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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)....


3

For the first, the answer is no and for the latter the answer is yes and your vague feeling is correct. Let me give you a simple example to illustrate why the first is false. Take $V$ to be the quadric defined by $xy-zw=0$ in projective 3-space. Take $L$ to be a line, say given by $x=z=0$ contained in $V$. One can prove (and not difficult) that there is no ...


3

This is a long and messy proof, and I don't think there is any easier way to deal with it unless some additional knowledge about commutative algebra is available. If I made a mistake somewhere please let me know... If you prefer to prove it yourself just read the claims and the lemma and try to prove them yourself without reading my proof (which is actually ...


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

The nilpotent elements of the product are obtained as the tuples of the nilpotent elements of single factors.


1

Hint: $R[x]/I \cong R$. 1- $I$ is prime iff $R$ is domain. 2- $I$ is maximal iff $R$ is field. 3- use (2) with the fact: $R$ is a commutative integral ring, $R[X]$ is a principal ideal domain imply $R$ is a field. 4- By the link above, $R$ is a field.


1

The following fact is useful and you should remember it: If an ideal $I \unlhd R$ contains $1 \in R$, then $I=R$. The proof is trivial: $1 \in I$ implies, by strong closure of multiplication, that $1 \cdot x \in I$ for any $x \in R$. But $1 \cdot x =x$, so $x \in I$. So $R \subset I$, and hence $I=R$. Now let $a$ be a unit, and define $I=(a)$. To ...


1

Consider an ideal $I=(f_1,\dotsc, f_s)$ in a noetherian factorial ring and let $f$ be the greatest common divisor of the $f_i$. Then $I$ is principal if and only if $f \in (f_1, \dotsc, f_s)$ holds. In practice, working in the polynomial ring, figuring out whether this holds, you need Groebner bases. Thus, in practice, you can also directly use the ...



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