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7

I've seen this question many times. The problem is that the second definition is strictly weaker than the first. Consider $(x^2,xy)$ in the ring $F[x,y]$ where $F$ is a field. According to the normal definition, it is not primary since it doesn't contain any powers of $y$ and doesn't contain $x$. However, it does satisfy the second definition. If $ab$ is ...


7

It is not prime since $$x_1(x_3x_5-x_2x_6)=x_3(x_1x_5-x_2x_4)-x_2(x_1x_6-x_3x_4)\in I$$ but $x_1\notin I$ because every nonzero element of $I$ has degree at least $2$, and $x_3x_5-x_2x_6\notin I$ because every element of degree $2$ in $I$ is a $k$-linear combination of $x_1x_5-x_2x_4$ and $x_1x_6-x_3x_4$ and so cannot involve either of the terms $x_3x_5$ or ...


6

If you compute a Gröbner basis for $I$, you will get that you need one additional generator, namely $x_1x_3x_5-x_1x_2x_6=x_1(x_3x_5-x_2x_6)$. Thus $x_1$ vanishes on the zero set of the ideal, but $x_1 \not \in I$, as in Alex Becker's anser above. Your ideal in fact decomposes as $I= I_{minors} \cap \langle x_1, x_4 \rangle$, where $I_{minors}$ is the ideal ...


4

The ideals of a quotient ring $R/I$ are of the form $J/I$ with $I\subseteq J\subseteq R$, $J$ ideal of $R$. In particular, the prime ideals are of the form $P/I$ with $I\subseteq P$, $P$ prime ideal of $R$. Now let $I=M^k$ with $M$ maximal. A prime ideal $P$ of $R$ containing $M^k$ contains $M$ (by the definition of prime ideals), so equals $M$. Conclusion: ...


4

Here is a simple example with $I$ not primary such that $Rad(I)$ is prime. Consider the ideal $(y^2,xy)\subset K[x,y]$. The radical of this ideal is $(y)$, i.e., it is prime. At the same time, clearly $x\cdot y\in I$. Also, $y\notin I$. Nevertheless $x^n\notin I$ for any $n$. Hence $I$ is not primary! Comment. As user26857 noted in a remark, one can even ...


3

As mentioned by user26857 above, this ideal is well-known to have arithmetic rank $3$, and is not a set-theoretic complete intersection. To see that $\text{ara}(I) \le 3$, notice that $I = (X_1X_3, X_1X_4, X_2X_3, X_2X_4) = \sqrt{(X_1X_3, X_2X_4, X_1X_4 + X_2X_3)}$, since $(X_1X_4)^2 = X_1X_4(X_1X_4 + X_2X_3) - (X_1X_3)(X_2X_4)$ (and similarly for ...


3

You can really solve this without using any higher principles. Here is an elementary solution: Well, what IS $I$? It is an ideal that consists of integer polynomials that have even constant term. $f = a_nx^n + \cdots + a_1x + a_0\in \mathbb{Z}[x]$, then $f\in I$ exactly means that $a_0$ is even. Okay, $I$ is prime: we need show $f,g\notin I \implies ...


3

You're doing well, but this isn't right: $\frac{\mathbb Z \times \mathbb Z}{M} = (\mathbb Z_{p}, \mathbb Z_{b})$ It's true that everything in the left side of the pairs is of the form $pa$ for $a$ ranging over $\Bbb Z$, and those are exactly the elements of $(p)\lhd \Bbb Z$, so the left part is indeed $\Bbb Z_p$. But in the right hand side, $b$ can be ...


3

$2-2Y=X^2 + (1-2Y+ Y^2)+(1-X^2-Y^2)$ Suppose $\langle f^2\rangle = \langle 1-Y\rangle$. Then $f(X,Y)^2=u(1-Y)$ for some $u\in\mathbb Q$. Note that elements of our ring are well-defined as functions on the unit circle, since $1-X^2-Y^2$ vanishes on it. Evaluating at $(3/5,4/5)$ gives $f(3/5,4/5)^2=u/5$, and evaluating at $(5/13,12/13)$ gives ...


2

An ideal is prime iff it is primary and radical (aka semiprime.) If we only talk about radical ideals, then such an ideal is prime iff it is primary. Remember that radical ideals have the property that $x^n\in I$ implies $x\in I$. Of course, if $r(\mathfrak{a})$ is prime, then it's primary. In the other direction, suppose $r(\mathfrak{a})$ is primary. If ...


2

Add an ordered pair $(c,d)$, where $c$ is not a multiple of $p$. Using the Bézout Identity we find that $(1,k)$ is in the resulting ideal for some $k$. But $(0,k)$ is in the ideal, so $(1,0)$ is. Since $(0,1)$ is in the ideal, we get all of $\mathbb{Z}\times \mathbb{Z}$.


2

Suppose $\;P\le R\;$ is prime $\;\iff R/P\;$ is an integer domain, and let $\;0\neq a+P\in R/P\;$ . We're given that $$\exists\,\,2<n_a\in\Bbb N\;\;s.t.\;\;a^{n_a}=a\implies (a+P)^{n_a}=a^{n_a}+P=a+P\implies$$ $$a(a^{n_a-1}-1)+P=P(=\overline 0)$$ and as $\;R/P\;$ is an integer domain we're done...


2

Define $$\phi : A/a \times A/b \rightarrow A/(a+b),\; \phi(x+a,y+b) = xy + (a+b).$$ Can you show $\phi$ is a well-defined bilinear map? Now if you use the universal property of tensor products to get a map $$\psi : A/a \otimes_A A/b \rightarrow A/(a+b),$$ can you show $\psi$ is injective? To show surjectivity, it will probably help you to show that an ...


2

It may be easier to prove the more general fact: if $R$ is any $A$-ring (i.e. we have a chosen homomorphism $\varphi : A \to R$), then we have $$ A/I \otimes_A R \cong R / IR $$ In the following, $a,a'$ are variables denoting elements of $A$, $i$ an element of $I$, and $r,r'$ elements of $R$. This can be done by explicitly writing down the maps: in the ...


2

You can go the abstract way directly using the first isomorphism theorems: $$\Bbb Z[x]/\langle x,2\rangle\cong\left(\Bbb Z[x]/\langle 2\rangle\right)/\left(\langle x,2\rangle/\langle2\rangle\right)\cong\Bbb F_2[x]/\langle x\rangle\cong\Bbb F_2$$ With $\;\Bbb F_2\cong\Bbb Z/2\Bbb Z=\;$ the field with two elements. Now just use the already mentioned ...


2

The statement that a maximal left ideal is the annihilator of a simple module is incorrect: indeed the annihilator of a module is a two-sided ideal, as you remark, so any maximal left ideal which is not two-sided is a counterexample. Matrix rings provide easy examples of maximal left ideals which are not two-sided. On the other hand it is true that the ...


2

You can actually prove this from elementary principles: Let $I$ be the ideal maximal with respect to not being finitely generated. Suppose $I$ is not prime. Then $\exists f,g\in R$ such that $f,g\notin I,\ fg\in I$. Then consider $J := I + (f)$. You can show that since $I$ not finitely generated, $J$ is not finitely generated. Moreover, $J \neq R$ ...


2

The product of ideals is an ideal because the operations are defined componentwise. It is completely straightforward to see that the product is closed under addition and left multiplication by elements in the product algebra. Because your indexing set is finite, every ideal is of this form. Given an ideal $J$ let $J_i$ be the set of elements in $B_i$ ...


2

You'll want to show this by induction on the number of generators of your ideal. I'll show this for two generators. If $I=(x,y)$, then certainly $(x+y+xy) \subseteq I$. Conversely, $x(x+y+xy) = x^2 +xy + x^2y = x+2xy = x \in (x+y+xy)$ (Assuming you've shown that $2x = 0\; \forall \; x$ Similarly, $(x+y+xy)y = y \in (x+y+xy)$ so $(x,y) \subseteq (x+y+xy)$ ...


2

Some hints: 1) First note that in a Noetherian ring every ideal contains a power of its radical. Hence there exists some $n>0$ such that $p^n \subset q$. Now if $x \in S \cap p$ then $x^n \in S \cap q$ because $S$ is multiplicatively closed. This implies that $S^{-1}q = S^{-1} A$. 2) If $A$ is an integral domain, i.e. it has no zero divisors, then the ...


1

Basically you need to know that when you contract $X$ and $Y$ to $A$, they will still share no common prime factors (so there are no ideals in each lying over the same prime). In the case you mention, all primes in $B$ lying over $p$ appear in the prime factorization of $pB$, so $F$ is not divided by any of these (being coprime). Therefore, when you ...


1

If $I$ is an ideal in $R$ and $e_i=(0,\dots,0,1,0,\dots,0)$ with the $1$ in the $i$th position, then $e_iI\subseteq I$ for all $I$. It follows that $I\supseteq\sum_ie_iI$. Now $e_1+\cdots+e_n=1_R$, so if $x\in I$, we have $$x=1_Rx=(e_1+\cdots+e_n)x=e_1x+\cdots+e_nx\in \sum_ie_iI.$$ This tells us that in fact $I=\sum_ie_iI$. You should have no trouble showing ...


1

Since $A/\mathfrak{a}$ is $A$-flat, the ideal $\mathfrak{a}$ is generated by a single idempotent element $e$ (so $e^2=e$). This is a consequence of Nakayama's lemma and uses that $\mathfrak{a}$ is finitely generated (a consequence of the assumption that $A$ is Noetherian). (I'm pretty sure this is proved somewhere on the site.) You can now check that ...


1

In general, if $A$ is any ring (not necessarily Noetherian), then $A/I$ is flat over $A$ iff $\text{Supp}(I) \cap \text{Supp}(A/I) = \emptyset$ iff $\text{Spec}(A) = \text{Supp}(I) \sqcup \text{Supp}(A/I)$. This follows from the fact that a flat local map is faithfully flat, hence injective, so if $p \in \text{Supp}(A/I) = V(I)$, then $A_p \to A_p/I_p$ is ...


1

This should be a comment, but reputation lacks. In general if $R$ is a ring and $r\in R$ then $(r)$ denotes the (principal) ideal generated by $r$. So in fact $I$ and $J$ mentioned in your question are ideals by definition. Maybe they want you to construct two ringhomomorphisms having $I$ and $J$ as kernels.


1

Careful: the only invertible elements of $\Bbb Z$ are $\pm 1$, so certainly if $n\neq \pm 1$, it cannot be invertible. I think you mean to write that $A$ is the set of integers coprime to $n$. Note that $A$ is a submonoid of $(\Bbb Z,\,\cdot\,)$. If $m/k\in\Bbb Z_{(6)}$ is invertible we can find $m'/k'$ for which $$\frac mk\frac{m'}{k'}=1$$ This means ...


1

As you already said, $p$ becomes the unique prime ideal of $A_p$, and hence the nilradical which consists precisely of all nilpotent elements of $A_p$. Since $A$ is reduced $A_p$ is reduced as well so the unique prime ideal of $A_p$ is the zero ideal. Since every non-unit of a commutative ring lies in a (maximal) prime ideal, every non-zero element of $A_p$ ...


1

Pretty much the same answer I gave to your other question (Maximal Ideal Must be Prime) This time, you'll need to prove that $J = I + (f)$ is not principal if $J$ is not principal. The method of proof is pretty much the same.


1

No. For instance, take $R = k[x^4,x^3y, xy^3, y^4]$. This is a $2$-dimensional domain which is not Cohen-Macaulay at the origin. In fact the depth of $R$ at the origin is $1$. Since $R$ is a domain, the ideal $0$ is unmixed. Let $f$ be a homogeneous non zero element in $R$. Then $R/(f)$ is $1$-dimensional with depth $0$. Therefore, the ideal $(0)$ in $R/(f)$ ...


1

It follows from the first part of the statement when the ring is local. Consider the following minimal free resolution for $R/I$: $$ 0 \to R^m \to R^n \to R \to R/I \to 0. $$ Since rank is additive, we see that $m + 1 = n$. Then the map $\phi: R^m \to R^n$ can be written as $m \times n$ matrix. The Theorem says $I = aI_n(\phi)$ where $a$ is an regular ...



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