01.02.17 · foundations / groups

Hilbert basis theorem; Noetherian rings and modules

shipped3 tiersLean: none

Anchor (Master): Hilbert 1890 Math. Ann. 36; Noether 1921 Math. Ann. 83; Atiyah-Macdonald Ch. 6--11

Intuition [Beginner]

Imagine a building under construction. Each floor represents an ideal — a special subset of a ring. You keep adding floors, one on top of the next. A Noetherian ring is a building that cannot keep growing forever: eventually the floors stop getting taller, and every new floor is identical to the one below it. The chain of ideals stabilises.

Why does this matter? In a Noetherian ring, every ideal can be described using finitely many generators. You never need an infinite list to pin down an ideal. This finite-generated property transfers to polynomial rings: if your base ring is Noetherian, then the ring of polynomials over it is also Noetherian. This is the Hilbert basis theorem, and it guarantees that any system of polynomial equations can be described by finitely many equations.

Why does this concept exist? It provides the finiteness condition that makes algebraic geometry work: every algebraic variety is cut out by finitely many equations.

Visual [Beginner]

A vertical stack of nested rectangles, each one contained inside the next, with an arrow showing that the containment eventually stops. The bottom rectangle is labelled $I_{1}$ , the next $I_{2}$ , and so on, with the final stabilised rectangle shaded to indicate that all subsequent ideals equal it.

The picture captures the ascending chain condition: after finitely many steps, the chain of nested sets stops growing.

Worked example [Beginner]

Consider the ring of integers $Z$ . An ideal in $Z$ is a set of the form $n Z$ = all multiples of $n$ , for some non-negative integer $n$ . For example, $6 Z = {\dots, - 12, - 6, 0, 6, 12, \dots}$ .

Step 1. Start an ascending chain: $6 Z \subset 3 Z \subset Z$ . The containment holds because every multiple of $6$ is also a multiple of $3$ , and every multiple of $3$ is an integer.

Step 2. Try to extend the chain. Since $Z$ is the entire ring, the next ideal must also be $Z$ . The chain has stabilised.

Step 3. More generally, any ascending chain in $Z$ must stabilise because the positive integers satisfy the well-ordering property: the generators $n$ form a decreasing sequence of positive integers, which cannot decrease indefinitely.

What this tells us: $Z$ is a Noetherian ring, and every ideal $n Z$ is generated by the single element $n$ .

Check your understanding [Beginner]

Formal definition [Intermediate+]

Throughout, $R$ is a commutative ring with identity. An ideal $I$ of $R$ is a subset satisfying: (i) $I$ is an additive subgroup, (ii) for every $r \in R$ and $a \in I$ , the product $r a \in I$ ^{[Dummit-Foote 2004]}.

Definition (Noetherian ring). A commutative ring $R$ is Noetherian if it satisfies any of the following equivalent conditions ^{[Atiyah-Macdonald 1969]}:

(ACC) Every ascending chain of ideals $I_{1} \subset I_{2} \subset I_{3} \subset \dots$ stabilises: there exists $N$ such that $I_{n} = I_{N}$ for all $n \geq N$ .

(FG) Every ideal of $R$ is finitely generated.

(MAX) Every nonempty set of ideals of $R$ has a maximal element (with respect to inclusion).

A module $M$ over $R$ is Noetherian if every ascending chain of $R$ -submodules of $M$ stabilises, equivalently if every submodule of $M$ is finitely generated.

Counterexamples to common slips [Intermediate+]

A subring of a Noetherian ring need not be Noetherian. The polynomial ring $k [x_{1}, x_{2}, \dots]$ in infinitely many variables over a field $k$ is not Noetherian (the chain $(x_{1}) \subset (x_{1}, x_{2}) \subset (x_{1}, x_{2}, x_{3}) \subset \dots$ never stabilises), but it is a subring of its field of fractions, which is Noetherian as a field.
Noetherian as a ring and Noetherian as a module are distinct. A Noetherian ring $R$ is always Noetherian as an $R$ -module. The converse is also true for the regular module $R$ . However, a ring can have Noetherian modules without being Noetherian itself.
Finitely generated does not mean generated by one element. The ideal $(x, y)$ in $k [x, y]$ requires two generators and cannot be generated by a single polynomial. The ring $k [x, y]$ is Noetherian by Hilbert's basis theorem, so every ideal is finitely generated, but the number of generators may exceed one.

Key theorem with proof [Intermediate+]

Theorem (Hilbert basis theorem). If $R$ is a Noetherian ring, then the polynomial ring $R [x]$ is Noetherian.

Proof. Let $I$ be an ideal of $R [x]$ . For each $n \geq 0$ , define $$ L_n(I) = {a \in R : a \text{ is the leading coefficient of some } f \in I \text{ of degree } n} \cup {0}. $$ Each $L_{n} (I)$ is an ideal of $R$ : if $a, b \in L_{n} (I)$ are leading coefficients of $f, g \in I$ with $de g f = de g g = n$ , then for appropriate choice of $λ \in R$ the polynomial $f + λ g$ has leading coefficient $a + b$ or lies in a lower degree, and in either case $a + b \in L_{n} (I)$ . Closure under multiplication by $R$ is immediate from multiplying polynomials by constants.

The ideals $L_{n} (I)$ form an ascending chain $L_{0} (I) \subset L_{1} (I) \subset L_{2} (I) \subset \dots$ because multiplying an element of $I$ of degree $n$ by $x$ produces an element of $I$ of degree $n + 1$ with the same leading coefficient.

Since $R$ is Noetherian, the chain stabilises: there exists $m$ such that $L_{n} (I) = L_{m} (I)$ for all $n \geq m$ . Each $L_{k} (I)$ for $0 \leq k \leq m$ is finitely generated as an ideal of $R$ ; choose generators $a_{k, 1}, \dots, a_{k, r_{k}}$ with $a_{k, j} = lc (f_{k, j})$ for some $f_{k, j} \in I$ of degree $k$ .

Set $S = {f_{k, j} : 0 \leq k \leq m, 1 \leq j \leq r_{k}}$ . This is a finite set of polynomials in $I$ . Let $J = (S)$ be the ideal generated by $S$ in $R [x]$ . We claim $J = I$ .

The inclusion $J \subset I$ is immediate since each $f_{k, j} \in I$ and $I$ is an ideal. For the reverse inclusion, suppose by contradiction that $I ∖ J \neq = \emptyset$ . Choose $f \in I ∖ J$ of minimal degree $d$ . If $d \leq m$ , then $lc (f) \in L_{d} (I)$ , so $lc (f) = \sum_{j} c_{j} a_{d, j}$ for some $c_{j} \in R$ . The polynomial $g = \sum_{j} c_{j} f_{d, j}$ has the same leading coefficient as $f$ and lies in $J$ , so $f - g \in I ∖ J$ has strictly smaller degree than $f$ , contradicting minimality.

If $d > m$ , then $L_{d} (I) = L_{m} (I)$ , so $lc (f) \in L_{m} (I)$ . Write $lc (f) = \sum_{j} c_{j} a_{m, j}$ . The polynomial $h = \sum_{j} c_{j} x^{d - m} f_{m, j}$ has degree $d$ with the same leading coefficient as $f$ and lies in $J$ . Then $f - h \in I ∖ J$ has degree strictly less than $d$ , again contradicting minimality.

Therefore $I = J$ , and $I$ is finitely generated. Since $I$ was arbitrary, $R [x]$ is Noetherian. $□$

Bridge. The Hilbert basis theorem builds toward 04.02.07 (Nullstellensatz and dimension theory), where finiteness of polynomial ideals translates into finiteness of the equation systems describing algebraic varieties, and this is exactly the mechanism that identifies algebraic sets with finitely generated ideals. The foundational reason is that the Noetherian property transfers from scalars to polynomials, and the central insight is that the degree-filtration argument extracts finite data from an arbitrary ideal by reading off leading coefficients. The bridge is between the ascending chain condition on $R$ and finite generation in $R [x]$ , and the pattern recurs in 01.02.10 (tensor products) through the finite-presentation properties of Noetherian modules.

Exercises [Intermediate+]

Exercise 3 (medium, symbolic).

Prove that the three characterisations of a Noetherian ring — ACC, finite generation, and maximal element — are equivalent.

Hint

Show (ACC) implies (MAX) by contrapositive, (MAX) implies (FG) by selecting a maximal element from the set of finitely generated subideals, and (FG) implies (ACC) by taking the union of a chain.

Answer

(ACC) $\Rightarrow$ (MAX). Suppose $S$ is a nonempty set of ideals with no maximal element. Pick $I_{1} \in S$ ; since $I_{1}$ is not maximal, there exists $I_{2} \in S$ with $I_{1} ⊊ I_{2}$ . Continue inductively to build a strictly ascending chain, contradicting ACC.

(MAX) $\Rightarrow$ (FG). Let $I$ be an ideal. Consider $S = {J \subset I : J is finitely generated}$ , a nonempty set (it contains the zero ideal). By MAX, $S$ has a maximal element $J_{0} = (a_{1}, \dots, a_{n})$ . If $J_{0} \neq = I$ , pick $a \in I ∖ J_{0}$ ; then $J_{0} + (a)$ is a finitely generated element of $S$ strictly containing $J_{0}$ , contradicting maximality. So $J_{0} = I$ .

(FG) $\Rightarrow$ (ACC). Given a chain $I_{1} \subset I_{2} \subset \dots$ , the union $I = ⋃_{n} I_{n}$ is an ideal. By FG, $I = (a_{1}, \dots, a_{n})$ . Each $a_{i} \in I_{k_{i}}$ for some $k_{i}$ . Setting $N = max_{i} k_{i}$ , all generators lie in $I_{N}$ , so $I = I_{N}$ and the chain stabilises at $N$ .

Exercise 5 (medium, symbolic).

Prove that a finitely generated module $M$ over a Noetherian ring $R$ is Noetherian.

Hint

A surjection $R^{n} \to M$ exists for some $n$ . Show that any submodule of $R^{n}$ is finitely generated by induction on $n$ , then use the correspondence theorem.

Answer

Since $M$ is finitely generated, there is a surjective $R$ -linear map $φ : R^{n} \to M$ for some $n$ . If $N$ is a submodule of $M$ , then $φ^{- 1} (N)$ is a submodule of $R^{n}$ , and $φ$ restricts to a surjection $φ^{- 1} (N) \to N$ . If $φ^{- 1} (N)$ is finitely generated, so is $N$ .

It remains to show that every submodule of $R^{n}$ is finitely generated. For $n = 1$ this is the hypothesis that $R$ is Noetherian. For $n > 1$ , consider the exact sequence $0 \to R^{n - 1} \to R^{n} \to R \to 0$ . Given a submodule $S \subset R^{n}$ , the image of $S$ in $R$ is an ideal (hence finitely generated), and the intersection $S \cap R^{n - 1}$ is finitely generated by induction. Choose finitely many generators for each; their union generates $S$ .

Exercise 6 (medium, symbolic).

Prove that if $0 \to M^{'} \to M \to M^{''} \to 0$ is a short exact sequence of $R$ -modules, then $M$ is Noetherian if and only if both $M^{'}$ and $M^{''}$ are Noetherian.

Hint

For the forward direction, ascending chains in $M^{'}$ or $M^{''}$ lift to $M$ . For the reverse, use the correspondence theorem for submodules.

Answer

Suppose $M$ is Noetherian. Any ascending chain of submodules of $M^{'}$ is also an ascending chain of submodules of $M$ , hence stabilises. Any ascending chain in $M^{''}$ pulls back to a chain in $M$ (take preimages under the surjection), which stabilises, pushing forward to stabilisation in $M^{''}$ .

Conversely, suppose $M^{'}$ and $M^{''}$ are Noetherian. Let $N_{1} \subset N_{2} \subset \dots$ be an ascending chain in $M$ . The images in $M^{''}$ form a stabilising chain: $\overline{N_{k}} = N_{k} + M^{'}$ stabilises at some $k_{0}$ . The intersections $N_{k} \cap M^{'}$ also form a stabilising chain in $M^{'}$ , stabilising at some $k_{1}$ . Setting $N = max (k_{0}, k_{1})$ , both the image and kernel stabilise, so the chain in $M$ stabilises.

Exercise 8 (hard, symbolic).

Prove the transfer principle for Noetherian modules: if $M$ is a Noetherian $R$ -module and $φ : R \to S$ is a ring homomorphism making $S$ into a finitely generated $R$ -module (via $φ$ ), then the $S$ -module $M \otimes_{R} S$ is Noetherian as an $S$ -module.

Hint

Show that any ascending chain of $S$ -submodules of $M \otimes_{R} S$ is in particular an ascending chain of $R$ -submodules, and use that $M \otimes_{R} S$ is Noetherian as an $R$ -module because $M$ is.

Answer

Since $S$ is finitely generated as an $R$ -module, say by $s_{1}, \dots, s_{m}$ , the module $M \otimes_{R} S$ is finitely generated as an $R$ -module: the elements $m_{i} \otimes s_{j}$ (where $m_{i}$ range over a finite generating set for $M$ ) generate $M \otimes_{R} S$ over $R$ .

Since $M$ is Noetherian over $R$ , so is $M \otimes_{R} S$ (being a quotient of a finite direct sum of copies of $M$ ). Any ascending chain of $S$ -submodules of $M \otimes_{R} S$ is in particular an ascending chain of $R$ -submodules. Since $M \otimes_{R} S$ is Noetherian as an $R$ -module, this chain stabilises. Hence $M \otimes_{R} S$ is Noetherian as an $S$ -module.

Advanced results [Master]

Theorem 1 (Iterated Hilbert basis). If $R$ is Noetherian, then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every finite $n$ .

Immediate by induction on $n$ using $R [x_{1}, \dots, x_{n}] ≅ (R [x_{1}, \dots, x_{n - 1}]) [x_{n}]$ and the Hilbert basis theorem. The case $n = \infty$ fails: $R [x_{1}, x_{2}, \dots]$ is never Noetherian for $R \neq = 0$ .

Theorem 2 (Noether normalisation lemma). Let $k$ be a field and $A = k [a_{1}, \dots, a_{n}]$ a finitely generated $k$ -algebra. There exist elements $y_{1}, \dots, y_{r} \in A$ , algebraically independent over $k$ , such that $A$ is integral over $k [y_{1}, \dots, y_{r}]$ .

The integer $r$ is the Krull dimension of $A$ . The proof is by induction on $n$ , distinguishing the case where $a_{1}, \dots, a_{n}$ are algebraically independent (then $r = n$ ) from the case where they satisfy an algebraic relation, in which case a change of variables produces a new presentation with fewer generators ^{[Eisenbud 1995]}.

Theorem 3 (Zariski's lemma). If $K$ is a field that is finitely generated as an algebra over a field $k$ , then $K$ is a finite extension of $k$ .

This follows from Noether normalisation: if $K = k [a_{1}, \dots, a_{n}]$ and $r = 0$ , then $K$ is integral over $k$ , hence a finite extension.

Theorem 4 (Hilbert's Nullstellensatz — weak form). If $k$ is algebraically closed and $m$ is a maximal ideal of $k [x_{1}, \dots, x_{n}]$ , then $m = (x_{1} - a_{1}, \dots, x_{n} - a_{n})$ for some $(a_{1}, \dots, a_{n}) \in k^{n}$ .

The field $k [x_{1}, \dots, x_{n}] / m$ is finitely generated as a $k$ -algebra, hence a finite extension of $k$ by Zariski's lemma. Since $k$ is algebraically closed, the quotient equals $k$ , and the evaluation homomorphism at $(a_{1}, \dots, a_{n})$ gives the desired identification.

Theorem 5 (Grothendieck's generic freeness). Let $R$ be a Noetherian domain, $S$ a finitely generated $R$ -algebra, and $M$ a finitely generated $S$ -module. There exists a nonzero $f \in R$ such that the localisation $M_{f}$ is a free $R_{f}$ -module.

This is the foundational result guaranteeing that finitely generated modules over Noetherian algebras become free after localisation at a suitable element. The proof proceeds by Noetherian induction on $Spec R$ ^{[Eisenbud 1995]}.

Theorem 6 (Exactness properties). In the category of finitely generated modules over a Noetherian ring, kernels, images, and cokernels of $R$ -linear maps are again finitely generated. Short exact sequences $0 \to M^{'} \to M \to M^{''} \to 0$ satisfy: $M$ is Noetherian if and only if both $M^{'}$ and $M^{''}$ are.

Theorem 7 (Eakin-Nagata theorem). If $S$ is a commutative ring that is finitely generated as a module over a subring $R$ , then $S$ is Noetherian if and only if $R$ is Noetherian.

The forward direction is due to Eakin (1968) and Nagata; the reverse follows from the fact that finitely generated modules over Noetherian rings are Noetherian.

Synthesis. The Hilbert basis theorem is the foundational reason that finiteness in commutative algebra transfers from coefficients to polynomials, and the central insight is that the degree filtration on an ideal extracts finite data from an infinite object. Putting these together with the Noether normalisation lemma, which provides a polynomial backbone for any finitely generated algebra, the bridge is between algebraic finiteness conditions and geometric dimension: the number of algebraically independent variables in the normalisation is the Krull dimension. This is exactly the structure that identifies the Nullstellensatz's point-correspondence with the maximal spectrum of $k [x_{1}, \dots, x_{n}]$ 04.02.07, and the pattern generalises to coherent sheaves 04.06.02, where Noetherianity of the structure sheaf guarantees that sheaf cohomology is finite-dimensional. Generic freeness closes the loop: after inverting a suitable element, every finitely generated module becomes free, which is the algebraic origin of locally free sheaves in algebraic geometry.

Full proof set [Master]

Proposition 1 (Noether normalisation — proof sketch). Let $A = k [a_{1}, \dots, a_{n}]$ be a finitely generated $k$ -algebra. There exist $y_{1}, \dots, y_{r} \in A$ , algebraically independent over $k$ , such that $A$ is integral over $B = k [y_{1}, \dots, y_{r}]$ .

Proof. If $a_{1}, \dots, a_{n}$ are algebraically independent, set $r = n$ and $y_{i} = a_{i}$ . Otherwise, there exists a nonzero polynomial $F \in k [x_{1}, \dots, x_{n}]$ with $F (a_{1}, \dots, a_{n}) = 0$ . Write $F = \sum_{α} c_{α} x^{α}$ where $α = (α_{1}, \dots, α_{n})$ is a multi-index.

Choose positive integers $d_{1} ≪ d_{2} ≪ \dots ≪ d_{n}$ (large gaps) and set $y_{i} = a_{i} - a_{n}^{d_{i}}$ for $1 \leq i \leq n - 1$ . Substituting $a_{i} = y_{i} + a_{n}^{d_{i}}$ into $F (a_{1}, \dots, a_{n}) = 0$ and examining the leading term in $a_{n}$ (which is unique for suitable choice of $d_{i}$ ), one obtains a monic polynomial equation for $a_{n}$ over $k [y_{1}, \dots, y_{n - 1}]$ . Hence $a_{n}$ is integral over $k [y_{1}, \dots, y_{n - 1}]$ , and therefore $A = k [a_{1}, \dots, a_{n}]$ is integral over $k [y_{1}, \dots, y_{n - 1}, a_{n}] = k [y_{1}, \dots, y_{n - 1}] [a_{n}]$ .

Since $k [y_{1}, \dots, y_{n - 1}] [a_{n}]$ is generated over $k [y_{1}, \dots, y_{n - 1}]$ by the integral element $a_{n}$ , it is finitely generated as a $k [y_{1}, \dots, y_{n - 1}]$ -module. By induction on $n$ , the result follows. $□$

Proposition 2 (Exactness for Noetherian modules). Let $0 \to M^{'} ι M π M^{''} \to 0$ be a short exact sequence of $R$ -modules. Then $M$ is Noetherian if and only if both $M^{'}$ and $M^{''}$ are Noetherian.

Proof. Suppose $M$ is Noetherian. Any ascending chain $N_{1}^{'} \subset N_{2}^{'} \subset \dots$ of submodules of $M^{'}$ satisfies $ι (N_{1}^{'}) \subset ι (N_{2}^{'}) \subset \dots$ in $M$ , which stabilises; hence $M^{'}$ is Noetherian. For $M^{''}$ , any chain $N_{1}^{''} \subset N_{2}^{''} \subset \dots$ in $M^{''}$ pulls back to $π^{- 1} (N_{1}^{''}) \subset π^{- 1} (N_{2}^{''}) \subset \dots$ in $M$ , which stabilises, pushing forward to stabilisation in $M^{''}$ .

Conversely, suppose $M^{'}$ and $M^{''}$ are Noetherian. Let $L_{1} \subset L_{2} \subset \dots$ be a chain in $M$ . The chain $π (L_{1}) \subset π (L_{2}) \subset \dots$ in $M^{''}$ stabilises at some index $k_{0}$ : $π (L_{k}) = π (L_{k_{0}})$ for all $k \geq k_{0}$ . The chain $L_{1} \cap M^{'} \subset L_{2} \cap M^{'} \subset \dots$ in $M^{'}$ stabilises at $k_{1}$ : $L_{k} \cap M^{'} = L_{k_{1}} \cap M^{'}$ for $k \geq k_{1}$ .

Set $N = max (k_{0}, k_{1})$ . For $k \geq N$ and $x \in L_{k}$ , there exists $y \in L_{N}$ with $π (x) = π (y)$ , so $x - y \in ker π = M^{'}$ . Since $x - y \in L_{k} \cap M^{'} = L_{N} \cap M^{'}$ , one has $x - y \in L_{N}$ . Then $x = (x - y) + y \in L_{N}$ . Hence $L_{k} = L_{N}$ for all $k \geq N$ . $□$

Connections [Master]

Nullstellensatz and dimension theory 04.02.07. The Hilbert basis theorem provides the finite-generation guarantee that makes the Nullstellensatz work: every ideal of $k [x_{1}, \dots, x_{n}]$ is finitely generated, so every algebraic set is cut out by finitely many polynomial equations. The Noether normalisation lemma then gives the bridge to dimension theory by exhibiting the Krull dimension as the number of algebraically independent variables in the normalisation.
Coherent sheaf 04.06.02. A Noetherian scheme is one whose structure sheaf has Noetherian stalks, and coherence of a sheaf is the geometric incarnation of finite generation. The Hilbert basis theorem transfers Noetherianity from affine patches to global sheaf cohomology, guaranteeing finite-dimensionality of cohomology groups on projective schemes. The Eakin-Nagata theorem governs when a subring of a Noetherian ring inherits the property.
Tensor product of modules 01.02.10. The transfer principle for Noetherian modules uses the tensor product to extend the Noetherian property across a base change: if $M$ is Noetherian over $R$ and $S$ is a finitely generated $R$ -algebra, then $M \otimes_{R} S$ is Noetherian over $S$ . Flatness, introduced in 01.02.10, interacts with Noetherianity through the generic freeness theorem: over a Noetherian domain, finitely generated modules become free after localisation.

Historical & philosophical context [Master]

Hilbert 1890 proved the basis theorem in Ueber die Theorie der algebraischen Formen ^{[Hilbert 1890]}, establishing that every ideal in a polynomial ring over a field or over $Z$ is finitely generated. The proof was nonconstructive — it showed existence without providing an algorithm — and provoked Gordan's famous exclamation "das ist nicht Mathematik, das ist Theologie." The abstract definition of a Noetherian ring, extracting the ascending chain condition from Hilbert's concrete setting, is due to Noether 1921 in Idealtheorie in Ringbereichen ^{[Noether 1921]}, which recast the entire subject in axiomatic terms and introduced the three equivalent characterisations used today.

The Noether normalisation lemma was stated and proved by Noether in her 1926 paper Abstrakter Aufbau der Idealtheorie in algebraischen Zahl- und Funktionenkoerpern, though the technique appears earlier in work of Hilbert and Hurwitz. Grothendieck's generic freeness theorem, generalising these finiteness results to arbitrary Noetherian bases, appears in EGA IV and remains the technical backbone of modern algebraic geometry's finiteness theorems.

Bibliography [Master]

@article{Hilbert1890,
  author = {Hilbert, David},
  title = {Ueber die Theorie der algebraischen Formen},
  journal = {Mathematische Annalen},
  volume = {36},
  year = {1890},
  pages = {473--534},
}

@article{Noether1921,
  author = {Noether, Emmy},
  title = {Idealtheorie in Ringbereichen},
  journal = {Mathematische Annalen},
  volume = {83},
  year = {1921},
  pages = {24--66},
}

@book{AtiyahMacdonald1969,
  author = {Atiyah, Michael F. and Macdonald, Ian G.},
  title = {Introduction to Commutative Algebra},
  publisher = {Addison-Wesley},
  year = {1969},
}

@book{Eisenbud1995,
  author = {Eisenbud, David},
  title = {Commutative Algebra with a View Toward Algebraic Geometry},
  publisher = {Springer},
  year = {1995},
  series = {Graduate Texts in Mathematics 150},
}

@book{DummitFoote2004,
  author = {Dummit, David S. and Foote, Richard M.},
  title = {Abstract Algebra},
  edition = {3rd},
  publisher = {Wiley},
  year = {2004},
}