Problem:
Solution:
Let's discuss how to we prove two finitely generated ideals are equal. By the previous problem, we can show one finitely generated ideal $ I $ to be a subset of another ideal $ J $ by simply showing its basis of $ I $ is in the ideal $ J $.
Therefore, to we prove two finitely generated ideals are equal, we simply check if the two sets of basis are included in each other. The rest is just algebra.
Part (a)
$ \begin{eqnarray*} x + y &=& 1x + 1y & \in & \langle x, y \rangle \\ x - y &=& 1x + (-1)y & \in & \langle x, y \rangle \end{eqnarray*} $
Therefore $ \langle x + y, x - y \rangle \subset \langle x, y \rangle $.
$ \begin{eqnarray*} x &=& \frac{1}{2}(x + y) + \frac{1}{2}(x - y) & \in & \langle x + y, x - y \rangle \\ y &=& \frac{1}{2}(x + y) + \frac{-1}{2}(x - y) & \in & \langle x + y, x - y \rangle \end{eqnarray*} $
Therefore $ \langle x, y \rangle \subset \langle x + y, x - y \rangle $.
Together we proved $ \langle x + y, x - y \rangle = \langle x, y \rangle $
Part (b)
$ \begin{eqnarray*} x + xy &=& (1)x + (x)y & \in & \langle x, y \rangle \\ y + xy &=& (y)x + (1)y & \in & \langle x, y \rangle \\ x^2 &=& (x)x + (0)y & \in & \langle x, y \rangle \\ y^2 &=& (x)x + (y)y & \in & \langle x, y \rangle \\ \end{eqnarray*} $
Therefore $ \langle x + xy, y + xy, x^2, y^2 \rangle \subset \langle x, y \rangle $.
$ \begin{eqnarray*} x &=& (1)(x+xy) + (-x)(y + xy) + (y)(x^2) + (0)(y^2) & \in & \langle x + xy, y + xy, x^2, y^2 \rangle \\ y &=& (-y)(x+xy) + (1)(y + xy) + (0)(x^2) + (x)(y^2) & \in & \langle x + xy, y + xy, x^2, y^2 \rangle \end{eqnarray*} $
Therefore $ \langle x, y \rangle \subset \langle x + xy, y + xy, x^2, y^2 \rangle $.
Together we proved $ \langle x + xy, y + xy, x^2, y^2 \rangle = \langle x, y \rangle $
Part (c)
$ \begin{eqnarray*} 2x^2 + 3y^2 - 11 &=& (2)(x^2 - 4) + (3)(y^2 - 1) & \in & \langle x^2 - 4, y^2 - 1 \rangle \\ x^2 - y^2 - 3 &=& (1)(x^2 - 4) + (-1)(y^2 - 1) & \in & \langle x^2 - 4, y^2 - 1 \rangle \end{eqnarray*} $
Therefore $ \langle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \rangle \subset \langle x^2 - 4, y^2 - 1 \rangle $.
$ \begin{eqnarray*} x^2 - 4 &=& (\frac{1}{5})(2x^2 + 3y^2 - 11) + (\frac{3}{5})(x^2 - y^2 - 3) & \in & \langle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \rangle \\ y^2 - 1 &=& (\frac{1}{5})(2x^2 + 3y^2 - 11) + (\frac{-2}{5})(x^2 - y^2 - 3) & \in & \langle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \rangle \end{eqnarray*} $
Therefore $ \langle x^2 - 4, y^2 - 1 \rangle \subset \langle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \rangle $.
Together we proved $ \langle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \rangle = \langle x^2 - 4, y^2 - 1 \rangle $
Solution:
Let's discuss how to we prove two finitely generated ideals are equal. By the previous problem, we can show one finitely generated ideal $ I $ to be a subset of another ideal $ J $ by simply showing its basis of $ I $ is in the ideal $ J $.
Therefore, to we prove two finitely generated ideals are equal, we simply check if the two sets of basis are included in each other. The rest is just algebra.
Part (a)
$ \begin{eqnarray*} x + y &=& 1x + 1y & \in & \langle x, y \rangle \\ x - y &=& 1x + (-1)y & \in & \langle x, y \rangle \end{eqnarray*} $
Therefore $ \langle x + y, x - y \rangle \subset \langle x, y \rangle $.
$ \begin{eqnarray*} x &=& \frac{1}{2}(x + y) + \frac{1}{2}(x - y) & \in & \langle x + y, x - y \rangle \\ y &=& \frac{1}{2}(x + y) + \frac{-1}{2}(x - y) & \in & \langle x + y, x - y \rangle \end{eqnarray*} $
Therefore $ \langle x, y \rangle \subset \langle x + y, x - y \rangle $.
Together we proved $ \langle x + y, x - y \rangle = \langle x, y \rangle $
Part (b)
$ \begin{eqnarray*} x + xy &=& (1)x + (x)y & \in & \langle x, y \rangle \\ y + xy &=& (y)x + (1)y & \in & \langle x, y \rangle \\ x^2 &=& (x)x + (0)y & \in & \langle x, y \rangle \\ y^2 &=& (x)x + (y)y & \in & \langle x, y \rangle \\ \end{eqnarray*} $
Therefore $ \langle x + xy, y + xy, x^2, y^2 \rangle \subset \langle x, y \rangle $.
$ \begin{eqnarray*} x &=& (1)(x+xy) + (-x)(y + xy) + (y)(x^2) + (0)(y^2) & \in & \langle x + xy, y + xy, x^2, y^2 \rangle \\ y &=& (-y)(x+xy) + (1)(y + xy) + (0)(x^2) + (x)(y^2) & \in & \langle x + xy, y + xy, x^2, y^2 \rangle \end{eqnarray*} $
Therefore $ \langle x, y \rangle \subset \langle x + xy, y + xy, x^2, y^2 \rangle $.
Together we proved $ \langle x + xy, y + xy, x^2, y^2 \rangle = \langle x, y \rangle $
Part (c)
$ \begin{eqnarray*} 2x^2 + 3y^2 - 11 &=& (2)(x^2 - 4) + (3)(y^2 - 1) & \in & \langle x^2 - 4, y^2 - 1 \rangle \\ x^2 - y^2 - 3 &=& (1)(x^2 - 4) + (-1)(y^2 - 1) & \in & \langle x^2 - 4, y^2 - 1 \rangle \end{eqnarray*} $
Therefore $ \langle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \rangle \subset \langle x^2 - 4, y^2 - 1 \rangle $.
$ \begin{eqnarray*} x^2 - 4 &=& (\frac{1}{5})(2x^2 + 3y^2 - 11) + (\frac{3}{5})(x^2 - y^2 - 3) & \in & \langle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \rangle \\ y^2 - 1 &=& (\frac{1}{5})(2x^2 + 3y^2 - 11) + (\frac{-2}{5})(x^2 - y^2 - 3) & \in & \langle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \rangle \end{eqnarray*} $
Therefore $ \langle x^2 - 4, y^2 - 1 \rangle \subset \langle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \rangle $.
Together we proved $ \langle 2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \rangle = \langle x^2 - 4, y^2 - 1 \rangle $
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