On the Wikipedia page below, it says "a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units."   On this Wolfram page it says, "An element $a$ of a ring which is nonzero, not a unit, and whose only divisors are the trivial ones."   Can someone explain the discrepancy between these two definitions of Wikipedia and Wolfram?  They are not the same definition, right? Is there actually a difference in definition of irreducible element of a general ring (i.e. no multiplicative identity, inverses, or commutativity) and that of a an integral domain?  And, if so, what is the difference(s)?  What is meant by "trivial ones", units?  I am being asked to explain how this question is different from a question asking about irreducible polynomials.  Irreducible polynomials are not a part of this question, clearly..
Edit:  I foolishly changed my example right before posting this. The following is what I meant to ask.  On the wikipedia page below, it says "a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units."   Then, on the wikipedia page below, it says "an irreducible polynomial is, roughly speaking, a non-constant polynomial that cannot be factored into the product of two non-constant polynomials."   The page then gives an example of a polynomial that is irreducible "over the rationals" (I assume it is implied that "the integers" is referring to the ring of polynomials over the field of rationals, right?)  $p(x) = x^2 + 2 = (x+\sqrt{2})(x-\sqrt{2})$  On the Wofram page below, it says "the irreducible polynomials are examples of irreducible elements."   If irreducible polynomials are a subset of irreducible elements, how is $x+\sqrt{2}$ a unit?  What can it be multiplied by to equal the multiplicative identity?  I think I have it right this time (sorry. long day)  ====================================================================  The following is the original post (for those who already responded and those looking at the corresponding responses)  On the wikipedia page below, it says "a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units."   Then, on the wikipedia page below, it says "an irreducible polynomial is, roughly speaking, a non-constant polynomial that cannot be factored into the product of two non-constant polynomials."   The page then gives an example of a polynomial that is reducible "over the integers" (I assume it is implied that "the integers" is referring to some sort of ring, right? due to the claim below)  $p(x) = x^2 + 4x + 4 = (x+2)(x+2)$  On the Wofram page below, it says "the irreducible polynomials are examples of irreducible elements."   If irreducible polynomials are a subset of irreducible elements, how is $x+2$ a unit?  What can it be multiplied by to equal the multiplicative identity?
What is the most "pythonic" way to iterate over a list in chunks? I have a Python script which takes as input a list of integers, which I need to work with four integers at a time.  Unfortunately, I don't have control of the input, or I'd have it passed in as a list of four-element tuples.  Currently, I'm iterating over it this way: for i in range(0, len(ints), 4):     # dummy op for example code     foo += ints[i] * ints[i + 1] + ints[i + 2] * ints[i + 3]  It looks a lot like &quot;C-think&quot;, though, which makes me suspect there's a more pythonic way of dealing with this situation.  The list is discarded after iterating, so it needn't be preserved.  Perhaps something like this would be better? while ints:     foo += ints[0] * ints[1] + ints[2] * ints[3]     ints[0:4] = []  Still doesn't quite &quot;feel&quot; right, though.  :-/ Related question: Sometimes it takes too long for the network manager to update to see new wireless AP. Is it possible to change the time interval between checks? Clicking on a spoiler tag to reveal the text shouldn't navigate you to another page This is an issue I noticed when reading . On multiple occasions I ended up inadvertently clicking a link that I could not see when I was just trying to reveal the spoiler text.         Can links be deactivated in spoilers until the text is revealed?  (Edit: this issue seems particularly to apply to those using touch screen devices with the "full site" theme) Let $X_1$ and $X_2$ are independent $N(0, \sigma^2)$ random variables. What is the distribution of $X_1^2 + X_2^2$? Let $X_1$ and $X_2$ are independent $N(0, \sigma^2)$ which means (mean = 0, variance = $\sigma^2$) random variables. What is the distribution of $X_1^2 + X_2^2$?  My approach is that  $X_1\sim N(0, \sigma^2)$ and $X_2\sim N(0, \sigma^2)$  Then $X_1^2$ and $X_2^2$ have chi-squared distribution with 1 degree of freedom. 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I have looked on the Immigration website and can't find the answers I need. Let $\mu(\Omega)$ be a probability measure (i.e. $\mu(\Omega) = 1$), and let $f,g$ be non-negative measurable functions on $\Omega$ such that $fg \geq 1$. Show that $1 \leq (\int f^p )(\int g^p)$ for all $0 &lt; p &lt; \infty$   I'm a bit at a loss of what to do. I know that since $fg \geq 1$, we may obtain:  $$\int_{\Omega} d\mu = \mu(\Omega) = 1 \leq \int_{\Omega} fg = \int_{\Omega} |fg|$$ The right-most integral is true since $fg$ is positive and it is equal to $||fg||_1$.  By Hölder's inequality, we obtain:  $$1 \leq \int |fg| \leq ||f||_p ||g||_q = (\int f^p)^\frac{1}{p}(\int g^q)^\frac{1}{q}$$  where $p$ and $q$ satisfy $\frac{1}{p} + \frac{1}{q} = 1$. Raising the left-most and right-most sides of this inequality to the power $p$, I may obtain:  $$1 \leq (\int f^p)(\int g^q)^\frac{p}{q}$$  I now wish to show that $(\int g^q)^\frac{p}{q} \leq \int g^p$, but I've been at a loss of how to do it. 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