“I don’t know why I should have to learn Algebra... I’m never likely to go there.”
―Billy Connolly
Definition8.13.Algebra.
Given a ring \(R\text{,}\) an \(R\)-algebra is a ring \(S\) equipped with a ring homomorphism \(\phi: R \to S\text{.}\) This defines an \(R\)-module structure on \(S\) given by restriction of scalars: for each \(r\in R\) and \(s\in S\text{,}\)\(rs := \phi(r)s\text{.}\) This \(R\)-module structure on \(S\) is compatible with the internal multiplication of \(S\) i.e.,
\begin{equation*}
r(st)=(rs)t=s(rt) \text{ for all } r\in R, s, t\in S.
\end{equation*}
We will call \(\phi\) the structure homomorphism of the \(R\)-algebra \(S\text{.}\)
Example8.14.Polynomial Rings are Algebras.
If \(A\) is a ring and \(x_1,\dots,x_n\) are indeterminates, the inclusion map \(A \hookrightarrow A[x_1,\dots,x_n]\) makes the polynomial ring into an \(A\)-algebra.
Example8.15.Inclusion Maps Give Algebra Structure.
More generally, any inclusion map \(R \subseteq S\) gives \(S\) an \(R\)-algebra structure. In this case the \(R\)-module multiplication coincides with the internal (ring) multiplication on \(S\text{.}\)
Example8.16.Rings Have Unique Structure as \(\Z\)-algebra.
Any ring comes with a unique structure as a \(\Z\)-algebra, since there is a unique ring homomorphism \(\Z \to R\text{:}\) the one given by \(n \mapsto n \cdot 1_R\text{.}\)
Definition8.17.Generated Algebra.
Let \(S\) be an \(R\)-algebra with structure homomorphism \(\varphi\) and let \(\Lambda \subseteq S\) be a set. The \(R\)-algebra generated by a subset \(\Lambda\) of \(S\text{,}\) denoted \(R[\Lambda]\text{,}\) is the smallest (with respect to containment) subring of \(S\) containing \(\Lambda\) and \(\varphi(R)\text{.}\) A set of elements \(\Lambda \subseteq S\)generates\(S\) as an \(R\)-algebra if \(S=R[\Lambda]\text{.}\)
Note that there are two different meanings for the notation \(R[S]\) for a ring \(R\) and set \(S\text{:}\) one calls for a polynomial ring, and the other calls for a subring of something. If \(S\) is a subset of elements of some other larger ring which is clear from context, then we are talking about the algebra generated by \(S\text{;}\) in contrast, if \(S\) is a set of indeterminates, then we are talking about a polynomial ring in those variables.
This can be unpackaged more concretely in a number of equivalent ways.
Lemma8.18.Generated Algebra Equivalencies.
The following are equivalent:
\(\Lambda\) generates \(S\) as an \(R\)-algebra.
Every element in \(S\) admits a polynomial expression in \(\Lambda\) with coefficients in \(\phi(R)\text{,}\) i.e.
Let \(S= \left\{\sum_{\mathrm{finite}} \phi(a) \lambda_1^{i_1} \cdots \lambda_n^{i_n} \mid a\in A, \lambda_j\in\Lambda, i_j\in \N\right\}\text{.}\)
For the equivalence between (2) and (3), we note that \(S\) is the image of \(\pi\text{.}\) In particular, \(S\) is a subring of \(R\text{.}\) It then follows from the definition that (1) implies (2). Conversely, any subring of \(R\) containing \(\phi(A)\) and \(\Lambda\) certainly must contain \(S\text{,}\) so (2) implies (1).
Remark8.19.
Let \(S\) be an \(R\)-algebra generated by \(\Lambda\text{,}\) let \(\pi\) be the surjective map in part (3) of Lemma 8.18, and let \(I := \ker \pi\text{.}\) By the [cross-reference to target(s) "thm-fit" missing or not unique], we have a ring isomorphism \(S\cong R[X]/I\text{.}\) The elements of \(I\) are the relations among the generators in \(\Lambda\text{.}\) If we understand the ring \(R\) and generators and relations for \(S\) over \(R\text{,}\) we can get a pretty concrete understanding of \(S\text{.}\)
Definition8.20.Free Algebra.
Suppose \(\Lambda\) generates \(S\) as an \(R\)-algebra. Let \(R[X]\) be a polynomial ring on a set of indeterminates \(X\) in bijection with \(\Lambda\text{,}\) and let \(\pi\) denote the \(R\)-algebra homomorphism
If the homomorphism \(\pi\) is injective (and thus an isomorphism) we say that \(S\) is a free algebra; a free algebra on \(R\) is isomorphic to a polynomial ring on \(R\text{.}\)
The ideal \(I = \ker(\pi)\) measures how far \(R\) is from being a free \(R\)-algebra and is called the set of relations on \(\Lambda\text{.}\)
Example8.21.\(\Z[\sqrt{d}]\) and \(\Z[\sqrt[3]{d}]\) are \(\Z\)-algebras.
You may have seen this used in \(\Z[\sqrt{d}]\) for some \(d \in \Z\) to describe the ring
\begin{equation*}
\{ a + b \sqrt{d} \mid a,b \in \Z\}.
\end{equation*}
The \(\Z\)-algebra generated by \(\sqrt{d}\) in the most natural place, the algebraic closure of \(\Q\text{,}\) is exactly the set above.
Solution.
The point is that for any power \((\sqrt{d})^n\text{,}\) we can always write \(n=2q+r\) with \(r\in \{0,1\}\text{,}\) so \((\sqrt{d})^n=d^q (\sqrt{d})^r\) is in the algebra generated by \(\mathbb{Z}\) and \(\sqrt{d}\text{.}\)
We can also write the one-generated \(\Z\)-algebra \(\Z[\sqrt{d}]\) as a quotient of a polynomial ring in one variable: if \(d\) is not a perfect square, the map \(\pi\) in part (3) of this Lemma 8.18 is
Unfortunately, Macaulay2 does not understand subalgebras directly, only quotient rings. But as we have discussed, any \(R\)-algebra can be thought of as a quotient of a polynomial ring over \(R\text{.}\) For example, the Veronese algebra \(V = \mathbb{Q}[x^2, xy, xz, y^2, yz, z^2]\) is a quotient of a polynomial ring over \(\mathbb{Q}\) in \(6\) variables, since it has \(6\) algebra generators. More precisely, \(V\) is the image of the map
so by the [cross-reference to target(s) "thm-fit" missing or not unique], \(V \cong \mathbb{Q}[w_1, \ldots, w_6]/ \ker \pi\text{.}\)
i4 : use R;
i5 : aux = QQ[w_1 .. w_6]
o5 = aux
o5 : PolynomialRing
i6 : p = map(R,aux,{x^2,x*y,x*z,y^2,y*z,z^2})
2 2 2
o6 = map (R, aux, {x , x*y, x*z, y , y*z, z })
o6 : RingMap R <--- aux
i7 : V = aux/ker p
o7 = V
o7 : QuotientRing
To do calculations with \(V\text{,}\) note that \(w_1\) is actually \(x_2\text{,}\)\(w_2\) is \(xy\text{,}\) and so on.
Definition8.22.Algebra-Finite.
We say that \(\varphi:R\to S\) is algebra-finite, or \(S\) is a finitely generated\(R\)-algebra, or \(S\) is of finite type over \(R\text{,}\) if there exists a finite set of elements \(f_1,\dots,f_t\in S\) that generates \(S\) as an \(R\)-algebra.
Remark8.23.
A better name might be finitely generatable, since we do not need to know an actual finite set of generators to say that an algebra is finitely generated.
Example8.24.Algebra Finite iff Quotient of Polynomial Ring.
From the discussion above, we conclude that \(S\) is a finitely generated \(R\)-algebra if and only if \(S\) is a quotient of some polynomial ring \(R[x_1,\dots,x_d]\) over \(R\) in finitely many variables.
If \(S\) is generated over \(R\) by \(f_1,\dots,f_d\text{,}\) we will use the notation \(R[f_1,\dots,f_d]\) to denote \(S\text{.}\) Of course, for this notation to properly specify a ring, we need to understand how these generators behave under the operations; this is no problem if \(R\) and \(\underline{f}\) are understood to be contained in some larger ring.
Conjecture8.25.Jacobian Determinant and \(\C\) Generation.
There are many basic questions about algebra generators that are surprisingly difficult. Let \(R=\C[x_1,\dots,x_n]\) and \(f_1,\dots,f_n\in R\text{.}\) When do \(f_1,\dots,f_n\) generate \(R\) over \(\C\text{?}\) It is not too hard to show that the Jacobian determinant
