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6

This ideal is in fact not prime: notice that $$ \begin{align*}(x_1^2 - x_4^2)(x_2^2 - x_3^2) &= x_1^2x_2^2 - x_1^2x_3^2 - x_4^2x_2^2 + x_3^2x_4^2 \\ &\equiv (x_1x_2)(x_3x_4) - (x_1x_3)(x_2x_4) - (x_4x_2)(x_1x_3) + (x_3x_4)(x_1x_2) \\ &= 0 \pmod I \end{align*}$$ If $I$ were prime, then one of $x_1 \pm x_4$, $x_2 \pm x_3 \in I$, but $I$ contains ...


5

A finite dimensional (commutative or not) algebra over a field is an artinian ring, because it certainly satisfies the descending chain condition on left or right ideals. Let $A$ be a commutative artinian ring and let $P$ be a prime ideal. Then $A/P$ is an artinian domain, which is a field: if $D$ is an artinian domain, consider $d\in D$ and the chain of ...


5

In the ring $\mathbb{Z}[X,Y]$, what happens if we take $U=V=(X,Y)$? Does $X^2 + Y^2$ lie in $U.V$?


5

These two pictures give a reasonable description of how cosets work, but they are not really suitable for explaining why normal subgroups and ideals are "analogous." These two diagrams say, in effect, "$G$ can be split up into chunks (cosets) and $gH$ is basically $H$ translated by $g$. These two translations are different in general, but when $H$ is ...


4

Your proof is not quite correct: $J^*$ may not be an element of the set $\Sigma$ under consideration (although $J^*$ is a prime ideal, since $S$ is totally ordered). Indeed, the claim is false: there exist rings with prime ideals of infinite height. If $p$ is such a prime ideal, then there exist prime ideals $p \supsetneq p_1 \supsetneq p_2 \supsetneq ...


4

The standard example is $R := k[x_1, x_2, \ldots]/(x_1^2, x_2^2, \ldots)$. Then $R$ has a single prime ideal $\mathfrak{m} = (x_1, x_2, \ldots)$ which is not finitely generated ($\mathfrak{m}$ is the only prime since it is maximal, but every prime ideal of $R$ must contain $x_1^2, x_2^2, \ldots$ and hence $\mathfrak{m}$).


3

You stated that $a=uva$ implies $uv = 1$. This is the cancellation law which holds in integral domains (but not for any ring). To prove this for integral domains, note that if $ca=cb$ for $c \neq 0$, then $c(a-b)=0$ so $a=b$. A counter-example for a ring which is not an integral domain is $\mathbb{Z}/4\mathbb{Z}$, where $2 \cdot 1 = 2 \cdot 3$.


3

Now that the mild ambiguity is resolved, let's address the question. We begin by showing maximality. Consider a non-trivial prime ideal, $\mathfrak{p}$ of $R$. If $x\not\in\mathfrak{p}$ is fixed, then considering the $\mod \mathfrak{p}$ reductions of $1, x,\ldots, x^{\dim_K(R)}=x^m$ we have a linear dependence by finite dimensionality of $R$. And since ...


3

How about you try something like this: let $ I $ be an ideal in $ M_n(K) $. Consider the set of $ e_1 I $ where $ e_1 = (1,0,\cdots) $. It is a vector space, say $ V $. Now you have to show that any matrix $ M $ in $ I $ has rows in $ V $. Well, to check the $ i$th row, look at $ e_i I = e_1 P I $ where $ P$ is some permutation matrix that takes $ 1 $ to $ ...


3

As pointed out in the comments, this is false in general. However, $\sqrt{I_1} = \sqrt{I_2}$ by Hilbert's Nullstellensatz.


3

Let $\;B\;$ be an ideal of $\;R\;$ containing $\;A\;$, and now define : $$B/A:= \{r+a\in R/A\;\;;\;\;r\in B\}$$ Prove that $\;B/A\;$ as defined above is an ideal of the factor ring $\;R/A\;$ . Important: don't forget that $\;A\le B\;$ ! Now, let $\;\overline B\le R/A\;$ (an ideal of $\;R/A\;$ ), and define $$B:=\{r\in R\;\;:\;\;r+A\in\overline B\}$$ ...


3

To solve the problem, consider $K=\mathbb{Z}[x,y]/\mathfrak{M}$ the residue field. If $\mathrm{char}(K)=p$ prime in $\mathbb{Z}$, then $[p]=[0]$, and $p\in \mathfrak{M}$. But on the other hand $K$ cannot contain $\mathbb{Q}$, because $\mathbb{Q}$ is not finitely generated as an algebra over $\mathbb{Z}$, while $\mathbb{Z}[x,y]$ is, and so also $K$.


3

Hints. $A/\Phi^{-1}(I)$ is isomorphic to a subring of $B/I$. The last one is a finite field. Finite integral domains are fields. An ideal is maximal iff its quotient ring is a field.


3

You can use part (a) to prove the second part. Let $x\in A\setminus M.$ Consider $N=M+xA=\{m+xa|\,m\in M, a\in A\}$. Clearly $N$ is an ideal of $A$ and contains $M.$ Thus it has be the whole ring $A.$ In particular $m+xa=1$ for $m\in M,\, a\in A$. thus $xa=xa=1-m.$ Now you can apply part (a);) it follows that $a$ is a unit. Done!


3

I assume that, for you, an associated prime $P$ of $R$ is a prime ideal that is the annihilator of some nonzero $m\in M$, and that you want to show that this implies that there is some submodule of $M$ isomorphic to $R/P$. Here is my hint: As part of the condition, we are given a nonzero $m\in M$. Using this, construct a homomorphism of modules $R\to M$, ...


2

Let $v_1=2x$ and $v_2=3x$ in $V$, then $v_1+v_2=5x \notin V$. The proof that $5x$ is not in $V$ is by the fact that $5$ is irreducible in $\mathbb Z$.


2

After "$1=b^{-1}b \in M'$" you go a bit off the rails, because you could simply conclude at this point "Thus $M'=A$, contradicting our hypothesis that $M'\neq A$. Earlier I misread something about the statement of (b) and tried an irrelevant example. Then BigM beat me to the most reasonable proof. I can't give a better one: so I recommedn you check that ...


2

Actually you want to prove the following: $$(X,Y)\cap(X-T,Z)=(XZ,YZ,X(X-T),Y(X-T)).$$ Let $f\in(X,Y)\cap(X-T,Z)$. Write $$f=Xf_1+Yf_2=(X-T)g_1+Zg_2\qquad(*)$$ Now sending $X$, $Y$, and $Z$ to $0$ obtain $g_1(0,0,0,T)=0$, so $g_1=Xh_1+Yh_2+Zh_3$. Plugging this in $(*)$ we get $$f=Xf_1+Yf_2=X(X-T)h_1+Y(X-T)h_2+Zk\qquad(**)$$ Sending $X$ and $Y$ to $0$ obtain ...


2

Suppose that $I := (xy,yz,zx)$ can be generated by two elements, i.e. $I = (f_1, f_2)$. If $m := (x,y,z)$ is the homogeneous maximal ideal, then $\overline{f_1}, \overline{f_2}$ generate $I/mI$ as a $k[x,y,z]/m \cong k$-module, so $\dim_k I/mI \le 2$. However, $\overline{xy}, \overline{yz}, \overline{zx}$ are $k$-linearly independent in $I/mI$: if $a ...


2

The radical ideals are precisely those which can be written as intersections of prime ideals. In a PID, the prime ideals are $\{0\}$ and $(p)$ for prime elements $p$. Hence, the radical ideals are $(0)$ and $\cap_i (p_i)$ for prime elements $p_i$. If there are infinitely many of them, the intersection is $(0)$. If not, we get $(\prod_i p_i)$. Hence, the ...


2

There are rings with ideals that are not definable (with parameters). The cardinality of the set of definable ideals is at most the cardinality of the ring (I am assuming the ring is infinite). Then for a counter example it suffices to take a ring $R$ that has $2^{|R|}$ ideals.


2

If you know some model theory of arithmetic, you may find the following example of a nondefinable ideal helpful. Let $\mathbb{Z}$ be the ordered ring of integers. Let $M$ be a proper elementary extension of $\mathbb{Z}$ such that $M$ contains an element which is divisible by every standard prime. (Such a model is a nonstandard model of $Th(\mathbb{Z})$ and ...


2

This is a possible answer that I have come up with. It can possibly be trimmed and edited for clarity. Let $R$ be a commutative domain and suppose that $A \subseteq R$ is an ideal maximal with respect to the property that $A^{-1} \not\subseteq R$. We show that $A$ must be prime. Take $r,s$ in $R$. Suppose that that their product $rs\in A$, but that ...


2

There are in fact many more non-principal height one primes in $R$: e.g. $(xw-y^2, yz-w^2)$ (recall that $(XW-Y^2, YZ-W^2, XZ-YW)$ is the ideal of the twisted cubic, and is a height $2$ prime in $k[X,Y,Z,W]$). One can see this example via algebraic geometry: $R$ is the coordinate ring of the affine cone over the quadric surface $Q = V(xz-yw) \subseteq ...


2

You concluded that $a,b \in X$, and therefore $z = a+b \in X$. This would not be true if $X$ were an arbitrary subset. For instance, in $\Bbb Z$, look at the ideals $(2)$ and $(3)$. $(2) + (3) = \Bbb Z$, but we could take $X = (2) \cup (3)$, which is not an ideal and does not contain $(2) + (3)$ (in particular, $1 = 3 - 2 \not\in X$. The proof you gave ...


2

Abstract hint: $\mathbb{K}$ field $\rightarrow \mathbb{K}[x] $ is a PID. Concrete hint: $\mathbb{Q}[x]$ is ED, so you can divide and you can look for a minimal polynomial.


1

Fix a point $c\in (0, 1)$. Define $I := \{ f \in C[0,1] : Z(f)$ contains an open nbd of $c \}$ where $Z(f)$ denotes the zero set of $f$. It can be easily checked that $I$ is a radical ideal and $I \subsetneq M_c := \{ f \in C[0,1] : f(c) = 0 \}$. Also note that $I \nsubseteq M_b$ for any $b \in [0, 1], b\neq c$. Now use the following fact: In a commutative ...


1

As Youngsu mentioned above, there is in general no relation between the primary components of $I$ and $J$, or the numbers $r$ and $s$ (and Youngsu has given the example of $I = m$ in a local ring for a case where $r < s$). For an example with $r > s$: take $R = k[x,y]$, $I = (x) \cap (x,y)^2 = (x^2,xy) \supseteq J = (x^2)$. On a positive note, there ...


1

When $m=3$ and $\mathbb F$ is infinite, there are no other obstructions besides the determinant. When $\mathbb F$ is finite, there are many others : for example if we put $\chi_{\mathbb F}(X)=\prod_{t\in {\mathbb F}^*} (X-t)$, $\chi(t)$ is zero iff $t$ is nonzero, so that the following $n$ polynomials are all obstructions : $$ ...


1

Picking $I \subseteq R$ to be maximal with respect to this condition, you would then have that for $0\neq r\in R-I$, $I\subsetneq (r)+I$. Thus we must have $((r)+I)^{-1}\subseteq R$. So what you're trying to show is a contradiction to the very condition you are starting with. Instead, consider that $((r)+I)^{-1}=R$. This means that the set ...



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