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5

Assuming $$\Bbb Z_7[i]=\Bbb F_{7^2}=\Bbb F_7[x]/\langle\,x^2+1\,\rangle =\text{ the field with $\;49\;$ elements}$$ Since a field only has the two trivial ideals (in fact, this characterizes fields among integer domains), and since clearly $\;\langle\,1+i\,\rangle\;$ is not the zero ideal, it then must be the whole ring, and finally $$\Bbb ...


5

Let $k$ be a field, let $A=k[x,y]$, let $B=k[z]$, and let $f:A\to B$ be defined by $f(x)=f(y)=z$. Consider the ideals $I=(x)$ and $J=(y)$ of $A$. Then $I\cap J=(xy)$ so $f(I\cap J)=(z\cdot z)=(z^2)$, whereas $f(I)\cap f(J)=(z)\cap (z)=(z)$. So in this case we have $$f(I\cap J)\subsetneq f(I)\cap f(J)$$


5

The ring $\mathbf Z[w]$ is isomorphic to the quotient ring $A=\mathbf Z[x]/(x^2+x+1)$. To show $2$ is prime, you have to show $A/2A$ is an integral domain. We have: $$A/2A \simeq \mathbf Z[x]/(2,x^2+x+1)\simeq \mathbf Z/2\mathbf Z[x]/(x^2+x+1)$$ so it is enough to show the polynomial $x^2+x+1$ is irreducible over $\mathbf Z/2\mathbf Z$, which means it has ...


5

First note that $J$ is the ideal generated by all idempotents in $P$, so $J\subset P$. Since $a^2-a\in J$ we have $a^2-a=a_1t_1+\cdots+a_nt_n$ with $t_i\in P$, $t_i^2=t_i$. Then $(1-t_1)\cdots(1-t_n)(a^2-a)=0$, so $(1-t)(a^2-a)=0$ for some $t\in P$, $t^2=t$. (In particular, $t\in J$.) Since $a^2-a\in P$ we have two possibilities: $(i)$ $a\in P$; from ...


5

Since $$\bigcap_{k \in \mathbb{N}_0} p^k \mathbb{Z}_p = 0,$$ for an arbitrary ideal $I \neq 0$ $$\bigcap_{k \in \mathbb{N}_0} I \cap p^k \mathbb{Z}_p = 0.$$ Since $I \cap p^0 \mathbb{Z}_p = I$, but the intersection over all $k \in \mathbb{N}_0$ is $0$, for some finite $n+1 \in \mathbb{N}$ the intersection must become $\subsetneq I$. Let this $n$ be ...


4

Hint: Remember that $m$ and $n$ can be negative... The general result that's relevant here is Bezout's identity (Wikipedia link).


4

If an element is not a unit, the ideal it generates is not all of $R$. Now, every proper ideal is contained in a maximal ideal.


3

You have almost solved it. $R/I$ is not even an integral domain. To see this use the factorization you got for $z = 5 + 3i.$ EDIT: We have the following factorization $5 + 3i = (4 - i)(1+i)$ in $R$ with $5 + 3i$ doesn't divide $4 - i$ and $1 + i.$ So in the quotient ring $R/I,$ the image $\overline{4 - i}$ and $\overline{1+i}$ are non-zero. But $\overline{4 ...


3

$I$ is a maximal ideal means if $J$ is any ideal of $R$ with $I \subseteq J$ then either $I = J$ or $J = R.$ In this case, take $J = I + (a).$ By the given condition, $J \neq I.$ So $J = R.$


3

A way to see this is that $K[x,y]/M \cong K$ is a field, while $K[x,y]/P \cong K[y]$ is not a field but an integral domain. This assumes you know the criterion or definition that $I$ is a maximal ideal of $R$ if and only of $R/I$ is a field, and that $I$ is a prime ideal of $R$ if and only of $R/I$ is an integral domain.


3

The Nullstellensatz says that $I(V(f)) = \sqrt{(f)}$, so $V(f) = V(g)$ implies (applying $I$ on both sides) that $\sqrt{(f)} = \sqrt{(g)}$, where I note $\sqrt{J}$ the radical of an ideal $J$.


3

First, to construct the quotient space, you want the relation $\sim$ defined by $x\sim y\iff x-y\in I$ to be an equivalence relation. Then you want to be able to define operations for the cosets so that $[x]+[y]=[x+y]$ and $[xy]=[x][y]$ are well defined and give $A/I$ the structure of a ring. From these requirements it follows that $I$ is an ideal. That is, ...


3

You can show that of the complement of $2 \mathbf{Z}_{(2)}$ is $\mathbf{Z}_{(2)}^{\times}$, showing thereby that $2 \mathbf{Z}_{(2)}$ is the unique maximal ideal of $\mathbf{Z}_{(2)}$. This show that $\mathbf{Z}_{(2)}$ is a local ring. "Alternatively", consider the "map" from $\mathbf{Z}_{(2)}$ to $\mathbf{F}_2$ defined by sending an element $x= ...


3

In a comment, you say that you took a question about $\Bbb F_5[i]$ and modified it. This is the core of our difficulty. By $\Bbb F_5[i]$ I’m sure the author meant $\Bbb F_5[X]/(X^2+1)$. But this ring is not a field. So the question makes very good sense: calling $i$ the image of $X$ in this ring, we see that $1+i$ generates a group of order $4$. Since $1+i$ ...


2

$\operatorname{Supp}M/xM=V(\operatorname{Ann}M/xM)$; $\operatorname{Supp}M/xM=\operatorname{Supp}(R/xR\otimes_R M)=V(x)\cap\operatorname{Supp}M=V(xR+\operatorname{Ann}M)$; $V(\operatorname{Ann}M/xM)=V(xR+\operatorname{Ann}M)$. Let $\bar R=R/\operatorname{Ann}M$, and $\bar x$ the residue class of $x$. Then $\bar R/\bar x\bar R=R/(xR+\operatorname{Ann}M)$, so ...


2

Suppose $a\in\operatorname{Rad}(\operatorname{Rad}(I))$. Then $a^m\in\operatorname{Rad}(I)$, for some $m$. Then $(a^m)^n\in I$ for some $n$.


2

Hints. $V(IJ)=V(I)\cup V(J)=V(0)$, so $\sqrt{IJ}=\sqrt{0}$. $R$ noetherian $\Rightarrow$ $IJ$ finitely generated, so $(IJ)^n=0$ for some $n$. Then use the CRT.


2

In order to better distinguish between elements and ideals I'll use Fraktur letters for the latter: $\mathfrak{b}$ and $\mathfrak{c}$. With $\mathfrak{c}/\mathfrak{b}$ the ideal $$ \{x+\mathfrak{b}:x\in \mathfrak{c}\} $$ is meant. You have to prove that $\mathfrak{c}/\mathfrak{b}$ is an ideal of $A/\mathfrak{b}$ Every ideal of $A/\mathfrak{b}$ is of this ...


2

Normally, one does not define algebraic sets associated to radical ideals only, but rather defines them for any ideal or even for any set, like $V(S)= \{p \ \colon p(s) = 0 \ \forall s \in S\}$. Then, one shows that every algebraic set is in fact already associated to the algebraic set given by some radical ideal. Since $V(S) = V(\langle S \rangle)$ and ...


2

(2) implies (1), for an ideal is maximal iff its quotient ring is a field. As for (2), the map $e_\alpha \colon X \to \mathbb{R}$, $e_\alpha(f) = f(\alpha)$, is a ring homomorphism with kernel $I$.


2

For $ab - 1 \in M$. The previous step was $(a + M)(b + M) = 1 + M$. Note that $(a + M)(b + M) = ab + M$ by definition so $ab + M = 1 + M$. This says in particular that $ab \in 1 + M$ so there is an $m \in M$ such that $ab = 1 + m$. But then $ab - 1 = m$ for some $m \in M$. For $1 \in I$ note that $a \in I$ and $I$ is an ideal so $ab \in I$. Also $m \in ...


2

It seems that you are trying to show that if $R/M$ is a field, then $M$ is maximal. So you assume that $I$ is another ideal of $R$ with $M \subset I$, and you want to show that $I = R$. Note that $(a+M)(b+M) = ab + M$ by definition of multiplication on the quotient ring, so $(a+M)(b+M) = 1 + M \Leftrightarrow ab+M = 1 + M \Leftrightarrow (ab - 1) + M = M ...


2

We first prove a separate result. Let $R$ be an integral domain which contains a field $k$. If $R$ has finite dimension as a vector space over $k$, then $R$ is a field. To see this, define, for each nonzero $v\in R$, a function $T_v:R\to R$ as $T_v(x)=vx$ for all $x\in R$. It is clear that $T$ is a linear transformation. Since $R$ is an integral domain, ...


2

0) I don't think there is a reasonable general, non tautological answer to this question. Here, however, are some generalities about your problem. 1) The basic tool is certainly to use the morphism of schemes $f:\mathbb A^n_R\to \operatorname {Spec} R$ associated to the inclusion $R\subset R[x_i,\dots, x_n]$. In some sense one can reduce the problem to ...


2

The only maximal ideal of $C[X]/X^2$ is the image of the ideal $(X) \subset C[X]$ in $C[X]/X^2.$ It is normally written as the ideal generated by $X + (X^2)$ or simply $(\bar X)$ where $\bar X$ denotes the image of $X$ in $C[X]/X^2$ (under the canonical projection). Sometimes people just write $(X)$ in $C[X]/X^2$ if there is no confusion. But one should keep ...


2

$m\mathbf{Z} + n\mathbf{Z} = d \mathbf{Z}$ where $d = \textrm{gcd}(m,n)$ (direct use of Bézout's theorem), so here you get $\mathbf{Z}$.


2

(i) looks correct. (ii): Consider $\varphi : \mathbb Z \to \mathbb Q$ the inclusion map. This is a ring homomorphism. $0$ is a maximal ideal in $\mathbb Q$, but $\varphi^{-1}(0) = 0$ is not a maximal ideal in $\mathbb Z$: pick any prime $p$ and notice that $0 \subsetneq (p) \subsetneq \mathbb Z$.


2

(ii): Take any integral domain $D$ which is not a field and let $k$ be it's field of fractions. Consider the natual inclusion $i :D[t] \to k[t].$ The ideal $M = tk[t]$ is a maximal ideal in $k[t].$ But $i^{-1}(M) = tD[t]$ is not a maximal ideal in $D[t].$


2

Consider the set of $x^{v}$ that are not in $in_{<}(I)$. Linear independence is easy - if a linear combination $F$ of the set is zero in $S/I$ then when considered as an element of $S$ we have $F\in I$ so one of them is in $in_{<}(I)$, contradiction. Spanning is harder. Consider the set $H$ of non zero $f\in S$ that are non zero in $S/I$ and are not ...


2

As you already told, $\operatorname{in}_{<}(I)$ is the ideal generated by $\operatorname{in}_{<}(f)$ for all $f\in I$. This means that an element of $\operatorname{in}_{<}(I)$ is a linear combination of the form $g_1\operatorname{in}_{<}(f_1)+\cdots+g_r\operatorname{in}_{<}(f_r)$ with $f_i\in I$ and $g_i\in K[x_1,\dots,x_n]$. If a monomial ...



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