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To find $h(X)$: $-3$ is a root of both $f(X)$ and $g(X)$ so $X+3$ divides both $f(X)$ and $g(X)$. Also you can write $x+3$ as a linear combination of $f(X)$ and $g(X)$. So let $h(X)=X+3$ then $(f)+(g)=(h)$. Note that $h$ is the greatest common divisor of $f$ and $g$. To find $k$, factor $f$ and $g$, then you get $f(X)=h(X)f_1(X)$ and $g(X)=h(X)g_1(X)$ (with ...

3

$I_a = (x,y) \cdot (z,x+a)$ and $(x,y) + (z,x+a) =(x,y,z,x+a) = (1)$, since it contains $a \in k \setminus \{0\}$. By the Chinese remainder theorem $$k[x,y,z]/I_a \cong k[x,y,z]/(x,y) \times k[x,y,z]/(z,x+a) \cong k[z] \times k[y]$$ is the product of two domains, hence reduced.

2

First, the assumption $(B/I)_t = (B/J)_t$ has to be made precise: I guess it's meant to say that the kernels of the projections $B_t \to (B/I)_t$ and $B_t \to (B/J)_t$ are equal, i.e. $IB_t = JB_t$. Now, suppose $i\in I$. Considering it as an element of $IB_t=JB_t$, there exist $j\in J$ and $k> 0$ such that $i = j/t^k$, i.e. $t^{k+l} i = t^l j$ in $B$ ...

2

For any $r\in R,$ and any $a\in I,$ we have $r.a\in I.$ Take $a=1.$ Then $r.1=r \in I;$ For any $r\in R.$

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It is more general: every ideal in $K[X]$ ($K$ a field) is principal, since on a polynomial ring over a field, you have a Euclidean division. Indeed, if $I\ne 0$ is an ideal, consider a polynomial $p$ in $I$ of smallest positive degree. For any polynomial $f\in I$, write the Euclidean division by $p$: there exist polynomials $q,r$ such that $$f=qp+r,\quad ... 1 Your question is equivalent to the following statement: Suppose deg f \geq deg g, f, g are homogeneous then f|_{x_i=1}=g|_{x_i=1} if and only if f=x_i^kg for some k\in \mathbb{N} Since we can divide f and g by x_i if necessary, we can assume both f and g are not multiples of x_i Then the statement can be proved by noticing ... 1 Suppose I is an ideal of \mathbb{R}[x] properly containing \langle x^2+1\rangle and let f(x)\in I\setminus\langle x^2+1\rangle. Then$$ f(x)=q(x)(x^2+1)+ax+b  for some $q(x)\in\mathbb{R}[x]$, $a,b\in\mathbb{R}$. Since $f(x)\notin\langle x^2+1\rangle$, you have $ax+b\ne0$, that is, not both $a$ and $b$ are zero. If $a=0$, then $b\ne0$ and so ...

1

Hint: $\mathbf R[x]$ is a P.I.D.. In a P.I.D., the ideal generated by an irreducible element is maximal.

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Hint: $R/I$ is isomorphic to the ring $S$ of $2\times2$ matrices with coefficients in $\mathbb{Z}/2\mathbb{Z}$. Can you find a surjective ring homomorphism $R\to S$? Can you tell what its kernel is?

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Hint: Ideals are closed under external multiplication. For all $r \in F$, $x \in I$, $rx \in I$. Just apply this to the elements you obtained. Complete Solution: Let $I$ be any non-zero ideal of a field $F$. Then, $I$ contains at least one non-zero element $x$. Since $F$ is a field, $x^{-1} \in F$, which implies that $x^{-1} x = 1 \in I$ (since $I$ is an ...

1

Hints, as requested: Let $I = \bigcap\limits_{j \in J} I_j$ be an intersection of ideals $I_j$ (where $J$ is an indexing set, finite or otherwise). Show that for any $x, y \in I$, $x - y \in I$. Use the fact that $x, y \in I_j$, $\forall j \in J$. Show that for any $r \in R$ and $x \in I$, $rx \in I$. Again, use the fact that $x \in I_j$, $\forall j \in ... 1 Let$I$be the greatest ideal which does not contain$X.$We shall show that$\forall A\subset X, A\in I \iff X\setminus A\not\in I.$So suppose$A\in I.$If$X\setminus A\in I,$then$X=I\cup (X\setminus A)\in I,$a contradiction. Thus$X\setminus A\not\in I.$Conversely, if$A\not\in I,$then adding$X\setminus A$to$I$generates an ideal containing$I.$... 1 Set$R=k[x_1,\ldots,x_n,y_1,\ldots,y_m]$. Denote the generators of$I$by$g_i = y_i - \psi_i$. Then, in the lexicographic order you chose, the leading monomial of$g_i$is$\operatorname{LM}(g_i) = y_i$. Because these monomials are pairwise coprime, the tuple$(g_1, \ldots, g_m)\$ is a Grobner basis.

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