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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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A single cancellation step lowers the degree of a polynomial in an ideal once its leading coefficient lies in a realised stage

Statement

Let R be a commutative ring, let a be an ideal of R[x], let n∈N, and let an be the stage ideal of The leading coefficients of the degree-n elements of an ideal of R[x], together with 0, form an ideal of R, and these ideals ascend with n. Suppose m∈N and g1,…,gm∈a are nonzero polynomials of degree n whose leading coefficients cj=lc⁡(gj) generate an, so that an=(c1,…,cm).

Let f∈a be nonzero of degree d with d≥n and lc⁡(f)∈an. Then there is a polynomial h in the ideal of R[x] generated by g1,…,gm such that

f−h=0orf−h≠0  and  deg⁡(f−h)<d.

At d=0 the second alternative is impossible, so there f=h. At m=0 the hypothesis on f cannot be met: an is then the zero ideal while a leading coefficient is nonzero, so the statement is not vacuously producing a degree drop out of nothing.

Facts & Assumptions

Given: A commutative ring R, an ideal a of R[x], an index n∈N, polynomials g1,…,gm∈a nonzero of degree n with leading coefficients c1,…,cm generating an, and a nonzero f∈a of degree d≥n with lc⁡(f)∈an.

[L1]

R[x] is the set of finitely supported functions a ⁣:N→R, with (a+b)i=ai+bi and (ab)i=∑j+k=iajbk; x is the sequence with coefficient 1R at index 1 and zero elsewhere (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[L2]

For 0≠f=∑iaixi∈R[x] the degree is deg⁡f=max⁡{i∈N:ai≠0} and the leading coefficient is lc⁡(f)=adeg⁡f; the zero polynomial has no degree and no leading coefficient (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L3]

In a commutative ring, (S) consists of finite sums ∑risi, and (a)=Ra; the empty sum is included and equals 0 (In a commutative ring, (S) consists of finite sums ∑risi, and (a)=Ra).

[L4]

For an ideal a of R[x] and n∈N, the set an of leading coefficients of the nonzero degree-n elements of a, together with 0, is an ideal of R, and an⊆an+1 (The leading coefficients of the degree-n elements of an ideal of R[x], together with 0, form an ideal of R, and these ideals ascend with n).

Proof

technique · direct
1.1L1L2given

Fix the data of the Given line and write c=lc⁡(f), a nonzero element of R lying in an; the exponent d−n is a natural number because d≥n.

2.1L3L4step 1.1

Since an is the ideal generated by c1,…,cm and c lies in it, the finite-sum description gives r1,…,rm∈R with c=∑j=1mrjcj. Any one such list may be taken; the construction below uses no property of it beyond this equation.

3.1L1L2L3step 2.1

Put h:=∑j=1mrjxd−ngj, which lies in the ideal of R[x] generated by g1,…,gm because each coefficient rjxd−n is an element of R[x]. By the convolution rule the coefficient of xi in rjxd−ngj is rj times the coefficient of xi−(d−n) in gj, read as 0 when i<d−n. For i>d that index exceeds n, so the coefficient vanishes; at i=d that index is exactly n, so the coefficient is rjcj. Summing over j, the polynomial h has zero coefficients at every index above d and coefficient ∑j=1mrjcj=c at index d.

4.1L2step 3.1

The polynomial f−h lies in a and has zero coefficient at every index i≥d: above d both f and h vanish, and at d the two coefficients are c and c. So either f−h=0, or f−h is nonzero and every index carrying a nonzero coefficient is below d, which by the definition of degree means deg⁡(f−h)<d.

5.1L2L3step 3.1step 4.1∎

Two extreme cases are worth recording. When d=n the exponent d−n is 0 and x0=1, so h=∑jrjgj and no shift occurs. When d=0 there is no natural number below d, so the second alternative of step 4.1 cannot occur and the conclusion is the equality f=h. When m=0 the sum defining an is empty, an=0, and no nonzero leading coefficient lies in it, so no f satisfies the hypothesis.

Remarks

  • Nothing is divided and no leading coefficient is inverted. The correction h is built by multiplying the given gj by ring elements and a power of x, so the argument runs over an arbitrary commutative ring rather than only over a field or a domain. That is exactly why the stage ideals are needed: over a field one generator per stage would do.

  • The gj must have degree exactly n. If some gj had degree below n the coefficient computation in step 3.1 would place rjcj at an index below d and the cancellation at xd would fail. This is what makes the exact-degree definition of the stage ideal the usable one.

  • The lemma produces one step, not a terminating procedure. Iterating it requires knowing that the leading coefficient of f−h again lies in a realised stage, which is what the ascending chain condition supplies in the finite-generation lemma that uses this one.

Depends on

Used by

Dependency tree · two levels

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Sources