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May 19

Lifting prime ideals to integral extensions

While working on homework I ran into the following obstacle: suppose A is an integrally closed integral domain, and B is a ring, integral over A. Suppose p is a prime ideal of A. I believe that it follows that $pB\neq B$ ; I am looking for a proof of this fact. Can anyone knowing of a proof provide a few small hints? I have Marcus's Number Fields, which has a proof under the assumption that A is a Dedekind domain, which unfortunately is not good enough. Eric. 131.215.159.91 (talk) 08:37, 19 May 2009 (UTC)[reply]

You might check if "lying over" proves this. If q is a prime ideal of B lying over p, then p is contained in q, so pB is contained in q, so pB is properly contained in B. JackSchmidt (talk) 14:18, 19 May 2009 (UTC)[reply]

Unfortunately my professor's proof that there exists a prime lying over p starts by assuming that pB is not B. The article Going up and going down is of interest but doesn't have a proof. Eric. 131.215.159.91 (talk) 15:41, 19 May 2009 (UTC)[reply]

Some course notes reduce to the case of local rings first. At any rate, it appears to have complete proofs so might be useful. You can also just exhaustively check commutative algebra books for proofs until you find one you like. I think Kaplansky's Commutative Rings is likely to have good proofs. JackSchmidt (talk) 16:04, 19 May 2009 (UTC)[reply]

Thanks for the link. It had not occurred to me to localize first; if I still have trouble then I'll read the rest of the proof that you linked to. Eric. 131.215.159.91 (talk) 19:38, 19 May 2009 (UTC)[reply]

Assume for contradiction that pB = B. Then there exist $a\in p$ and $b\in B$ such that ab = 1. As B is integral over A, b is a root of a monic polynomial with coefficients from A. However, b = a⁻¹ is in the fraction field of A, hence $b\in A$ as A is integrally closed. Thus p = A, a contradiction. — Emil J. 16:13, 19 May 2009 (UTC)[reply]

Do you think it is still true if A is not integrally closed? JackSchmidt (talk) 16:23, 19 May 2009 (UTC)[reply]

I don't think so. The argument goes both ways, doesn't it? If

a\in A

is not invertible in A and a⁻¹ is integral over A, we can find a maximal ideal p containing a, and put B = A[a⁻¹]. Then p is a prime ideal of A, and pB = B. — Emil J. 16:37, 19 May 2009 (UTC)[reply]

I'm talking nonsense: whenever a has an inverse in an integral extension of A, it has an inverse in A, regardless of whether A is integrally closed (it does not even have to be a domain). If ab = 1 and bⁿ + c_n−1bⁿ⁻¹ + … + c₀ = 0 with c_i from A, we can multiply by aⁿ to obtain 1 + c_n−1a + … + c₀aⁿ = 0, hence ad = 1 where d = −c_n−1 − … − c₀aⁿ⁻¹ is in A. — Emil J. 10:44, 20 May 2009 (UTC)[reply]

Thanks for the clarification. I only have a limited repertoire of domains, and I determined that such an a must be a unit in the integral closure. However, looking at specific units in quadratic integers, it was always annoyingly simple to convert from a to a⁻¹, usually no harder than to convert from i to 1/i = −i. The reason I asked is because the Lying Over proof does not require (as far as I can see) that A be integrally closed, only that B be integral over A (both domains). I can't tell if "lying over" is much too heavy of machinery for the problem though. JackSchmidt (talk) 12:33, 20 May 2009 (UTC)[reply]

I'm not sure what you mean by "the Lying Over proof". Also, it turns out that my assumptions of integrally closed and integral domain were unnecessesary. Eric. 131.215.159.91 (talk) 06:11, 21 May 2009 (UTC)[reply]

The proof I gave using LO. JackSchmidt (talk) 13:58, 21 May 2009 (UTC)[reply]

I also have a counterexample (courtesy of a friend of mine) for the case B not integral over A. Take

A=\mathbb {C} [x,y],p=(x,y),B=\mathbb {C} [x,y,z]/(xz+y-1)

. Then the minimal polynomial for z is

xZ+y-1\in A[Z]

which is not monic. (Or simpler yet, take A any integrally closed integral domain that has a nonzero maximal ideal p, and B the fraction field of A.) Eric. 131.215.159.91 (talk) 19:48, 19 May 2009 (UTC)[reply]

I should clarify that by pB I mean the smallest ideal of B containing p. It is not necessarily true that all elements of pB are of the form ab with a in p and b in B; in general they are sums of elements of such form. Eric. 131.215.159.91 (talk) 19:27, 19 May 2009 (UTC)[reply]

That's a good point, I forgot that one cannot express the ideal so easily. This is a serious problem, I don't know how to fix the argument. — Emil J. 10:44, 20 May 2009 (UTC)[reply]

For future reference, here is an outline of the solution. Localize A and B at $A\setminus p$ to make $A_{p}$ and $B_{p}$ . Then p is the unique maximal ideal of A_p. Let m be any maximal ideal of B_p. Then m lies above a maximal ideal of A_p, which must be p (we have used that B / A is integral). Then the assumption pB = B gives a contradiction. Eric. 131.215.159.91 (talk) 06:11, 21 May 2009 (UTC)[reply]

Name that theorem...

"A sequentially weakly lower semicontinuous coercive functional on a reflexive Banach space has a minimizer." A slight generalization appears in Giusti (2003), Direct methods in the calculus of variations, World Scientific. I'd like something to link from an article, so a theorem name or article that contains the theorem would be welcome. Otherwise, I would be happy to add the theorem to an article if a suitable target can be found. Sławomir Biały (talk) 20:49, 19 May 2009 (UTC)[reply]

What do you want exactly? I'm not sure if one can link this classic theorem to a name of an author. Usually, it is called the direct method in the calculus of variation: proving existence of a minimizer by showing that a minimizing sequence is compact in some topology for which the functional is sequentially lower-semicontinuous. A reference for it is more easily found in books rather than articles: e.g. Giaquinta & Hildebrandt "Calculus of Variations I", or Struwe's "Variational Methods", &c. --pma (talk) 22:10, 19 May 2009 (UTC)[reply]

Yes, I'm aware that one form or another of this result appears in books. And, yes I'm aware that it is absolutely fundamental in the calculus of variations. However, our article calculus of variations is primarily devoted to Euler-Lagrange type calculations, and addresses neither necessary nor sufficient conditions for the existence of a minimizer. Would it be reasonable, then, to start an article direct methods in the calculus of variations? Or has anyone on Wikipedia already done this, and put it in a less obvious place? Sławomir Biały (talk) 22:55, 19 May 2009 (UTC)[reply]

Exponential functions multi-valued?

My math teacher says the following is wrong, but can't explain why. I'd appreciate it if someone here could point out the error for me... For g(x) = x^1/2, R -> R, y = g(x) -> x = g^-1(y), so y = x^1/2 -> y^2 = x, which is multi-valued at least for positive real x. Any exponential function a^x, for x = 1/2, is a^1/2. Therefore, exponential functions are multi-valued at at least one point. (More than that, actually, because of x^1/4, etc., at least a countable infinity number of times.) (I'm only in Pre-Calc, by the way, sorry if this should be obvious.) —Preceding unsigned comment added by 68.46.107.173 (talk) 22:50, 19 May 2009 (UTC)[reply]

The question is not quite clear (which may be because I'm not aware of the term "multi-valued"). First, you are certainly not talking about exponential functions (which are expressed by x:->a^x, i.e. the parameter is the exponent, not the base). I think you are looking at the inverse "function" of a parabola, and expect it to have two branches, one above the x-axis and one below, so that sqrt(4) should be both 2 and -2. Well, what is wrong is the assumption that the inverse of a function is a function. A function by definition only has one output value for every input value. That implies that only strictly monotonic functions have an inverse that is also a function - for other function, the inverse is only a relation. --Stephan Schulz (talk) 23:12, 19 May 2009 (UTC)[reply]

Yeah, I was looking at the inverse of y = x^2 there in the beginning.

Also, oops, I didn't mean to say function. I knew that was a relation.

What I was saying was that if all for all positive x > 0, y = x^1/2 is multi-valued (by that I mean it maps to more than one output) then so should exponentials. For example, when x = e in y = x^1/2, y = ±√e. (right?) I meant that there would be a point in e^x that is the same relation, y = e^1/2, when x = 1/2. Does x^1/2's not being a function make the argument invalid? I'm trying to say that because exponentials share the same things that make x to some fraction with an even base not functions, it seems like they shouldn't be functions either. 68.46.107.173 (talk) 23:52, 19 May 2009 (UTC)[reply]

Exponentials are single-valued, the inverse of an exponential (a logarithm) is multi-valued. The notation isn't ideal, but in this context e^1/2 means the positive square root, so it is single-valued. --Tango (talk) 00:01, 20 May 2009 (UTC)[reply]

Actually, that isn't quite right. Logarithms are only multivalued when you are working with complex numbers, any positive real number has a unique real logarithm (in a given base). Exponentials where the base in a complex number is also multi-valued. See Exponential function for a rather better explanation that I seem to have managed! --Tango (talk) 00:06, 20 May 2009 (UTC)[reply]

Alright, then. So you're saying that the reason e^x only yields positive values is because it's conventionally defined that all roots in the exponent of an exponential function are the principal root -- not based on what the combination of symbols 'e^x' taken by itself would entail, but an external definition? 68.46.107.173 (talk) 03:45, 20 May 2009 (UTC)[reply]

Ah, perhaps I misunderstood what you were asking before (when I wrote my reply below). Take

e^{1/2}

as a concrete example. This expression is meaningless until we define what a fractional exponent means. By definition,

e^{1/2}={\sqrt {e}}

, which is the positive square root. Likewise, by definition,

e^{1/3}={\sqrt[{3}]{e}}

, which is the principal cube root. I don't think this is an "external definition"—it's the definition of fractional exponents, as far as I am aware. You argue that

y=x^{1/2}

is equivalent to

y^{2}=x

. But this isn't true, because that's not how fractional exponents are defined. [For example,

(-2)^{2}=4

, but

4^{1/2}\not =-2

.] I suppose fractional exponents could be defined in this way, and then fractional exponents would be multi-valued. This would complicate things, though, so the "principal root" definition of fractional exponents is used instead of your definition. This is a subtle point—either way you go, you have to define what a fractional exponent means separately, because the "repeated multiplication" definition of positive integer exponents doesn't make sense for a fraction. —Bkell (talk) 06:44, 20 May 2009 (UTC)[reply]

Another way to answer this question is to note that the "squaring" function,

f(x)=x^{2}

, is a function, meaning that each input x has exactly one output, but it is not one-to-one, meaning that several different inputs can map to the same output. (It is also not onto, meaning that there are some real numbers that are never produced as output—namely, the negative numbers—assuming that we are restricting the domain of the function to real numbers only.) So, when we look at it from the other direction, mapping outputs of the squaring function back to the inputs that produced them, we find that some outputs (the positive real numbers) correspond to more than one input, and other outputs (the negative real numbers) correspond to no inputs at all. The number zero is the only output that corresponds to exactly one input. So the "inverse" of the squaring function is not a function. This is because the squaring function is not both one-to-one and onto. The properties of injectivity ("one-to-one-ness") and surjectivity ("onto-ness") are not necessarily preserved under the operation of finding an inverse to a relation. You might find it interesting to work out the conditions needed for a one-to-one relation to have a one-to-one inverse, and other similar questions. It might help to draw pictures like the ones in the injective function article. —Bkell (talk) 03:11, 20 May 2009 (UTC)[reply]

See Exponentiation#Roots of arbitrary complex numbers and multivalued function. An expression such as z^1/2 where z is a complex number is defined either as the multiset of solutions to the equation x²=z, or as a so-called principal value. Both options are found in the litterature, but none of them are satisfactory, because the multiset of solutions is usually not considered to be a number, and the principal value of z^1/2 is not a continuous function of z. The usual rescue is to avoid using noninteger powers of (complex) numbers which are not positive reals. Bo Jacoby (talk) 09:54, 20 May 2009 (UTC).[reply]

You can define the unambiguously single-valued exponential function

\exp x=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=\lim _{n\rightarrow \infty }\left(1+{\frac {x}{n}}\right)^{n}

where n is an integer and the superscript means repeated multiplication, and then define

y^{x}=\exp(x\log y)

, where log is the real log, the multivalued complex log, or a branch of the multivalued log. So there are at least four different meanings of the superscript notation in common use: repeated multiplication (also used in discrete algebras where the others don't make sense), the single-valued continuous one where y is a positive real (also used in things like e^A where A is a matrix), the multivalued continuous one where y is any nonzero complex number, and (at least in numerical computing) the discontinuous complex one using the standard branch of the log function with imaginary part in (−π,π]. Fortunately it's usually obvious which one is meant. -- BenRG (talk) 12:44, 20 May 2009 (UTC)[reply]