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If a maximal subalgebra $L\\subset W_n(K)$ has rank $n$ and is a submodule of $W_n(K)$ then $L$ is not simple. Moreover, $L$ is of the form $L=\\{ D\\in W_n(K) \\ | \\ D(I)\\subsete"},"verification_status":{"content_addressed":true,"pith_receipt":true,"author_attested":false,"weak_author_claims":0,"strong_author_claims":0,"externally_anchored":false,"storage_verified":false,"citation_signatures":0,"replication_records":0,"graph_snapshot":true,"references_resolved":false,"formal_links_present":false},"canonical_record":{"source":{"id":"2605.22970","kind":"arxiv","version":1},"metadata":{"license":"http://creativecommons.org/licenses/by/4.0/","primary_cat":"math.RA","submitted_at":"2026-05-21T18:58:16Z","cross_cats_sorted":[],"title_canon_sha256":"eb531299415435faa9d1cb6e7b201577a384c59b93f9573f1d7dff8287b11f22","abstract_canon_sha256":"4c1c47c36d1cfd3973b49bb7419128c5c9f0da6f4eaa5382835bd2414ae84cd3"},"schema_version":"1.0"},"receipt":{"kind":"pith_receipt","key_id":"pith-v1-2026-05","algorithm":"ed25519","signed_at":"2026-05-25T02:01:32.676590Z","signature_b64":"H063u2rRhRB2FPDE3SRqNplIpo1w2kNs3oSHw8kVy6Z780/j3KUem/zF7/VSU1gl7KMw8mz7l3rJcJ/+WRQXBw==","signed_message":"canonical_sha256_bytes","builder_version":"pith-number-builder-2026-05-17-v1","receipt_version":"0.3","canonical_sha256":"7917f92fc64259e56b88ed92ecaa50b510993df6bd581e2f4d7fc83836d6adf0","last_reissued_at":"2026-05-25T02:01:32.675813Z","signature_status":"signed_v1","first_computed_at":"2026-05-25T02:01:32.675813Z","public_key_fingerprint":"8d4b5ee74e4693bcd1df2446408b0d54"},"graph_snapshot":{"paper":{"title":"Maximal subalgebras of the Lie algebra $W_n(\\mathbb{K})$","license":"http://creativecommons.org/licenses/by/4.0/","headline":"","cross_cats":[],"primary_cat":"math.RA","authors_text":"A.Petravchuk, O.Tyshchenko, Y.Chapovskyi","submitted_at":"2026-05-21T18:58:16Z","abstract_excerpt":"Let $K$ be an algebraically closed field of characteristic zero, $A= K[x_1, \\dots, x_n]$ the polynomial ring in $n$ variables, and let $W_n(K)$ be the Lie algebra of all $K$-derivations of $A.$ This Lie algebra also is the free $A$-module of rank $n$ over the ring $A,$ so every subalgebra of $W_n(K)$ has a rank $\\leq n$ over $A.$ We prove that every maximal subalgebra of rank $\\leq n$ of $W_n(K)$ is a simple Lie algebra. If a maximal subalgebra $L\\subset W_n(K)$ has rank $n$ and is a submodule of $W_n(K)$ then $L$ is not simple. 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