## 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.