|
Hi I don't know how to show that $\langle 2,x \rangle$ is not principal and the definition of a principal ideal is unclear to me. I need help on this, please. The ring that I am talking about is $\mathbb{Z}[x]$ so $\langle 2,x \rangle$ refers to $2g(x) + xf(x)$ where $g(x)$, $f(x)$ belongs to $\mathbb{Z}[x]$. |
|||||
|
|
I think it's relatively easy to see that $I=\langle 2,x \rangle = \{a_nx^n+\dots+a_1x+a_0; a_0\text{ is even}\}$. Now, suppose that $I=\langle f(x) \rangle$ for some $f(x)\in I$. If $f(x)$ is a constant polynomial, then $\langle f(x) \rangle$ contains only polynomials with even coefficients, and we do not get $x$. If $f(x)$ is of degree at least 1, then $\langle f(x) \rangle$ contains only polynomials of degree at least 1, and we do not get $2$. So $I$ is not of the form $\langle f(x) \rangle$. |
|||||||||||||||||
|
|
I want to record a somewhat less elementary, but perhaps more conceptual answer. Note first that $\langle 2 \rangle$, $\langle x \rangle$ and $\langle 2, x \rangle$ are all prime ideals of $\mathbb{Z}[x]$. Indeed, the quotients by these ideals are isomorphic, respectively, to $(\mathbb{Z}/2\mathbb{Z})[x]$, $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$, which are all integral domains. So in particular we have a proper inclusion of nonzero prime ideals $0 \subsetneq \langle x \rangle \subsetneq \langle x, 2 \rangle$ in which the smaller ideal is principal. Now let $R$ be any integral domain and let $I \subset J$ be a proper inclusion of nonzero prime ideals, with $I$ a principal ideal. Then $J$ cannot be principal. Indeed, $I = \langle x \rangle$ with $x$ a prime element. Suppose that also $J = \langle y \rangle$. Then $x \in J$, so that there exists $a \in R$ with $x = ay$. Since $ay = x \in I$ and $I$ is prime, we have either $a \in I$ or $y \in I$. If $a \in I$, then $a = bx$, so $x = byx$ or $x(1-by) = 0$ in the domain $R$; since $x \neq 0$ we conclude $by = 1$, i.e., $y$ is a unit and therefore $J = R$, contradiction. Similarly if $y \in I$, then $y = bx$, so $x = abx$ and we conclude that $a$ is a unit and thus $I = J$, contradiction. Added: A variant on the above argument is: if $0 \subsetneq I = \langle x \rangle \subsetneq \langle y \rangle = J$ with both $I$ and $J$ prime, then $x$ and $y$ are both irreducible elements and $y$ properly divides $x$, contradiction. This is technically a stronger fact because in an arbitrary domain a generator of a principal prime ideal is necessarily an irreducible element but the converse generally does not hold. However, the easiest way to show that an element $x \in R$ is irreducible is to show that $\langle x \rangle$ is prime, or equivalently that $R/\langle x \rangle$ is a domain. To show that a nonprime element $x$ is irreducible is much more delicate! Remark: If $R$ is a commutative Noetherian ring, then if $J$ is any nonzero principal prime ideal, there cannot be any nonzero prime ideal $I$ -- principal or otherwise -- with $0 \subsetneq I \subsetneq J$. This is a special case of Krull's Principal Ideal Theorem. |
|||||||
|
|
Below is a $\:$ complete $\:$ proof from first principles - easily comprehensible to a high-school student. In particular there is absolutely no handwaving - as there is elsewhere in some other answers. We show $\rm\:(2,x)\ =\ (f)\ $ in $\rm\:\mathbb Z[x]\: $ yields a parity contradiction, by simply evaluating polynomials. $\rm\ \ f\ \in\ (2,x)\ \Rightarrow\ f\ =\ 2\ G + x\ H\:.\: $ Eval at $\rm\: x = 0\ \Rightarrow\ f(0)\ =\ 2\ G(0)\ =\ 2\:n\:,\:$ for some $\rm\: n\in \mathbb Z\:.$ $\rm\ \ 2\ \in\ (f)\ \Rightarrow\ 2\ =\ f\: g\:\ \Rightarrow\ deg\ f\ =\: 0\ \:$ hence $\rm\ f\ =\ f(0)\ =\ 2\:n\:.$ $\rm\ \ x\ \in\ (f)\ \Rightarrow\ x\ =\ f\: h\ =\ 2\:n\:h\:.\ $ Eval at $\rm\ x = 1\ \Rightarrow\ 1\: =\ 2\:n\:h(1)\ \Rightarrow\ 1\:$ is even $\ \Rightarrow\Leftarrow$ REMARK $\ $ The proof still works if we replace $\rm\:\mathbb Z\:$ by any domain where $\rm\:2\:$ is not a unit. i.e. $\rm\:2\nmid 1\:.\:$ In particular, it works over any domain with a sense of parity, i.e. having $\rm\:\mathbb Z/2\:$ as ring image, e.g. the Gaussian integers, or the rationals writable with odd denominator - see this post. Conversely, the result is clearly false if $\rm\:2\:$ is a unit since then $\rm\:(2,x) = (1)\:$ is trivially a principal ideal. Furthermore, the proof still works if we replace $2$ by any nonunit of the coefficient domain. |
|||||||||||||||
|
|
One way to see that $\langle 2,x \rangle$ is not principal is to note that $\mathbb{Z}[x]$ is a UFD(See example 3 in the wiki page), and both $2$ and $x$ are primes. So if the ideal is principal, then $2$ and $x$ will share a common divisor. Contradiction. It is not as down to earth as Martin's solution, but it is a way to look at the problem. |
|||||||||||
|
|
An ideal $\langle a_1, \dots, a_k \rangle$ is the smallest ideal containing these elements, explicitly the set of all linear combinations $r_1a_1+\dots + r_ka_k$ where the $r_i$ are arbitrary elements from the ring. A principal ideal is an ideal that can be generated by a single element. So first of all, you have to say which ring you are looking at to have a definite question. Now, you could comment on whether you understand this definition of principal ideal. If the ideal $\langle 2,x \rangle$ were principal, the generator would have to divide 2. What are the integer polynomial divisors of 2? |
|||||||||||
|
|
Suppose to the contrary that $(2, x) = \{2p(x) + xq(x) : p(x), q(x) \in \mathbb Z\}$ is a principal ideal where $(2, x) = (a(x))$ for some $a(x) \in \mathbb Z[x]$. Observe that $(2, x)$ is proper since the constant term must be even. Moreover it is immediate that $2 \in (a(x))$ and by definition there exists $p(x) \in \mathbb Z[x]$ such that $2 = p(x)a(x)$. But observe that $0 =\deg p(x)a(x) = \deg p(x) + \deg a(x)$ which implies that $\deg p(x) = \deg a(x) = 0$. Since 2 is prime, we it must follow that $a(x), p(x) \in \{\pm1, \pm2\}$. But if $a(x) = \pm1$ then $(a(x)) = R$ which is a contradiction to $(a(x))$ being a proper ideal. Hence it must follow that $a(x) = \pm2$. So $(a(x)) = (2) = (-2)$. But by construction it must also follow that $x \in (a(x)) = (2)$ so there must exist $q(x) \in \mathbb Z[x]$ such that $x = 2q(x)$. The only way this can happen is if $q(x) = \frac12 x$ which is impossible since $q(x)$ can only have integer coefficients. Hence we have arrived at a contradiction to our hypothesis that $(2, x)$ is a principal ideal. Note: This immediately implies that $\mathbb Z[x]$ is not a PID. |
|||
|
|
