At 18:50, The idea of the proof is that: Suppose $I$ can not be written as an n-1 ideal. Then we can create $y = x_1x_2...x_n + x_1x_3...x_n + x_1x_2...x_{n-1}$ such that $y \in I$ but $y in P_j$ for any $j$. So, ours assumption that "$I$ can not be written as an n-1 ideal" is false i.e. $I$ can be written as an n-1 ideal. Then, By assumption, any if $I$ can be written as an n-1 ideal $I$ must be contained in some of those (n-1) ideal.
Sir I have some doubt about the proof of last result, i.e. if P = I1 intersection I2, then P=I1 or P=I2. P subset of I1 int. I2 implies P is subset of one of I1 or I2; and I1 int. I2 subset of P implies one of I1 or I2 is a subset of P. So there is a possibility of P subset of I1 and I2 subset of P. Then how can we say P=I1 or P=I2 ?
If P ⊆ I_1 int I_2, then P is a subset of BOTH I_1 and I_2. Then if I_1 int I_2 ⊆ P, then either I_1 ⊆ P or I_2 ⊆ P by primeness of P. So we have either I_1 = P or I_2 = P. Does this make sense?
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At 18:50, The idea of the proof is that: Suppose $I$ can not be written as an n-1 ideal. Then we can create $y = x_1x_2...x_n + x_1x_3...x_n + x_1x_2...x_{n-1}$ such that $y \in I$ but $y
in P_j$ for any $j$. So, ours assumption that "$I$ can not be written as an n-1 ideal" is false i.e. $I$ can be written as an n-1 ideal. Then, By assumption, any if $I$ can be written as an n-1 ideal $I$ must be contained in some of those (n-1) ideal.
Sir I have some doubt about the proof of last result, i.e. if P = I1 intersection I2, then P=I1 or P=I2. P subset of I1 int. I2 implies P is subset of one of I1 or I2; and I1 int. I2 subset of P implies one of I1 or I2 is a subset of P. So there is a possibility of P subset of I1 and I2 subset of P. Then how can we say P=I1 or P=I2 ?
If P ⊆ I_1 int I_2, then P is a subset of BOTH I_1 and I_2. Then if I_1 int I_2 ⊆ P, then either I_1 ⊆ P or I_2 ⊆ P by primeness of P. So we have either I_1 = P or I_2 = P. Does this make sense?