# coprime ideals in $K[X]$

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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• 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. – Robert Shore Feb 10 at 4:36
• still need help.... – Kevin Feb 10 at 5:03
• Can you get any non-zero constant polynomial into $I+J$? If so, what consequences follow? – Robert Shore Feb 10 at 5:04
• not sure what do you mean – Kevin Feb 10 at 5:32
• @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. – user26857 Feb 10 at 18:33

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.

• Isn't pretty clear that $I+J$ contains a non-zero element from $K$, that is, a unit, and therefore equals the whole ring? – user26857 Feb 10 at 18:32
• @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! – Robert Lewis Feb 10 at 18:39
• Yes, this is what I meant. (I let it as a hint for the OP.) – user26857 Feb 10 at 18:40

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

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