proof

Statements (1) and (2) are clear: the constant polynomial $0$ vanishes everywhere, whereas the constant polynomial $1$ vanishes nowhere.

For (3), take a point in the union, say $P\in V(\mathfrak a_1)$. Then $f(P)=0$ for every $f\in\mathfrak a_1$. Every element of the product ideal $\mathfrak a_1\cdots\mathfrak a_k$ has the form

$$h=\sum_{j=1}^m r_j f_{1j}f_{2j}\cdots f_{kj},$$

with $f_{ij}\in\mathfrak a_i$. Since $f_{1j}(P)=0$ in every term, we get $h(P)=0$. Thus $P$ belongs to the zero locus on the right.

Conversely, suppose that $P$ does not belong to the union on the left. Then $P\notin V(\mathfrak a_i)$ for every $i=1,\ldots,k$. For each $i$ there is an $f_i\in\mathfrak a_i$ with $f_i(P)\ne0$. Since $K$ is a field,

$$(f_1f_2\cdots f_k)(P)\ne0,$$

while $f_1f_2\cdots f_k\in\mathfrak a_1\cdots\mathfrak a_k$. Therefore $P$ cannot belong to the zero locus on the right.

For (4), take $P\in\mathbb A_K^n$. The point $P$ belongs to $V(\mathfrak a_i)$ for every $i\in I$ precisely when $f(P)=0$ for every $f\in\mathfrak a_i$ and every $i\in I$. This holds precisely when $f(P)=0$ for every $f$ in the sum of these ideals. $\square$

Source navigation: course · Lecture 1 · Lecture 3 (source) · Worksheet 2

Worksheet 2

Practice exercises

Exercise 2.1

Consider the two planes in $\mathbb A_K^3$,

$$E_1=V(3x+4y+5z) \qquad\text{and}\qquad E_2=V(2x-y+3z).$$

Find a parametrisation of the intersection $E_1\cap E_2$.

Exercise 2.2 ★

Find the intersection points of the unit circle and the line through the points $(-1,1)$ and $(4,-2)$.

Exercise 2.3

Find the coordinates of the two intersection points of the line $G$ and the circle $K$, where $G$ is given by the equation

$$3y-4x+2=0,$$

and $K$ has centre $(2,5)$ and radius $7$.

Exercise 2.4

Calculate the intersection points of the two curves

$$C=V(x^2+2y^2+3xy+x-2) \qquad\text{and}\qquad L=V(4x+3y-5).$$

Exercise 2.5

Prove that the intersection of two distinct circles in the affine plane is the intersection of one of the circles with a line.

Exercise 2.6 ★

Find the intersection points of the unit circle $E$ and the circle $K$ with centre $(1,0)$ and radius $2$.

Exercise 2.7 ★

Find the intersection points of the two ellipses

$$\{(x,y)\in\mathbb R^2\mid x^2+xy+3y^2=3\}$$

and

$$\{(x,y)\in\mathbb R^2\mid 2x^2-xy+y^2=4\}.$$

Exercise 2.8 ★

Find the intersection points of the unit circle and the standard parabola.

Exercise 2.9 ★

Let

$$P=\{(x,y)\in\mathbb R^2\mid y=x^2\}$$

be the standard parabola, and let $K$ be the circle with centre $(0,1)$ and radius $1$.

1. Sketch $P$ and $K$.
2. Write down an equation for $K$.
3. Determine the intersection $P\cap K$.
4. Describe the lower semicircle as the graph of a function from $[-1,1]$ to $\mathbb R$.
5. Describe the position of the parabola relative to the lower semicircle.

Exercise 2.10

Find all simultaneous solutions of the two equations

$$x^3+y^2=2 \qquad\text{and}\qquad 2xy=3$$

over the fields $K=\mathbb Z/(3)$, $\mathbb Z/(5)$, and $\mathbb Z/(7)$.

Exercise 2.11 ★

Consider the variety of commuting $2\times2$ matrices, that is, the set of pairs of matrices

$$V=\{(A,B)\mid A,B\in\operatorname{Mat}_2(K),\ AB=BA\} \subseteq\operatorname{Mat}_2(K)\times\operatorname{Mat}_2(K) \cong\mathbb A_K^8.$$

1. Prove that $V$ is an affine variety, and find equations describing it that are as simple as possible.

2. Prove that the map

   $$\pi:V\longrightarrow\operatorname{Mat}_2(K), \qquad(A,B)\longmapsto A,$$

   is surjective.

3. Determine the inverse image of

   $$\begin{pmatrix}1&1\\0&0\end{pmatrix}$$

   under $\pi$.

Exercise 2.12

Prove that, for a point

$$P=(a_1,\ldots,a_n)\in\mathbb A_K^n,$$

the corresponding ideal

$$(X_1-a_1,X_2-a_2,\ldots,X_n-a_n)$$

is maximal.

Exercise 2.13

Let $K$ be an algebraically closed field and $n\ge2$. Prove that a point $P\in\mathbb A_K^n$ cannot be the zero set of a single polynomial.

Exercise 2.14

Let $K$ be a finite field, and let

$$M=\{P_1,P_2,\ldots,P_m\}\subseteq\mathbb A_K^n$$

be a finite set of points. Prove that $M$ is the zero set of a single polynomial.

Exercise 2.15 ★

Let $K$ be a field. Prove that the following three ideals in $K[X,Y,Z,W]$ are equal:

$$\mathfrak a=(Z^2+W^2-1,\ X-2Z^2+1,\ Y-2ZW),$$

$$\mathfrak b=(Z^2+W^2-1,\ X^2+Y^2-1,\ (1-X)Z-YW,\ YZ-(1+X)W),$$