Rewrite the polynomial \[ 4x^6 + 3x^4y^3z + 2x^5yz^2 + x^4y^2z^2 \] using each of the lex, grlex, and grevlex orders.
lex: \(4 x^{6} + 2 x^{5} y z^{2} + 3 x^{4} y^{3} z + x^{4} y^{2} z^{2}\)grlex: \(2 x^{5} y z^{2} + 3 x^{4} y^{3} z + x^{4} y^{2} z^{2} + 4 x^{6}\)grevlex: \(3 x^{4} y^{3} z + 2 x^{5} y z^{2} + x^{4} y^{2} z^{2} + 4 x^{6}\)Compute the quotient and remainder of \(4x^4-5x+3\) on division by \(2x^2-x+1\).
We find a quotient of \(2 x^{2} + x - \frac{1}{2}\) and remainder \(-\frac{13}{2} x + \frac{7}{2}\).
Consider the following polynomials: \[\begin{align*} f_1 &= x^4 + 4x^3 + 5x^2 + 4x + 1\\ f_2 &= 2x^4 + 7x^3 + 4x^2 - 2x - 1\\ f_3 &= x^3 + 6x^2 + 10x + 3 \end{align*}\] Compute \(\gcd(f_1,f_2,f_3)\).
First we compute the gcd of \(f_1\) and \(f_2\) via the Euclidean algorithm. The remainder of \(f_2\) on division by \(f_1\) is \[ r_1 = -x^3 - 6x^2 - 10x - 3.\] The remainder of \(f_1\) on division by \(r_1\) is \[ r_2 = 7x^2 + 21x + 7.\] Note that \(r_2 = 7(x^2+3x+1)\). The remainder of \(r_1\) on division by \(r_2/7\) is \(0\). Thus \[\gcd(f_1, f_2)=x^2+3x+1.\] Now we compute \(\gcd(f_3,x^2+3x+1)\). The remainder of \(f_3\) on division by \(x^2+3x+1\) is \(0\). Thus \[\gcd(f_1,f_2,f_3) = x^2+3x+1.\]
Consider the following polynomials: \[\begin{aligned}[c] f &= x^4 + x^3 - x^2 + 2x - 6 \end{aligned} \quad\quad\quad\quad \begin{aligned}[c] g_1 &= x^3-3x^2+2x-6\\ g_2 &= x^3-2x^2+2x-4 \end{aligned}\] Determine whether \(f\) is in the ideal \(\langle g_1, g_2 \rangle\).
First we compute \[\gcd(g_1,g_2)\] via the Euclidean algorithm. The remainder of \(g_1\) on division by \(g_2\) is \(-(x^2+2)\), which divides \(g_2\). Thus \(\gcd(g_1,g_2)=x^2+2\) or, equivalently, \(\langle g_1,g_2 \rangle = \langle x^2+2\rangle\). The remainder of \(f\) on division by \(x^2+2\) is \(0\) so \(f\) is in the ideal.
Consider the following polynomials: \[\begin{aligned}[c]
f &= x^3y^2-xy^3+3xy\\
\end{aligned}
\quad\quad\quad\quad
\begin{aligned}[c]
g_1 &= xy^2+2y\\
g_2 &= x^2y-xy\\
g_3 &= 3y^2-4
\end{aligned}\] Compute the remainder of \(f\) on division by the ordered set \(G=(g_1,\ g_2,\ g_3)\) using the division algorithm and the lex order.
We apply the division algorithm, indicating the state after each iteration through the main loop in the following table:
| \(p\) | \(q_1\) | \(q_2\) | \(q_3\) | \(r\) |
|---|---|---|---|---|
| \(x^3y^2-xy^3+3xy\) | \(0\) | \(0\) | \(0\) | \(0\) |
| \(-2x^2y-xy^3+3xy\) | \(x^2\) | \(0\) | \(0\) | \(0\) |
| \(-xy^3+xy\) | \(x^2\) | \(-2\) | \(0\) | \(0\) |
| \(xy+2y\) | \(x^2-y\) | \(-2\) | \(0\) | \(0\) |
| \(2y^2\) | \(x^2-y\) | \(-2\) | \(0\) | \(xy\) |
| \(\frac{8}{3}\) | \(x^2-y\) | \(-2\) | \(\frac{2}{3}\) | \(xy\) |
| \(0\) | \(x^2-y\) | \(-2\) | \(\frac{2}{3}\) | \(xy+\frac{8}{3}\) |
Thus, the remainder is \(xy+\frac{8}{3}\).
Determine whether the following ideals are equal: \[\begin{align*} I &= \langle x, y^2 \rangle\\ J &= \langle x^3+y^2, 2x+y^2, xy^3-y^3 \rangle \end{align*}\]
From the equations \[\begin{align*} x^3+y^2 &= (x^2)(x) + (1)(y^2)\\ 2x+y^2 &= (2)(x) + (1)(y^2)\\ xy^3-y^3 &= (y^3)(x)+(-y)(y^2) \end{align*}\] we conclude that \(J \subseteq I\).
For the converse, note that \(x+\frac{1}{2}y^2 \in J\). Thus if \(xf+g \in J\) then \[xf+g-f\left(x+\frac{1}{2}y^2\right)=-\frac{1}{2}y^2f+g \in J.\] Thus \[xy^3-y^3 \in J \implies -\frac{1}{2}y^5-y^3 \in J\] and \[\begin{align*} & x^3+y^2 & \in J\\ \implies -\frac{1}{2}x^2y^2+y^2 & \in J\\ \implies \frac{1}{4}xy^4+y^2 & \in J\\ \implies -\frac{1}{8}y^6+y^2 & \in J. \end{align*}\] We see that \(\gcd(-\frac{1}{2}y^5-y^3,-\frac{1}{8}y^6+y^2)=y^2\) either via the Euclidean algorithm or by inspection since \[\begin{align*} -\frac{1}{2}y^5-y^3 &= \left(-\frac{1}{2}y^2-1\right)y^3\\ -\frac{1}{8}y^6+y^2 &= \left(-\frac{1}{8}y^4+1\right)y^2 \end{align*}\] Thus \(y^2 \in J\). Thus \(2x+y^2-y^2=2x \in J\). We conclude that \(I \subseteq J\).
Thus the two ideals are equal.
Let \(I=\langle f_1, \ldots, f_s \rangle\) be an ideal of \(k[x_1,\ldots, x_n]\). Prove that \(I=k[x_1,\ldots,x_n]\) if and only if there exist polynomials \(h_1, \ldots, h_s \in k[x_1,\ldots,x_n]\) such that \(h_1f_1 + \cdots + h_sf_s = 1\).
Suppose \(I=k[x_1,\ldots,x_n]\). By definition of generating set, every element of \(I\) can be written in the form \(h_1f_1 + \cdots + h_sf_s\) for some set of polynomials $h_1, , h_s. Since \(I\) contains every polynomial, this is true in particular for \(1\).
For the converse, suppose \[1=h_1f_1 + \cdots + h_sf_s\] for some polynomials \(h_1, \ldots, h_s\). We have \(I \subseteq k[x_1,\ldots, x_n]\) from the definition of ideal. Now suppose \(g \in k[x_1,\ldots,x_n]\). Multiplying both sides of the equation above by \(g\) we obtain \[g=(h_1f_1 + \cdots + h_sf_s)g=(h_1g)f_1 + \ldots (h_sg)f_s.\] Thus \(g\) is a polynomial linear combination of \(f_1,\ldots, f_s\); in other words \(g \in I\). Thus \(k[x_1,\ldots,x_n] \subseteq I\). Thus \(I=k[x_1,\ldots,x_n]\) as desired.
Let \(I,J\) be ideals in \(k[x_1,\ldots,x_n]\). Prove that \(I \cap J\) is also an ideal.
Recall that \(I\) is an ideal if and only if it satisfies
Since \(0 \in I\) and \(0 \in J\), we see that \(0 \in I \cap J\).
Suppose \(f,g \in I \cap J\). Then \(f,g \in I\) and so \(f+g \in I\). Similarly, \(f+g \in J\). Thus \(f+g \in I \cap J\).
Suppose \(f \in I \cap J\) and \(h \in k[x_1,\ldots,x_n]\). Then \(f \in I\) and so \(hf \in I\). Similarly, \(hf \in J\). Thus \(hf \in I \cap J\).
Thus \(I \cap J\) is an ideal as desired.