Monthly Archives: November 2010

Exhibit an infinite family of prime ideals in R[x,y] when R is an integral domain

Let R be an integral domain and let a and b be relatively prime integers greater than 0. Prove that the ideal (x^a - y^b) is prime in R[x,y].

We will show that R[x,y]/(x^a - y^b) is an integral domain by demonstrating that it is isomorphic to a subring of the integral domain R[t]. To that end, define a ring homomorphism \varphi : R[x,y] \rightarrow R[t] by extending x \mapsto t^b, y \mapsto t^a, r \mapsto r for r \in R homomorphically. Certainly then \varphi(x^a - y^b) = t^{ab} - t^{ab} = 0, so that (x^a - y^b) \in \mathsf{ker}\ \varphi.

Now let \alpha \in R[x,y], and consider the coset \overline{\alpha} \in R[x,y]/(x^a-y^b). We may eliminate powers of \overline{y} greater than b-1 in favor of powers of \overline{x} in the representative \alpha, so that in fact \overline{\alpha} = \overline{\sum_{i=0}^{b-1} \alpha_i y^i}, where \alpha_i \in R[x]. Thus every element of R has the form \alpha = \left( \sum_{i=0}^{b-1} \alpha_i y^i\right) + \alpha^\prime, where \alpha^\prime \in (x^a-y^b). Suppose now that \alpha \in \mathsf{ker}\ \varphi. Then we have \varphi(\alpha) = \sum_{i=0}^{b-1} \varphi(\alpha_i) t^{ai} = 0. Now letting \alpha_i(x) = \sum_{j=0}^m r_{i,j} x^j, we have \varphi(\alpha) = \sum_{i=0}^{b-1} \sum_{j=0}^m r_{i,j} t^{bj + ai}.

We claim that the exponents bj + ai are pairwise distinct. To see this, suppose that bj + ai = bj^\prime + ai\prime. Mod b, we have ai \equiv ai^\prime. Since a is relatively prime to b, it is a unit mod b. Thus i \equiv i^\prime \mod b. However, since 0 \leq i,i^\prime < b, this yields i = i^\prime. Then bj = bj^\prime, so that j = j^\prime. Thus we have r_{i,j} = 0 for all i,j, so \alpha_i = 0 for all i, and thus \alpha \in (x^a - y^b).

So \mathsf{ker}\ \varphi = (x^a - y^b), and by the first isomorphism theorem for rings, R[x,y]/(x^a - y^b) is isomorphic to a subring of the integral domain R[t]. Thus R[x,y]/(x^a - y^b) is an integral domain, and so the ideal (x^a - y^b) is prime.

Demonstrate that two rings are not isomorphic

Let F be a field. Prove that F[x,y]/(y^2-x) and F[x,y]/(y^2-x^2) are not isomorphic.

First we show that F[x,y]/(y^2-x) is an integral domain. To that end, define a ring homomorphism \varphi : F[x,y] \rightarrow F[t] by extending \varphi(x) = t^2, \varphi(y) = t, and \varphi|_R = \mathsf{id}_R homomorphically. Certainly \varphi(y^2 - x) = t^2 - t^2 = 0, so that (y^2 - x) \subseteq \mathsf{ker}\ \varphi. Now let \alpha \in R[x,y] and consider the coset \alpha + (y^2 - x) = \overline{\alpha}. Since \overline{y^2} = \overline{x}, we may eliminate all powers of y greater than 1 in the representative of \alpha + (y^2 - x) in favor of powers of x, and so write \alpha + (y^2 - x) = \alpha_0 + \alpha_1y + (y^2 - x), where \alpha_0, \alpha_1 \in F[x]. That is, every element of F[x,y] can be written as \alpha_0 + \alpha_1y + (y^2 - x)\alpha_2, where \alpha_0,\alpha_1 \in F[x] and \alpha_2 \in F[x,y]. Suppose now that \alpha \in \mathsf{ker}\ \varphi; then we have \varphi(\alpha) = \varphi(\alpha_0) + \varphi(\alpha_1)t = 0. Note that every term in \varphi(\alpha_0) has even degree in t, since x \mapsto t^2, and that similarly every term in \varphi(\alpha_1)t has odd degree. Thus in fact all coefficients of \varphi(\alpha_0) and of \varphi(\alpha_1) are zero, so that \alpha = (y^2 - x)\alpha_2 \in (y^2 - x). So \mathsf{ker}\ \varphi = (y^2 - x). By the first isomorphism theorem for rings, we have F[x,y]/(y^2 - x) isomorphic to a subring of the integral domain F[t]; hence F[x,y]/(y^2 - x) is an integral domain.

On the other hand, F[x,y]/(y^2 - x^2) is not an integral domain since (\overline{y+x})(\overline{y-x}) = 0, but neither \overline{y+x} nor \overline{y-x} is zero considering degrees.

Thus F[x,y]/(y^2 - x) and F[x,y]/(y^2 - x^2) are not isomorphic for any fields F.

Exhibit a prime ideal whose square is not primary

Let R = \mathbb{Q}[x,y,z] and let bars denote passage to \mathbb{Q}[x,y,z]/(xy - z^2). Prove that \overline{P} = (\overline{x},\overline{z}) is a prime ideal. Show that \overline{xy} \in \overline{P} but that no power of \overline{y} is in \overline{P}^2; that is, \overline{P} is a prime ideal whose square is not primary.

Note that (R/(xy - z^2)/((x,z,xy-z^2)/(xy-z^2)) \cong R/(x,z,xy-z^2) = R/(x,z) \cong \mathbb{Q}[y] is a domain, so that \overline{P} is a prime ideal.

We have \overline{x}\overline{y} = \overline{z^2} \in \overline{P}. Now suppose \overline{y}^n \in \overline{P}. That is, y^n + (xy - z^2) \in (x,z,xy-z^2)/(xy - z^2) = (x,z)/(xy-z^2). But then y^n \in (x,z); this is a contradiction considering the x (or z) degrees. So no power of \overline{y} is in \overline{P}.

Exhibit a primary ideal which is not a power of a prime ideal

Show that the radical of the ideal I = (x,y^2) in \mathbb{Q}[x,y] is (x,y). Deduce that I is a primary ideal that is not a power of a prime ideal.

We discussed some basic properties of radical ideals and primary ideals here and here, respectively.

Recall that \mathsf{rad}(I) = \{ r \in R \ |\ r^n \in I\ \mathrm{for\ some}\ n \in \mathbb{N}^+ \}. Certainly we have (x,y) \subseteq \mathsf{rad}\ (x,y^2). Recall that (x,y) \subseteq \mathbb{Q}[x,y] is maximal; thus \mathsf{rad}\ (x,y^2) \subseteq \mathsf{Jac}\ (x,y^2) \subseteq (x,y), using this prior exercise about Jacobson radicals. Thus \mathsf{rad}\ (x,y^2) = (x,y).

Next we show that (x,y^2) is primary; recall that an ideal P is called primary if whenever ab \in P and a \notin P, then b^n \in P for some n \geq 1. We saw previously that an ideal P \subseteq R is primary if and only if every zero divisor in R/P is nilpotent. Note that \mathbb{Q}[x,y]/(x,y^2) \cong \mathbb{Q}[y]/(y^2). Let \alpha(y) + (y^2) \in \mathbb{Q}[y]/(y^2) be a zero divisor and say \alpha(y) = \alpha_0 + \alpha_1y; then we have \alpha(y)\beta(y) \in (y^2) for some \beta \notin (y^2). Without loss of generality, let \beta(y) = \beta_0 + \beta_1y. Now \alpha_0\beta_0 + (\alpha_0\beta_1 + \alpha_1\beta_0)y = 0, and thus \alpha_0\beta_0 = 0 and \alpha_0\beta_1 + \alpha_1\beta_0 = 0. Suppose \alpha_0 \neq 0; then \beta_0 = 0 and so \beta_1 = 0. But then \beta \in (y^2), a contradiction. Thus \alpha_0 = 0. Thus \alpha^2 \in (y^2), so that \alpha is nilpotent. So (y^2) is primary in \mathbb{Q}[y], and thus (x,y^2) is primary in \mathbb{Q}[x,y].

Now suppose (x,y^2) = P^n, where n \geq 1 and P is a prime ideal. Now P cannot contain a constant, as in this case we have (x,y^2) = \mathbb{Q}[x,y], a contradiction. So every element of P has degree at least 1, and thus every element of P^n has degree at least n. Since x \in P^n, we have n = 1. But then (x,y^2) = P is prime; this to is a contradiction since \mathbb{Q}[x,y]/(x,y^2) \cong \mathbb{Q}[y]/(y^2) contains a zero divisor (namely y + (y^2)) and thus is not an integral domain. So (x,y^2) is not a power of any prime ideal.

Exhibit a ring with infinitely many minimal prime ideals

Prove that the ring \mathbb{Z}[x_1,x_2,\ldots]/(x_1x_2,x_3x_4,x_5x_6,\ldots) has infinitely many \subseteq-minimal prime ideals.

Let R = \mathbb{Z}[x_1,x_2,\ldots] and let K = (x_1x_2,x_3x_4,\ldots).

Let X = \{ \{2k+1,2k+2\} \ |\ k \in \mathbb{N} \}, and let Y denote the set of all choice functions on X. (That is, each element of Y is a function that, given k \in \mathbb{N}, chooses one element from the pair \{2k+1,2k+2\}.) For each \lambda \in Y, define I_\lambda = (\lambda(0),\lambda(1),\ldots). For example, one such ideal is that generated by all the x_i with odd indices. Certainly there are infinitely many such ideals, each distinct. Also, we have that K = (x_1x_2,x_3x_4,\ldots) \subseteq I_\lambda for each \lambda, since each x_ix_{i+1} is in I_\lambda by construction.

For each \lambda, by the third isomorphism theorem for rings we have (R/K)/(I_\lambda/K) \cong R/I_\lambda, and R/I_\lambda is isomorphic to R. Since R is an integral domain, I_\lambda/K is a prime ideal of R/K.

We claim also that each I_\lambda is minimal. To see this, suppose J/K \subseteq I_\lambda/K is a prime ideal. Let (i,i+1) be a pair (with i odd) such that (without loss of generality) x_i \in I. Now x_ix_{i+1} \in K \subseteq J. Since J is prime and x_{i+1} \notin J (as x_{i+1} \notin I), we have x_i \in J– a contradiction. So in fact J = I_\lambda, and hence J/K = I_\lambda/K.

Note that R/K is not an integral domain (since (x_1+K)(x_2+K) = 0), so that the zero ideal is not prime. Thus each I_\lambda is an inclusion-minimal prime ideal.

A polynomial ring in infinitely many variables contains ideals which are not finitely generated

Let R be a commutative ring with 1. Prove that R[x_1,x_2,\ldots] contains an ideal which is not finitely generated.

Consider the ideal I = (x_1,x_2,\ldots), and suppose I = (A) is finitely generated. Since A is finite, and every element of A is a finite sum of terms in finitely many variables, there exists an index k such that x_k does not appear as a factor of any term in A. Moeroever, A does not contain any elements with nonzero constant term. This implies that any element of (A) which has a term with x_k as a factor necessarily has a term with degree at least 2; in particular, x_k \notin (A), a contradiction. So I is not finitely generated.

Show that two given rings have isomorphic fields of fractions

Let F be a field and let R = F[x,x^2y,x^3y^2,\ldots,x^{n+1}y^n,\ldots] be considered as a subring of F[x,y]. More precisely, let R_0 = F, R_{k+1} = R_k[x^{k+1}y^k], and R = \bigcup R_k.

  1. Prove that R and F[x,y] have isomorphic fields of fractions.
  2. Prove that R contains an ideal that is not finitely generated.

We begin with some lemmas.

Lemma 1: R = \left\{ \displaystyle\sum_{i,j} \alpha_{i,j} x^i y^j \ |\ \mathrm{if}\ j \neq 0\ \mathrm{then}\ i > j \right\}. Proof: (\subseteq) If p \in R, then in each term of p we can collect the x and y variables. Certainly in every nonconstant term the exponent on x is greater than that on y. (\supseteq) If p = \sum \alpha_{i,j} x^i y^j and \alpha_{i,j} = 0 whenever i < j, we have p = \sum \alpha_{i,j}(x^{i-j-1})(x^{j+1}y^j) \in R. \square

Lemma 2: Let R be an integral domain and S \subseteq R a subring. If for all r_1,r_2 \in R there exist s_1,s_2 \in S such that r_1s_2 = r_2s_1, then (where \mathsf{Fof} denotes field of fractions) we have \mathsf{Fof}(S) \cong \mathsf{Fof}(R). Proof: By the universal property of fields of fractions, the natural injection \Phi : \mathsf{Fof}(S) \rightarrow \mathsf{Fof}(R) given by \Phi(\frac{a}{b}) = \frac{a}{b} is well-defined, and moreover is an injective ring homomorphism. Now suppose \frac{r_1}{r_2} \in \mathsf{Fof}(R); by our hypothesis, there exist s_1,s_2 \in S such that r_1s_2 = r_2s_1, and thus \frac{r_1}{r_2} = \frac{s_1}{s_2}. So \Phi(\frac{s_1}{s_2}) = \frac{r_1}{r_2}, and \Phi is surjective. \square

Lemma 3: Let \alpha \in F[x,y], and let k be greater than the y degree of \alpha. Then x^k\alpha \in R. Proof: Certainly, in each term of x^k \alpha, the exponent on x exceeds that of y. Thus x^k \alpha \in R. \square

Now let \alpha,\beta \in F[x,y], and let k be the larger of the y-degrees of \alpha and \beta. Certainly x^k\alpha\beta = x^k\alpha\beta, and moreover x^k\alpha,x^k\beta \in R. Thus \mathsf{Fof}(R) \cong \mathsf{Fof}(F[x,y]).

Now consider the ideal I = (x^{t+1}y^t \ |\ t \in \mathbb{N}) in R. We claim that I is not finitely generated. To see this, suppose to the contrary that I = (A), where A is finite. Now each element of A is a finite sum of polynomials in x^{t+1}y^t; thus there exists a natural number k such that A \subseteq (x^{t+1}y^t \ |\ 0 \leq t \leq k). In particular, we have I = (x^{t+1}y^t \ |\ 0 \leq t \leq k). We claim now that x^{k+2}y^{k+1} \notin (x^{t+1}y^t \ |\ 0 \leq t \leq k). To see this, suppose to the contrary that x^{k+2}y^{k+1} = \sum_{t=0}^k p_t(x,y) x^{t+1}y^t, where p(x,y) \in R. Note that every term on the right hand side having y-degree k+1 must have x-degree at least k+3, a contradiction. Thus I is not finitely generated.

A polynomial ring in more than one variable is not a PID

Let R be a commutative ring with 1. Prove that a polynomial ring in more than one variable over R is not a principal ideal domain.

[Short, Nice Proof]

If R[x_1,\ldots,x_n,x_{n+1}] = R[x_1,\ldots,x_n][x_{n+1}] is a PID, then by Corollary 8 on page 281 of D&F, R[x_1,\ldots,x_n] is a field- a contradiction.

[Long, Complicated Proof]

We begin with a lemma.

Lemma: Let R be a commutative ring with 1. If R[x] is a principal ideal domain, then every element of R is either 0, a unit, or a zero divisor. Proof: Let r \in R be nonzero. The ideal (r,x) = (d) is principal. Since r = dp for some p, the constant coefficient of d is nonzero. Now x = dq for some q. Let q_0, d_0, and p_0 be the constant terms of q, d, and p, respectively. If q_0 \neq 0, then d_0 is a zero divisor since d_0q_0 = 0. Thus r = d_0p_0 is a zero divisor. If q_0 = 0, then letting d_1 and q_1 be the linear coefficients of d and q, respectively, we have d_0q_1 + d_1q_0 = 1, so that d_0q_1 = 1. In particular, d_0 is a unit. Thus we in fact have (r,x) = (1 + xe(x)) for some e \in R[x]. Now (1 + xe(x))t(x) = x for some t, so that t(x) + xt(x)e(x) = x. Note that xe(x)t(x) has no constant term, so that comparing coefficients, the constant term t_0 of t is zero. Write t(x) = xt^\prime(x). Then xt^\prime(x) + x^2e(x)t^\prime(x) = x, so that x(t^\prime(x) + xe(x)t^\prime(x)) = x. Note that x is not a zero divisor, so that t^\prime(x)(1 + xe(x)) = 1. That is, we have (r,x) = R[x]. Using a lemma proved in this previous exercise, we have 0 \cong R[x]/((r)+(x)) \cong R/(r), so that (r) = R, and thus r is a unit. \square

Now let R be a commutative ring with 1 and consider R[x,y]. If R[x,y] = (R[x])[y] is a principal ideal domain, then every element of R[x] is either zero, a unit, or a zero divisor. However, x \in R[x] is not zero and not a zero divisor. Moreover, x is not a unit since xp(x) has constant term 0 for all p. Every polynomial ring in two or more variables is isomorphic to R[x,y] for some R, and thus the result is proved.

This is a handy lemma that we ended up not using here, but don’t want to throw out.

Lemma: Let R be a commutative ring with 1, and suppose a \in R is irreducible and not a unit. Then (a,x) \subseteq R[x] is not a principal ideal. Proof: Suppose to the contrary that (a,x) = (d). Since a = du for some u and a is irreducible, there are two cases. If u is a unit, then in fact x = ap for some p \in R[x]. Comparing degrees, we see that p(x) = c_0 + c_1x, so that ac_1 = 1. This is a contradiction since a is not a unit. If d is a unit, then (a,x) = R[x]. In particular, ap + xq = 1 for some p,q. Comparing constant coefficients, we have that a is a unit, a contradiction. Thus (a,x) is not principal. \square.

The ideal (x,y) is not principal in QQ[x,y]

Prove that (x,y) \subseteq \mathbb{Q}[x,y] is not a principal ideal.

Suppose (x,y) = (p(x,y)) is a principal ideal. Then we have pq = x for some q \in \mathbb{Q}[x,y]. Since \mathbb{Q} is a domain, we have \mathsf{deg}_y(p) + \mathsf{deg}_y(q) = 0, so that \mathsf{deg}_y(p) = 0. Similarly, \mathsf{deg}_x(p) = 0, considering y \in (p). But then p \in \mathbb{Q}, so that (x,y) = \mathbb{Q}[x,y]. In particular, there exist a,b \in \mathbb{Q}[x,y] such that xa + yb = 1. If a \neq 0, then the x degree of xa + yb is at least 1, while the x degree of 1 is 0. Thus a = 0. Similarly, b = 0, so that 0 = 1, a contradiction. Thus (x,y) is not principal.

Two prime ideals in ZZ[x,y]

Prove that (x,y) and (2,x,y) are prime ideals in \mathbb{Z}[x,y] but that only (2,x,y) is maximal.

In this previous exercise, we saw that \mathbb{Z}[x,y]/(x,y) \cong \mathbb{Z}. Since this quotient is a domain but not a field, (x,y) is a prime ideal but not maximal.

On the other hand, we have \mathbb{Z}[x,y]/(2,x,y) \cong \mathbb{Z}[x][y]/((2,x) + (y)) \cong \mathbb{Z}[x]/((2)+(x)) \cong \mathbb{Z}/(2), which is a field. Thus (2,x,y) is maximal and hence prime.