Today we discuss something on polynomials.<\/p>\n
<\/p>\n<\/div>\n
Let A<\/span> be a commutative unital ring, and A[x]<\/span> the polynomial ring over A<\/span>. In particular, these implies the nilradical\u00a0\\mathfrak{N}=\\mathfrak{R}<\/span><\/span>\u00a0in polynomial rings. If f\\in\\mathfrak{R}<\/span>, then\u00a01+xf<\/span><\/span>\u00a0is invertible. This means a_0,...,a_n<\/span> are all nilpotent, hence f<\/span> nilpotent. In the proof of the\u00a0Hilbert Nullstellensatz<\/b><\/i>, we will see that this is valid also in prime quotients of polynomial rings.<\/p>\n If f<\/span> is a zero-divisor<\/b><\/span>, then a_0,..,a_n<\/span> are all zero-divisors. Indeed, if fg=0<\/span>, then a_n b_r=0<\/span>, and f a_n g =0<\/span>, with \\mathrm{deg} a_n g<\\mathrm{deg} g<\/span>. Repeating this, eventually a_n g<\/span>=0. This yields (f-a_n x^n) g=0<\/span>. Then a_i g=0,a_i b_n=0,\\forall i<\/span>.<\/p>\n A general version of Gauss’s lemma<\/b><\/i> holds: if (a_0,...,a_n)=(1)<\/span>, then f<\/span> is said to be primitive. If f,g<\/span> are primitive, then so is f g<\/span>. The proof is analogous: If (c_0,...,c_n)\\in\\mathfrak{p}<\/span> for some maximal p<\/span>, then in (A\/\\mathfrak{p}[x]<\/span>, we have f g=0<\/span>. Since this is a domain, either f,g<\/span> is 0<\/span>, a contradiction.<\/p>\n The above is easily generalized to several variables (actually arbitrarily many, since a polynomial always involves only finite terms)<\/b>, keeping in mind A[X_1,...,X_n]=A[X_1,...,X_{n-1}][X_n]<\/span>.<\/p>\n The case of power series is different in many aspects. First<\/span><\/b>, if f=a_0+a_1 x+...<\/span>, then f<\/span> is invertible if and only if a_0<\/span> is. This is because suppose g=b_0+b_1 x+...<\/span>, then f g=a_0 b_0\u00a0+ (a_0 b_1+a_1 b_0)x+(a_0 b_2+a_1 b_1+a_2 b_0)x^2+...<\/span> where a_i<\/span> can be solved inductively as long as a_0 b_0=1<\/span>. Second<\/b><\/span>, although f<\/span> nilpotent implies a_i<\/span> nilpotent for all i<\/span>, via some similar induction focusing on the lowest degree term, the converse is not<\/b><\/span> true. In fact, there are some restrictions on the vanishing degree: if f^s=0<\/span>, then a_0^s=0<\/span>, so (f-a_0)^{2s}=0<\/span>; then a_1^{2s}=0<\/span>, so (f-a_1 x)^{4s}=0<\/span>. In general a_i^{2^i s}=0<\/span>. If the least s_i<\/span> for a_i^{s_i}=0<\/span> increases rapidly<\/span><\/b>, making 2^{-i} s_i\\rightarrow\\infty,i\\rightarrow \\infty<\/span>, then f<\/span> is not nilpotent. For example take s_i=3^i,A=\\prod_{i\\in\\mathbb{Z}^+}\\mathbb{C}[x_i]\/(x_i^{s_i}),a_i=x_i<\/span>. The argument also applies in the polynomial case, but then n<\/span> is finite.<\/p>\n If 1+g f<\/span> is invertible iff 1+a_0 b_0<\/span> is invertible. So f\\in\\mathfrak{R}(A[[x]])<\/span> iff a_0\\in\\mathfrak{R}(A)<\/span>.<\/p>\n The ideal F(\\mathfrak{I})<\/span> of f<\/span> with a_0\\in \\mathfrak{I}<\/span> is an ideal of A[[x]]<\/span>. Moreover A\/\\mathfrak{I}\\cong A[[x]]\/F(\\mathfrak{I})<\/span>. So if \\mathfrak{I}<\/span> is prime<\/b><\/span>, so is F(\\mathfrak{I})<\/span>; same for maximality<\/b><\/span>. In fact, the same holds in A[x]<\/span>.<\/p>\n The above topic is from\u00a0Atiyah, M. F.; MacDonald, I. G. (February 21, 1994). “Chapter 1: Rings and Ideals”. Introduction to Commutative Algebra. Westview Press. p. 11. ISBN 978-0-201-40751-8.<\/u><\/p>\n The case of countable variables is also of interest. We will discuss this in later posts.<\/p>\n Thursday, August 1, 2013<\/p>\n <\/p>","protected":false},"excerpt":{"rendered":" Sorry, this entry is only available in \u4e2d\u6587. For the sake of viewer convenience, the content is shown below in the alternative language. You may click the link to switch the active language.\u4eca\u5929\u6211\u4eec\u8ba8\u8bba\u591a\u9879\u5f0f \u4ea4\u6362\u73af\u4e0a Nilpotents and units are closely related. In a commutative unital ring , if nilpotent, unit, then is again a unit. 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\nLet f=a_0+a_1x+...+a_n x^n<\/span>. If a_1,a_2,...,a_n<\/span> are nilpotent<\/b>, so will be f-a_0<\/span>. If moreover a_0<\/span> is invertible<\/span><\/b>, f<\/span> will be invertible; if instead a_0<\/span> is nilpotent<\/span><\/b>, f<\/span> is nilpotent. The converses are both true<\/b>. For nilpotency<\/b><\/span>, the highest degree term of f^m<\/span> is a sole<\/b><\/span> a_n^m x^m<\/span>, if f<\/span> is nilpotent, a_n<\/span> is forced to be; but then f-a_n x^n<\/span> is again nilpotent. For invertibility<\/b><\/span>, immediately a_0<\/span> is invertible; Suppose fg=1<\/span> with g=b_0+b_1x+...+b_r x^r<\/span>. Then a_n b_r=0,a_n b_{r-1}+a_{n-1} b_r=0,...<\/span>. Multiplying the second by a_n<\/span>, we get a_n^2 b_{r-1}=0<\/span>; repeating this yields a_n^{r+1} b_0=0<\/span>, and b_0<\/span> is invertible so a_n<\/span> is nilpotent.<\/p>\n