Elementary Algebraic Geometry: Exercise 23, Sect 15.1 Dummit & Foote

[Math Amateur]

I am trying to gain an understanding of the basics of elementary algebraic geometry and am reading Dummit and Foote Chapter 15: Commutative Rings and Algebraic Geometry ...

At present I am focused on Section 15.1 Noetherian Rings and Affine Algebraic Sets ... ...

I need help to get started on Exercise 23 of Section 15.1 ...Exercise 23 of Section 15.1 reads as follows:View attachment 4762***NOTE***

I do not really fully understand the nature of a k-algebra or an R-algebra ... so any help in making the notion of an algebra clearer will help ... as well as a significant start on the exercise ...
Hope someone can help ...

Peter

 

[mathbalarka]

Let's talk a bit about algebras first : Given a commutative unital ring $R$, an $R$-algebra $(A, f)$ is simply a ring $A$ with a ring homomorphism $f : R \to A$. Note that you have a natural $R$-module structure on $A$ defined by $ra = f(r)a$. Sometimes we just say, by abuse of notation, that $A$ is an $R$-algebra, forgetting about the equipped homomorphism.

For example, a $k$-algebra, for a field $k$, is a ring $A$ with a ring homomorphism $k \to A$. As fields don't have nontrivial ideals, the homomorphism must be an injection. Thus, equivalently, a $k$-algebra is a ring $A$ containing a subring isomorphic to $k$.

Note that for any ring $A$, there is natural ring homomorphism $\Bbb Z \to A$ given by sending $1 \in \Bbb Z$ to the identity $1_A \in A$ and then linearly extending. This makes any ring into a $\Bbb Z$-algebra.

If $(A, f)$ and $(B, g)$ are $R$-algebras, then an $R$-algebra homomorphism $A \to B$ is a ring homomorphism $h : A \to B$ such that $h \circ f = g$. It's called an $R$-algebra isomorphism if $h$ is an isomorphism.

Now note that coordinate ring of affine varieties are naturally $k$-algebras, since coordinate rings are of the form $k[x_1, \cdots, x_n]/\mathcal{I}$, and the embedding $k \hookrightarrow k[x_1, \cdots, x_n]/\mathcal{I}$ can be defined by $x \mapsto x \pmod{\mathcal{I}}$. In particular, the $k$-algebras arising as coordinate ring of affine varities are called affine $k$-algebras, and it can be proved that these are precisely the finitely generated reduced $k$-algebras. In fact, if $V \to W$ is a regular map between varieties over $k$, then you can show without much difficulty that you have a $k$-algebra homomorphism $k[W] \to k[V]$. This gives you a contravariant correspondence between affine varieties over $k$ and affine $k$-algebras. In fact, this is an equivalence of categories, which is actually a corollary of Nullstellensatz.

---

OK, so now let's get to that exercise. $V$ here is the variety over $\Bbb R$ sketched by the equation $aX + bY + c = 0$. The coordinate ring $\Bbb R[V]$ of this is $\Bbb R[X, Y]/(aX + bY + c)$. Can you now show that this is $\Bbb R$-algebra isomorphic to $\Bbb R[T]$? Start by constructing a ring isomorphism $\Bbb R[X, Y]/(aX + bY + c) \to \Bbb R[T]$ such that the embedded copy of $\Bbb R$ is preserved under this isomorphism.

If you can do this one, you should be able to write down the corresponding regular map $V \to \Bbb R[T]$ which is an isomorphism.

 

[Deveno]

One can proceed by cases:

Suppose $b = 0$, so that we have the line $V = ax + c$. Note that to actually have a line in the first place, this entails $a \neq 0$.

Consider the homomorphism $\phi: \Bbb R[x] \to \Bbb R$ given by $\phi(f(x)) = f\left(\dfrac{-c}{a}\right)$.

Show this induces a homomorphism $\psi:\Bbb R[x,y] \to \Bbb R[y]$, by considering $\Bbb R[x,y]$ as $(\Bbb R[x])[y]$, and writing:

$\psi(g(x,y)) = \psi(f_0(x) + f_1(x)y + \cdots + f_n(x)y^n) = \phi(f_0(x)) + \phi(f_1(x))y + \cdots + \phi(f_n(x))y^n$.

Use the fact that $\phi$ is onto $\Bbb R$ (why?) to conclude $\psi$ is onto $\Bbb R[y]$.

Now show the kernel of $\psi$ consists precisely of those elements of $\Bbb R[x,y]$ that, when viewed as elements of $(\Bbb R[x])[y]$ (strictly speaking, we are employing a "unnamed" isomorphism between $\Bbb R[x,y]$ and $(\Bbb R[x])[y]$), have coefficients that lie in the kernel of $\phi$, which is to say they all have a common factor of $\left(x + \dfrac{c}{a}\right)$.

Conclude that the kernel of $\psi$ is indeed the ideal generated by $ax + c$.

Case 2 is somewhat easier (and shows why we treated case 1 separately): Suppose $b \neq 0$, and consider the homomorphism:

$\phi: \Bbb R[x,y] \to \Bbb R[x]$ given by $\phi(g(x,y)) = g\left(x,\dfrac{-a}{b}x - \dfrac{c}{b}\right)$

Show that $g \in \text{ker }\phi \iff (ax + by + c)|g$, and that $g$ maps constant polynomials to their constant terms (that is, acts as the identity on $\Bbb R$).

Note: although it "looks" different, we are actually using the same trick of considering $\Bbb R[x,y]$ as $(\Bbb R[x])[y]$, and "evaluating" $g(x,y)$ at the ring element $f(x) = \dfrac{-a}{b}x - \dfrac{c}{b}$ in $\Bbb R[x]$.

Some random thoughts: what is a line, really? Let's be a *bit* more specific, and consider a line embedded in a Euclidean plane. Normally, we think of it as points satisfying an equation:

$L = \{(x,y) \in \Bbb R^2: Ax + By + C = 0\}$.

But this equation depends on our "coordinate system", we have to "locate" the $x$ and $y$ axes to come up with the numbers $A,B,C$. That works fine for a *vector space*, where we have a distinguished origin. But we actually think of the plane as being "undifferentiated", any part of it looks like any other. In an affine space, there isn't any special point of origin. So what remains the same? The relationship between the points is "linear", that is, the locus of the line represents the zero-set of a linear polynomial in $\Bbb R[x,y]$. In other words we have moved away from any *specific* $A,B,C$, and have characterized a line as being determined by the DEGREE of a polynomial in two variables. This characterization is now independent of a coordinate-system, as it should be, in our understanding of just "an abstract line" (which ought to be a line no matter what coordinates we are using).

I urge you to consider what "affine varieties" we obtain by considering the zero-sets of degree TWO polynomials in $\Bbb R[x,y]$ (hint: you already know these-the parabola is one example).

Degree three and higher (what are commonly called "algebraic curves") polynomial zero-sets for $\Bbb R[x,y]$ in the plane start getting a tad complicated. Go figure.

Mathbalarka: if I've made any silly mistakes, please point them out-my mind seems to be slipping, lately :).

 

[Math Amateur]

One can proceed by cases:

Suppose $b = 0$, so that we have the line $V = ax + c$. Note that to actually have a line in the first place, this entails $a \neq 0$.

Consider the homomorphism $\phi: \Bbb R[x] \to \Bbb R$ given by $\phi(f(x)) = f\left(\dfrac{-c}{a}\right)$.

Show this induces a homomorphism $\psi:\Bbb R[x,y] \to \Bbb R[y]$, by considering $\Bbb R[x,y]$ as $(\Bbb R[x])[y]$, and writing:

$\psi(g(x,y)) = \psi(f_0(x) + f_1(x)y + \cdots + f_n(x)y^n) = \phi(f_0(x)) + \phi(f_1(x))y + \cdots + \phi(f_n(x))y^n$.

Use the fact that $\phi$ is onto $\Bbb R$ (why?) to conclude $\psi$ is onto $\Bbb R[y]$.

Now show the kernel of $\psi$ consists precisely of those elements of $\Bbb R[x,y]$ that, when viewed as elements of $(\Bbb R[x])[y]$ (strictly speaking, we are employing a "unnamed" isomorphism between $\Bbb R[x,y]$ and $(\Bbb R[x])[y]$), have coefficients that lie in the kernel of $\phi$, which is to say they all have a common factor of $\left(x + \dfrac{c}{a}\right)$.

Conclude that the kernel of $\psi$ is indeed the ideal generated by $ax + c$.

Case 2 is somewhat easier (and shows why we treated case 1 separately): Suppose $b \neq 0$, and consider the homomorphism:

$\phi: \Bbb R[x,y] \to \Bbb R[x]$ given by $\phi(g(x,y)) = g\left(x,\dfrac{-a}{b}x - \dfrac{c}{b}\right)$

Show that $g \in \text{ker }\phi \iff (ax + by + c)|g$, and that $g$ maps constant polynomials to their constant terms (that is, acts as the identity on $\Bbb R$).

Note: although it "looks" different, we are actually using the same trick of considering $\Bbb R[x,y]$ as $(\Bbb R[x])[y]$, and "evaluating" $g(x,y)$ at the ring element $f(x) = \dfrac{-a}{b}x - \dfrac{c}{b}$ in $\Bbb R[x]$.

Some random thoughts: what is a line, really? Let's be a *bit* more specific, and consider a line embedded in a Euclidean plane. Normally, we think of it as points satisfying an equation:

$L = \{(x,y) \in \Bbb R^2: Ax + By + C = 0\}$.

But this equation depends on our "coordinate system", we have to "locate" the $x$ and $y$ axes to come up with the numbers $A,B,C$. That works fine for a *vector space*, where we have a distinguished origin. But we actually think of the plane as being "undifferentiated", any part of it looks like any other. In an affine space, there isn't any special point of origin. So what remains the same? The relationship between the points is "linear", that is, the locus of the line represents the zero-set of a linear polynomial in $\Bbb R[x,y]$. In other words we have moved away from any *specific* $A,B,C$, and have characterized a line as being determined by the DEGREE of a polynomial in two variables. This characterization is now independent of a coordinate-system, as it should be, in our understanding of just "an abstract line" (which ought to be a line no matter what coordinates we are using).

I urge you to consider what "affine varieties" we obtain by considering the zero-sets of degree TWO polynomials in $\Bbb R[x,y]$ (hint: you already know these-the parabola is one example).

Degree three and higher (what are commonly called "algebraic curves") polynomial zero-sets for $\Bbb R[x,y]$ in the plane start getting a tad complicated. Go figure.

Mathbalarka: if I've made any silly mistakes, please point them out-my mind seems to be slipping, lately :).

Deveno, Mathbalarka,

Thanks so much for the posts ... just working through them now and reflecting on what you have said ...

The posts are helping a lot with my effort to understand the algebraic basics of elementary algebraic geometry ...

Thanks again,

Peter