TUT ALG3 3

Problem 2

Exercise 301.1(Exercise).

Let $R=\mathbb{C}[x,y,z]$ and $S=\mathbb{C}[t]$. Let $\varphi :R\to S$ be a homomorphism defined by $\varphi (x)=t^3$, $\varphi (y)=t^4$, and $\varphi (z)=t^5$. Show that $\ker \varphi =(x^3-yz,y^2-xz,z^2-x^{2}y)$.

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Let $I=(x^3-yz,y^2-xz,z^2-x^{2}y)$. We will prove that

$$\frac{R}{I}\cong \mathbb{C}[t^3,t^4,t^5].$$

Define the map $\phi :R/I\to \mathbb{C}[t^3,t^4,t^5]$ by $p(x,y,z)+I\mapsto \varphi (p(x,y,z))$. It is easily verified that this is a well defined homomorphism. It is also clearly surjective. It only remains to show injectivity.

Let $p(x,y,z)\in R$. Note that the following rewrite rules when applied on $p(x,y,z)$ preserve membership in $p(x,y,z)+I$:

$$\mathfrak{R}=\{\begin{array}{ll}(1)& x^3\to yz,\\ (2)& x^{2}y\to z^2,\\ (3)& y^2\to xz.\end{array}$$

Claim 301.2(Claim).

Any sequence of applications of the above rules will terminate, that is, a polynomial on which the rules cannot be applied will be obtained in a finite number of steps.

This is clear, since rules $(1)$ and $(2)$ reduce the degree of the polynomial, and while rule $(3)$ does not alter the degree, its applications are limited by the number of $y$‘s in the polynomial and the degree will have to be sacrificed eventually to get more $y$‘s.

It can be easily deduced that for any polynomial $p(x,y,z)\in R$, any sequence of applications of the rules in $\mathfrak{R}$ must terminate in a polynomial in the set

$$S=\begin{array}{r}\{x,y,x^2,xy,1\}\{z^k|k\in {\mathbb{Z}}_{\ge 0}\}.\end{array}$$

Next, consider the map $\psi :S\to \mathbb{C}[t^3,t^4,t^5]$ given by

$$\psi (s(x,y,z))=\phi (s(x,y,z)+I).$$

Claim 301.3(Claim).

$\psi$ is injective.

Proof.

We will prove injectivity for monomials; the general case follows. Let $x^i{y}^j{z}^k$ and $x^a{y}^b{z}^c$ be two monomials in $S$ such that $\psi (x^i{y}^j{z}^k)=\psi (x^a{y}^b{z}^c)$. Then,

$$3i+4j+5k=3a+4b+5c.$$

Assume $j\ne b$. The only possible cases are $j=1,b=0$, or $j=0,b=1$. WLOG, if $j=1$, we have

$$4=5(c-k)+3(a-i),$$

which requires $a-i\le -2$ or $a-i\ge 3$, which together with $j=1$ imply $x^i{y}^j{z}^k\notin S$. Thus, $j=b$. $i=a$ and $k=c$ immediately follow.□

Claim 301.4(Claim).

For a given polynomial $p(x,y,z)$, any sequence of applications of the rules in $\mathfrak{R}$ will terminate in the same polynomial.

Proof.

We have established that any polynomial obtained by application of the rules on$p(x,y,z)$ stays in $p(x,y,z)+I$. Since $\psi$ is injective, no two elements of $S$ lie in the same coset of $I$, so there is only one element of $S$ that $p(x,y,z)$ can reduce to.□

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For $p(x,y,z)\in R$, denote the unique reduced polynomial by $\tilde{p}(x,y,z)$.

The same argument as in Claim 4 can be used to show that if $p(x,y,z)$ and $q(x,y,z)$ lie in the same coset of $I$, then $\tilde{p}(x,y,z)=\tilde{q}(x,y,z)$. Thus, we have a bijection between the cosets of $I$ and $S$ given by

$$p(x,y,z)+I\leftrightarrow \tilde{p}(x,y,z).$$

Let $\alpha (t^3,t^4,t^5)\in \mathbb{C}[t^3,t^4,t^5]$. Since $\phi$ is a surjective homomorphism, there exists $p(x,y,z)+I$ such that $\phi (p(x,y,z)+I)=\alpha (t^3,t^4,t^5)$. It follows that $\psi (\tilde{p}(x,y,z))=\alpha (t^3,t^4,t^5)$, so $\psi$ is surjective, and hence bijective.

Claim 301.5(Claim).

$\phi$ is injective, and hence a bijection.

Proof.

$$\begin{array}{rl}& \phi (p(x,y,z)+I)=\phi (q(x,y,z)+I)\\ & \phi (\tilde{p}(x,y,z)+I)=\phi (\tilde{q}(x,y,z)+I)\\ & \psi (\tilde{p}(x,y,z))=\psi (\tilde{q}(x,y,z))\\ & \tilde{p}(x,y,z)=\tilde{q}(x,y,z)\\ & \\ ⟹& p(x,y,z)+I=q(x,y,z)+I.\end{array}$$ □

Thus, $R/I\cong \mathbb{C}[t^3,t^4,t^5]$. Together with the observation that $I\subseteq \ker \varphi$, this immediately implies $I=\ker \varphi$.

For my future self: yes, this is immediate. By the mapping property of quotient groups, there exists a unique homomorphism $R/I\to R/\ker \varphi$ which makes this diagram commute:

Note that by construction, $\phi$ makes this diagram commute.

Alternate solution

Let I = (x³-yz, y²-xz, z²-x²y). Its easy to see I is contained in kernel. Now if you have any monomial x^i y^j z^k, you can write it (mod I) as x^i y^j z^k with j+k ⇐1. Point is replace y² with xz, z² with x²y, and yz with x^3(since these are equivalent mod I) to reduce powers of y,z.

So any polynomial in x,y,z can be written as p(x) + yq(x) + z r(x) modulo I. If you take the image under tbe homomorphism of this, you get p(t³) + t⁴q(t³) +t⁵r(t³). If the origianl polynomial was in the kernel, then the above polynomial in t is zero (modulo I does not change this as I is contained in kernel). Checking the coefficients of powers of t of the form 3k, 3k+1, 3k+2, gives you p=q=r=0. So the polynomial is 0 mod I i.e. it is contained in I.

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Problem 8

Exercise 301.6(Exercise).

Suppose $I_1+\cdots +I_n=R$ and for each $1\le k\le n$, $I_k\cap (I_1+\cdots +I_{k-1}+I_{k+1}+\cdots +I_n)=(0)$. Then show that there is an isomorphism $R\cong I_1\times \cdots \times I_n$.

There must exist elements $a_1,\dots ,a_n$, $a_i\in I_i$, such that $a_1+\cdots +a_n=1$ in $R$. Further, these elements are unique: If some $a_i^{\prime }$‘s satisfied the same property, then

$$\sum _{i=1}^n(a_i-a_i^{\prime })=0,$$

so for each $k$, $(a_k-a_k^{\prime })={\sum }_{i\ne k}(a_i-a_i^{\prime })$. Note that the LHS is a member of $I_k$, and the RHS is a member of $(I_1+\cdots +I_{k-1}+I_{k+1}+\cdots +I_n)$. Thus, we have $a_k=a_k^{\prime }$ for all $k$. The same method can be used to prove that if $e_1+\cdots +e_n=e_1^{\prime }+\cdots +e_n^{\prime }$ and $e_i,e_i^{\prime }\in I_i$, then $e_i=e_i^{\prime }$; we will be using this shortly.

Before we proceed, we have to show that $I_1\times \cdots \times I_n$ is indeed a ring. The only it is missing is an identity, and $(a_1,\dots ,a_n)$ does the job. Indeed, for $b_i\in I_i$,

$$\begin{array}{rl}(b_1+\cdots +b_n)& =(b_1+\cdots +b_n)(a_1+\cdots +a_n)\\ & =b_1{a}_1+\cdots +b_n{a}_n,\end{array}$$

since all the cross terms must lie in $I_i\cap I_j$ for some $i\ne j$, and thus must be $0$. Hence, $b_i=b_i{a}_i$, and we have

$$(b_1,\dots ,b_n)(a_1,\dots ,a_n)=(b_1{a}_1,\dots ,b_n{a}_n)=(b_1,\dots ,b_n)$$

for all $(b_1,\dots ,b_n)\in I_1\times \cdots \times I_n$.

Define a map $\phi :R\to I_1\times \cdots \times I_n$ by

$$r\mapsto (r{a}_1,\dots ,r{a}_n).$$

This is clearly a well defined map which respects addition. To see that it respects multiplication, pull the same trick as before: write

$$r_1{r}_2=(r_1{a}_1+\cdots +r_1{a}_n)(r_2{a}_1+\cdots +r_2{a}_n)=r_1{a}_1{r}_2{a}_1+\cdots +r_1{a}_n{r}_2{a}_n,$$

which then is equal to $r_1{r}_2(a_1+\cdots +a_n)$, so we have $r_1{r}_2{a}_i=r_1{a}_i{r}_2{a}_i$. Multiplication is okay!

All that’s left to prove is surjectivity. Take $(b_1,\dots ,b_n)\in I_1\times \cdots \times I_n$, and observe that

$$\phi (b_1+\cdots +b_n)=(b_1{a}_1,\dots ,b_n{a}_n)=(b_1,\dots ,b_n).$$

Exercise 301.7(Exercise).

If $I_1,\dots ,I_n$ are ideals in $R$, then show that there is an injective map $\phi :R/(I_1\cap \cdots \cap I_n)\to (R/I_1)\times \cdots \times (R/I_n)$.

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Define the obvious map

$$\phi \left(r+\bigcap I_i\right):=(r+I_1,\dots ,r+I_n).$$

That $\phi$ is a homomorphism is immediate. If $r_1,r_2$ are such that $\phi \left(r_1+\bigcap I_i\right)=\phi \left(r_2+\bigcap I_i\right)$, then $r_1-r_2\in I_i$ for all $i$, so $r_1-r_2\in \bigcap I_i$, so $r_1+\bigcap I_i=r_2+\bigcap I_i$.

Exercise 301.8(Exercise).

Suppose $I_i+I_j=R$ for all $i\ne j$. If $b_1,\dots ,b_n\in R$, then show that there exists $b\in R$ such that $b\cong b_i\mathrm{mod}{I}_i$ for all $i=1,\dots ,n$.

We just have to show that the homomorphism in Exercise 7 is surjective given these hypotheses. For the pair $I_i,I_j$, let $e_{i_j}\in I_i$ and $e_{j_i}\in I_j$ be the elements such that $e_{i_j}+e_{j_i}=1$. Then, we have

$$\begin{array}{rl}1& =(e_{1_2}+e_{2_1})(e_{1_3}+e_{3_1})\dots (e_{1_n}+e_{n_1})\\ & ={\alpha }_1+e_{2_1}{e}_{3_1}\dots e_{n_1},\end{array}$$

where ${\alpha }_1\in I_1$. Thus, we have shown that $I_1+I_2{I}_3\dots I_n=R$. Generally, we have $I_k+{\prod }_{i\ne k}{I}_i=R$ for all $k$. For $I_k$, let $v_k$ be an element in ${\prod }_{i\ne k}{I}_i$ such that $u_k+v_k=1$, where $u_k\in I_k$. Note that $v_k\in 1+I_k$ and $v_k\in I_i$ for $i\ne k$.

Now, consider $(a_1,\dots ,a_n)\in (R/I_1)\times \cdots \times (R/I_n)$. We have

$$\phi (a_1{v}_1+\cdots +a_n{v}_n)=(a_1,\dots ,a_n).$$

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Problem 11

Exercise 301.9(Exercise).

Let $\phi :R\to S$ be a surjective homomorphism. Let $I\in R$ and $J\in S$ be corresponding ideals. Show that $I$ is prime iff $J$ is prime.

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Since $I$ and $J$ are corresponding ideals, we know $J=\phi (I)$, $I={\phi }^{-1}(J)$, and $\ker \phi \subseteq I$.

Assume $I$ is prime. Suppose $x,y\in S$ such that $xy\in J$. Since $\phi$ is surjective, $x=\phi (a)$ and $y=\phi (b)$ for some $a,b\in R$. Then $\phi (a)\phi (b)=\phi (ab)\in J$, so $ab\in {\phi }^{-1}(J)=I$, so $a\in I$ or $b\in I$, hence $\phi (a)\in J$ or $\phi (b)\in J$.

Assume $J$ is prime. Suppose $a,b\in R$ such that $ab\in I$. Then, $\phi (ab)=\phi (a)\phi (b)\in J$, so either $\phi (a)\in J$ or $\phi (b)\in J$, so $a\in I$ or $b\in I$.

Exercise 301.10(Exercise).

Describe all prime ideals in $\mathbb{Z}[x]$.

Let $\mathfrak{P}$ be a prime ideal of $\mathbb{Z}[x]$. Then $\mathfrak{P}\cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}$ (this holds whenever $R\subseteq S$ are commutative rings). We have two possibilities: $\mathfrak{P}\cap \mathbb{Z}=(0)$, or $\mathfrak{P}\cap \mathbb{Z}=(p)$ for some prime $p$.

Case 1: $\mathfrak{P}\cap \mathbb{Z}=(0)$.

Stuff on localization

Lemma 301.11(Milne (2020) 1.11).

Consider a (commutative) integral domain $A$ and a multiplicative subset $S$ of $A$. For an ideal $\mathfrak{a}$ of $A$, let ${\mathfrak{a}}^e$ denote the ideal it generates in the localization $S^{-1}A$; for an ideal $\mathfrak{a}$ of $S^{-1}A$, write ${\mathfrak{a}}^c$ for $\mathfrak{a}\cap A$. Then:

1. ${\mathfrak{a}}^{ce}=\mathfrak{a}$ for all ideals $\mathfrak{a}$ of $S^{-1}A$.
2. ${\mathfrak{a}}^{ec}=\mathfrak{a}$ if $\mathfrak{a}$ is a prime ideal of $A$ disjoint from $S$.

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Basically, we have maps in two directions between the set of ideals of $A$ and the set of ideals of $S^{-1}A$ called the expansion and contraction maps given by¹

$$\begin{array}{ccc}A& & S^{-1}A\\ \mathfrak{a}& \mapsto & {\mathfrak{a}}^e:=(\mathfrak{a})(S^{-1}A)\\ {\mathfrak{b}}^c:=\mathfrak{b}\cap A& \leftarrow & \mathfrak{b}\end{array}$$

Contraction followed by expansion is the identity on the set of ideals of $S^{-1}A$. Expansion followed by contraction is the identity when restricted to prime ideals of $A$ disjoint from $S$.

Lemma 301.12(Lemma).

Let $A$ be a (commutative) integral domain and let $S$ be a multiplicative subset of $A$. The map $\mathfrak{p}\mapsto {\mathfrak{p}}^e$ is a bijection from the set of prime ideals in $A$ such that $\mathfrak{p}\cap S=\varnothing$ to the set of prime ideals in $S^{-1}A$; the inverse map is $\mathfrak{p}\mapsto \mathfrak{p}\cap A$.

Proof.

Let$\mathfrak{p}\subset A$ be a prime ideal such that $\mathfrak{p}\cap S=\varnothing$. It is easy to see that elements of ${\mathfrak{p}}^e$ are of the form $p/s$, where $p\in P$ and $s\in S$. Let $a_1/s_1,a_2/s_2\in S^{-1}A$ such that $a_1{a}_2/s_1{s}_2=p/s\in {\mathfrak{p}}^e$. Then, $a_1{a}_{2}s=s_1{s}_{2}p$. Since $s_1{s}_{2}p\in \mathfrak{p}$ and $s\notin \mathfrak{p}$, we have $a_1\in \mathfrak{p}$ or $a_2\in \mathfrak{p}$, so $a_1/s_1\in {\mathfrak{p}}^e$ or $a_2/s_2\in {\mathfrak{p}}^e$. Hence, ${\mathfrak{p}}^e$ is a prime ideal.

Let $\mathfrak{p}\subset S^{-1}A$ be a prime ideal. It is clear that $\mathfrak{p}\cap A$ is a prime ideal in $A$. $\mathfrak{p}$ cannot contain any elements of $S$, since if it did it would also contain $1$.

Lemma 11 shows that the two maps are inverses.□

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This provides the necessary context to understand Magidin (2012) .

Back to the problem at hand

If $\mathfrak{P}=(0)$, we are done; otherwise, let $S=\mathbb{Z}\setminus \{0\}$. $S$ is a multiplicative set and $S\cap \mathfrak{P}=\varnothing$. Localize $\mathbb{Z}[x]$ at $S$ to obtain $\mathbb{Q}[x]$. Since $\mathbb{Q}[x]$ is a PID and $S^{-1}\mathfrak{P}$ is a prime ideal in $\mathbb{Q}[x]$ (by Lemma 12), $S^{-1}\mathfrak{P}$ is of the form $(q(x))$ for some irreducible² polynomial $q(x)$. We may assume that $q(x)$ has integer coefficients and that the gcd of its coefficients is $1$ (!).

From Lemma 12, we know that $(S^{-1}\mathfrak{P})\cap \mathbb{Z}[x]=(q(x))(\mathbb{Q}[x])\cap \mathbb{Z}[x]=\mathfrak{P}$. This is precisely $(q(x))(\mathbb{Z}[x])$. Thus, $\mathfrak{P}$ is of the form $(q(x))$ for some irreducible polynomial $q(x)\in \mathbb{Z}[x]$.

Case 2: $\mathfrak{P}\cap \mathbb{Z}=(p)$.

Consider the image of $\mathfrak{P}$ in $\mathbb{Z}[x]/(p)\cong {\mathbb{F}}_p[x]$. Since the map is onto, by Exercise 9 the image is prime. The prime ideals of ${\mathbb{F}}_p[x]$ are of the form $(q(x))$ where $q(x)$ is monic and irreducible over ${\mathbb{F}}_p[x]$. If the image is $(0)$, then $\mathfrak{P}=(p)$, and we are done.

Otherwise, let $p(x)$ be a polynomial in $\mathbb{Z}[x]$ that reduces to $q(x)$ modulo $p$ and that is monic. Note that $p(x)$ must be irreducible in $\mathbb{Z}[x]$, since any nontrivial factorization in $\mathbb{Z}[x]$ would induce a nontrivial factorization in ${\mathbb{F}}_p[x]$.

By the correspondence theorem, $\mathfrak{P}$ is the inverse image of $(q(x))$ under the projection homomorphism $\pi :\mathbb{Z}[x]\to \mathbb{Z}[x]/(p)$. We claim that $\mathfrak{P}=(p,p(x))$. It is clear that $(p,p(x))\subseteq \mathfrak{P}$ since $\pi ((p,p(x)))\subseteq (q(x))$. Conversely, let $r(x)\in \mathfrak{P}$. Then, there exists $s(x)\in {\mathbb{F}}_p[x]$ such that $s(x)q(x)=\overline{r}(x)$. If $t(x)$ is a polynomial in $\mathbb{Z}[x]$ such that it reduces to $s(x)$ mod $p$, then $t(x)p(x)-r(x)\in (p)$. Thus, there exists a polynomial $u(x)\in \mathbb{Z}[x]$ such that $r(x)=t(x)p(x)+pu(x)$. Therefore, $r(x)\in (p,p(x))$, hence $\mathfrak{P}\subseteq (p,p(x))$, giving equality.

Footnotes

1. Katex does not support \mapsfrom :( ↩

2. Recall that prime and maximal ideals coincide for PIDs. ↩

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References

Magidin, A. (2012). Answer to “Classification of Prime Ideals of $\mathbb{\vphantom}Z\vphantom{}[X]$.” In Mathematics Stack Exchange. https://math.stackexchange.com/a/174713/1677240

Milne, J. S. (2020). Algebraic Number Theory. https://www.jmilne.org/math/CourseNotes/ANT.pdf