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If $K$ is a field, $A=K[X]$, take $m,n \in K$ such that $m \ne n$. Prove that the ideals $I=(X-m)$ and $J=(X-n)$ are coprime.

I know the regular definition of coprime. But here, should we prove $I + J = A$ or $K$? And what are the units in $K[X]$?

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    $\begingroup$ You want to prove $I+J=K[X],$ and the unit of $K[X] \text{ is } 1.$ With these clarifications, let us know if you still need help. $\endgroup$ – Robert Shore Feb 10 at 4:36
  • $\begingroup$ still need help.... $\endgroup$ – Kevin Feb 10 at 5:03
  • $\begingroup$ Can you get any non-zero constant polynomial into $I+J$? If so, what consequences follow? $\endgroup$ – Robert Shore Feb 10 at 5:04
  • $\begingroup$ not sure what do you mean $\endgroup$ – Kevin Feb 10 at 5:32
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    $\begingroup$ @Kevin He means that $(X-m)+(n-X)=n-m\in I+J$. And an ideal which contain an invertible element equals the whole ring. $\endgroup$ – user26857 Feb 10 at 18:33
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What is the unit in $K[X]$? A (multiplicative) unit in $A = K[X]$ is an element $1_A$ such that

$1_A a = a 1_A = a \tag 1$

for any

$a \in A = K[X]; \tag 2$

it is clear from the ordinary definition of multiplication in $K[X]$ that $1_K \in K$ fulfills this requirement; here of course I have invoked a slight abuse of notation insofar as I have tacitly identified $K$ with the set of $0$-degree polynomials or constants in $K[X]$; technically I suppose we should write $1_KX^0$ for the unit in $K[X]$ but as is conventional I will simply refer to $0$-degree elements in $K[X]$ as being elements of $K$; so the unit in $K[X]$ is $1_K$ or simply $1$.

We want to show that

$(X - m) + (X - n) = A = K[X], \tag 3$

since by definition, two ideals

$I, J \subset K[X] \tag 4$

are co-prime if and only if

$I + J = (1_K) = K[X]; \tag 5$

this is what we need to prove with

$I = (X - m), \; J = (X - n), \tag 6$

where

$m, n \in K, \; m \ne n. \tag 7$

Now, polynomials of degree $1$ such as $X - m$ and $X - n$ are clearly irreducible in $K[X]$, since if

$rX + s = a(x)b(x), \; a(x), b(x) \in K[X], \tag 8$

then

$\deg a(x) + \deg b(x) = \deg (rX + s) = 1; \tag 9$

this implies that at precisely one of $a(x)$, $b(x)$ is of degree $1$, and the other is of degree $0$; thus $rX + s$ cannot be factored into two polynomials of positive degree, so it is indeed irreducible. Thus we see that $X - m$, $X - n$ are irreducible over$K$; this in turn implies that

$d(X) = \gcd(X - m, X - n) = 1, \tag{10}$

for it is clear from the above that $\deg d(X) = 0$ or $\deg d(X) = 1$; in the latter case

$d(X) = pX + q \mid X - m$ $\Longrightarrow \exists b \in K, \; b(pX + q) = bpX + bq = X - m \Longrightarrow bp = 1, bq = -m; \tag{11}$

and likewise,

$d(X) = pX + q \mid X - n$ $\Longrightarrow \exists c \in K, \; c(pX + q) = cpX + cq = X - m \Longrightarrow cp = 1, cq = -n; \tag{12}$

now the equations

$bp = 1 = cp \tag{13}$

imply

$p \ne 0, \; b = c; \tag{14}$

but then

$m = -bq = -cq = n, \tag{15}$

contradicting our assumption that $m \ne n$; therefore,

$p = 0 \tag{16}$

and any common divisor of $X - m$, $X - n$ must be an element $0 \ne q \in K$; since $d(X)$ is defined up to multiplication by a unit, we have established (10). Now since $K[X]$ is a principle ideal domain, there exist

$f(X), g(X) \in K[X] \tag{17}$

with

$f(X)(X - m) + g(X)(X - n) = d(X) = 1; \tag{18}$

thus for any $r(X) \in K[X]$

$r(X) = r(X)(1)$ $= r(X)f(X) (X - m) + r(X) g(X)(X - n) \in (X - m) + (X - n), \tag{19}$

and thus we see that

$K[X] \subset (X - m) + (X - n) \Longrightarrow K[X] = (X - m) + (X - n) \tag{20}$

proved as per request.

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    $\begingroup$ Isn't pretty clear that $I+J$ contains a non-zero element from $K$, that is, a unit, and therefore equals the whole ring? $\endgroup$ – user26857 Feb 10 at 18:32
  • $\begingroup$ @user26857: you mean as in $(X - m) - (X - n) = n - m \ne 0$? Nice catch, maybe you ought to write it up as an answer! $\endgroup$ – Robert Lewis Feb 10 at 18:39
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    $\begingroup$ Yes, this is what I meant. (I let it as a hint for the OP.) $\endgroup$ – user26857 Feb 10 at 18:40
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Note that, if $I + J = k[X]$, then there exist $p,q$ polynomials with $p(X-n) + q(X-m) = s$ for some $s \in k$ (actually, we can do this for any $s$ in the field).

Reciprocally, if $s \in I+J$, for some $s$ in $k$, then $1 \in s^{-1}(I+J) = I+J$ and so $k[X] = I+J$. Thus our problem reduces to showing that there exist $p,q \in k[X]$ with $p(X-n) + q(X-m) = s$ for some nonzero $s$. But this is rather direct, as

$$ (X-n) - (X-m) = m-n \neq 0. $$

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