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8

In your general case, we have $I+J=\langle f_1,\ldots,f_i,g_1,\ldots,g_j\rangle$. In this case, that means we have the ideal $$I+J=\langle x_0+x_1,\ldots,x_0+x_n,x_0x_1\cdots x_n\rangle$$Now note that $$x_0x_1\cdots x_n+(x_0+x_1)x_0x_2x_3\cdots x_n= x_0^2x_2x_3\cdots x_n\in I+J$$(remember that our coefficients are in $\Bbb Z_2$, so $+$ and $-$ are the same)....


4

In an integral domain, you have the following four equivalent definitions for a nonzero nonunit $a$ to be irreducible. $a = bc \Rightarrow (a) = (b)$ or $(a) = (c)$. $a = bc \Rightarrow a$ is a unit multiple of $b$ or $c$. $(a)$ is maximal among the proper principal ideals. $a = bc \Rightarrow b$ or $c$ is a unit. However, in commutative rings in general,...


3

Consider the map $f:k[u,v]\rightarrow k[x,y]$ sending $u\mapsto x^2$, $v\mapsto xy$. Note that $f(u)$ and $f(v)$ are homogeneous polynomials of degree $2$. Also $f$ maps the monomials $u^av^b$ to distinct monomials in $k[x,y]$, so $f$ is injective. Thus $I=0$ and every point is a common zero of $I$. However there are no $x,y\in k$ for which $(x^2,xy)=(0,1)$.


3

Ok Slup, here goes. Let $R$ be any commutative ring and let $A$ be a polynomial ring over $R$. Let $P$ be any projective module over $R$. Then Quillen (and Suslin a bit later in this generality) proved that if for every maximal ideal $\mathfrak{m}$ of $R$, $P_{\mathfrak{m}}$ is of the form $Q\otimes_{R_{\mathfrak{m}}} A_{\mathfrak{m}}$ for some projective $...


2

Finding a free resolution can be a difficult task by hand, because the simple linear algebra task of finding the kernel of a map given by a matrix is no longer simple when the base ring isn't a field. That said, modulo the problem of being able to say for sure that you have found a complete set of generators for the kernel of a map (which the computer does ...


2

Pick any element $f$ in Jacobson radical but not in nil radical (say, $(\bar2,\bar2,\bar2,\dots)$), the preimage in $R$ of any maximal ideal of $R_f$ will do. It seems impossible to give a concrete construction. A maximal ideal of $R$ containing a non-maximal prime ideal mustn't be of the form $\left<(\dots, \bar1, \bar 2, \bar1, \dots)\right>$. So ...


2

Theorem: If Spec$(R)$ is Noetherian, then so is Spec$(R[X])$. [Theorem 2.5 in ``Rings with Noetherian spectrum'' by Ohm and Pendleton] So for the example, take $R[X]$, where $R$ is any non-Noetherian ring with Noetherian spectrum.


2

Let us say that a ring $A$ satisfies condition $(S)$ if every finite type projective $A$-module is free. Clearly a necessary condition for $(S)$ to hold is $K_0(A)=\mathbf{Z}$. Example. (i) If $A$ is local Noetherian then $(S)$ holds. (ii) If $A$ is a Dedekind domain then $K_0(A)=\mathbf{Z}\oplus\mathrm{Pic}(A)$, and thus a necessary condition for $(S)$ is $...


2

In my opinion, the exercise's approach is going about the problem backwards. (or, the exercise just wanted you to do something trivial to obtain $I+J$ rather than do any calculation) The ring $R/J$ is so simple that it is much easier to compute $$ R / (I+J) \cong (R / J) / I $$ and then if you want to, find $I+J$ from that. On the right hand side, by $(R/...


2

A polynomial in $n$ variables is symmetric exactly when you can do any permutation on the variables and leave the polynomials unchanged. In other words, you must have $$f(x_1, \dots, x_n) = f(x_{\sigma(1)}, \dots, x_{\sigma(n)})$$ for any permutation $\sigma \in \mathfrak{S}_n$. The standard symmetric polynomial verify this, so a linear combination of them ...


1

Yes. In any commutative ring, the set of zero-divisors is the union of the prime ideals in $\operatorname{Ass}A$. An artinian ring has Krull dimension $0$, hence a local artinian ring has only $1$ prime ideal, and $\operatorname{Ass}A=\operatorname{Max}A=\operatorname{Spec}A$ is the set of zero-divisors in $A$.


1

The same logic from this solution to a special case applies here, except that you disregard the comments about $F_2$ and $F_3$ and just settle for all the quotients by prime ideals being fields. The battle plan is, briefly: The intersection of all prime ideals is the zero ideal The quotient by any prime ideal is in fact a field. The ring embeds into a ...


1

If the ring is finite, then it is an artinian reduced ring and therefore splits as the product of artinian local rings. Every term of the product must be reduced and an artinian local reduced ring is a field, proving your claim.


1

Hint. Show that $p_2p_3 \dotsb p_n$ is not contained in $p_1$ and take some element according to this fact.


1

For a vector space, having finite dimension is the same as being finitely generated. One direction is obvious: if the vector space has a finite basis it clearly is finitely generated. Suppose $V$ is finitely generated. From any finite spanning set $S$, one can extract a minimal spanning set $B$, in the sense that $B\subseteq S$ is a spanning set and no ...


1

It should be true for any PID. See the book by Lam, Serre's Conjecture


1

Since $1 \in A_0$, every $\delta_n$ with $n \ge 1$ must be zero. The highest degree $\delta_i$ will be the product of the highest degree $a_i$ and the highest degree $b_i$, and since $A$ is a domain this is non-zero. Therefore since the degrees add the highest $a_i$ and $b_i$ must both be in degree $0$, i.e. $a$ and $a^{-1}$ are homogeneous degree $0$.


1

Consider the exact sequence $0\to m/m^2\to A/m^2\to A/m=k\to 0$. Use $\nu$ to get a push out $0\to k\epsilon\to k[\epsilon]\to k\to 0$. Check whatever is required for this map $A\to k[\epsilon]$.


1

Here's another hint. After lots of work (at least for me), you find that you don't really need the last generator of the ideal, because $$ \frac{y}{3}\left(x^2+2y-3\right) + \left(\frac{y+x}3 \right)(y^2-yx) = y^3-y=y(y-1)(y+1). $$ Hence your ideal has two generators $$ I = (x^2+2y^2-3,y^2-yx). $$ The first factor is irreducible, but the second factor is ...


1

Here's a tool that may be helpful for this kind of stuff. Proposition: Let $I,J,K$ be ideals of a ring $R$ with $J + K = R$. Then $$I + JK = (I + J)(I + K) = (I + J) \cap (I + K).$$ Proof: $(I+J)(I+K) = I^2 + I(J+K) + JK = I + JK$. Edit: I probably should mention that the last equality is simply a special case of the fact that $IJ = I \cap J$ if $I + ...


1

Here's how in Macaulay2. Let's say your ideal is $I=(x^2+2y^2-3,y^2-yx)$. Then I claim that $y^3-y$ is in the ideal. Then we can write i59 : f 2 2 o59 = x + 2y - 3 o59 : R i60 : g 2 o60 = - x*y + y o60 : R i61 : f = x^2+2*y^2-3 2 2 o61 = x + 2y - 3 o61 : R i62 : g = y^2-x*y 2 o62 = - x*y + ...


1

The answer is negative. Let $R$ be a Noetherian UFD such that there is a rank 2 non-free finitely generated projective module $M$; various examples of such can be found in the answers to this MO question. Further, let $Q$ be the field of fractions of $R$. As $M$ is of rank 2, it embeds into $Q\oplus Q$. Let $N$ be the intersection of $M$ with the direct ...



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